[PHP] Running several applications involving PHP on virtual private server - optimization possible?
I currently have 1 domain on a VPS with 1 instance of Moodle 2.3.2. To minimize the danger of overtaxing the resources (I share a processor :-( but have 2GBytes of guaranteed RAM) I was wondering if there were anything I could do if I wanted to put, say, Wordpress or another domain on the same server with another instance of Moodle in terms of PHP tuning. For the moment all is well, but I prefer to avoid problems in future if possible. Ah yes, my PHP version is 5.3.3.
[PHP] Question about PHP-CGI, mysqld and sudden spiking average load readings on Linux Virtual Dedicated Server running CentOS 6.3
I have noticed when doing a top via SSH on my GoDaddy virtual dedicated server that every time a user logs in in the COMMAND column I see the PHP-CGI process start to run, followed by mysqld. Often when these two processes run I start to see a spike in average load, and sometimes this spike spins out of control to the point that when trying to log on to my Moodle 2.3.2 LMS I get an 500 Internal Server Error from Apache. Now, what I'd like to know given the fact that I have only 2GBytes of RAM and am sharing a processor with other users :-( is if there is any way to make sure my PHP 5.3.3 is properly configured. I am given, for example, a choice between CGI, fastCGI and Apache in my Plesk panel, and I am not quite sure how to optimize things for my environment. I went to the Moodle.org site to try and see what's the right setting, but it's not clear if the persons writing there have a different setup or not. Unfortunately, it's impossible to get specs on my VDS's processor from the fine folks at GoDaddy, and perhaps there is just a simple processor power issue going on here, but I only have 30 students and only a handful going in simultaneously. And at times my average load can move from .08 to 1.5+ and all the way up to 2+ within the space of a few seconds. I should point out that at one point it was discovered by GoDaddy technicial people that a user was running a rogue process that overtaxing resources. But perhaps I also need to tweak my installation despite this incident. Any help here would be appreciated.
[PHP] Problem with PHP in Moodle LMS
Has anyone ever heard of a problem with the Moodle LMS when trying to edit an activity (quiz, resource) in which there is a timeout after only a few seconds and a message saying something to the effect of that the system is sorry but there was a time out, please retry. When I look in the URL I see the following after my domain name: .../course/modeit.php Is this purely a Moodle thing, or something someone can help me with. (I've already searched the Moodle site for information about this error, but can't see anything related to version 2.2.3, which I am using.) Any help would be appreciated. I thought I might want to up the timout figure in the php5.ini file in the root directory of my shared Linux hosting with Godaddy, but I don't think that's really the problem. Regards, Gary
Re: [PHP] Depreciation message I can't make out....
By the way, I told GoDaddy about the # deprecation issue and they told me that I could simply change my root directory php5.ini file. But the error message is related to the PHP.ini file in /web/config, as I said. I could almost hear them shrugging their shoulders, what-me-worry style. If anyone with more might than i could explain this error might affect others when multiplied hundreds of times it could be useful. Am 30.06.2012 19:23 schrieb Daniel Brown danbr...@php.net: On Thu, Jun 28, 2012 at 8:37 PM, tamouse mailing lists tamouse.li...@gmail.com wrote: Just a quick followup -- # comments are deprecated for .ini files, only, correct? They are still full citizens in php source, aren't they? Correct, Tam. -- /Daniel P. Brown Network Infrastructure Manager http://www.php.net/ -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] Depreciation message I can't make out....
Hi, I am running Moodle 2.2.3 and using PHP 5.3 on a Linux server with GoDaddy and am getting the message about depreciation after having typed the following command into their cron job manager: /web/cgi-bin/php5_3 $HOME/html/moodle223a/admin/cli/cron.php PHP Deprecated: Comments starting with '#' are deprecated in /web/conf/php5.ini on line 1256 in Unknown on line 0 PHP Deprecated: Comments starting with '#' are deprecated in /web/conf/php5.ini on line 1257 in Unknown on line 0 br / bDeprecated/b: Directive 'magic_quotes_gpc' is deprecated in PHP 5.3 and greater in bUnknown/b on line b0/bbr / Content-type: text/html GoDaddy says that I am using a command that is not adapted to version 5.3 of PHP, but they suggested I contact you to fix it. Also, I do not believe I have access to php5.ini; only they do. But I would need to tell them how to fix this issue. And no one at Moodle.org seems to be able to help. Perhaps it's a Moodle thing; perhaps not, but I thought it wise to contact you. By the way, this message is only sent to me when using version 5.3 of PHP; I have another Moodle site using 5.2 and things go perfectly well, without this message. Regards,
Re: [PHP] PHP cron job optimization
I want to design an application that reads news from RSS sources. I have about 1000 RSS feed to collect from. I also will use Cron jobs every 15 minutes to collect the data. the question is: Is there a clever way to collect all those feed items without exhausting the server any Ideas Thank you in advance Just watch your memory and you'll be fine. As was stated fetching an rss feed is fast and cheap, so I would contrariwise think about paralleling if I would you. signature.asc Description: OpenPGP digital signature
Re: [PHP] Re: Dynamically Created Checkboxes
Steve Staples sstap...@mnsi.net wrote in message
news:1298568238.14826.2431.camel@webdev01...
On Thu, 2011-02-24 at 11:42 -0500, Gary wrote:
Steve Staples sstap...@mnsi.net wrote in message
news:1298492194.14826.2425.camel@webdev01...
On Wed, 2011-02-23 at 14:56 -0500, Gary wrote:
Steve Staples sstap...@mnsi.net wrote in message
news:1298490417.14826.2418.camel@webdev01...
On Wed, 2011-02-23 at 14:17 -0500, Gary wrote:
Jim Lucas li...@cmsws.com wrote in message
news:4d653673.7040...@cmsws.com...
On 2/23/2011 4:35 AM, Gary wrote:
Pete Ford p...@justcroft.com wrote in message
news:62.c1.32612.d49d4...@pb1.pair.com...
This bit?
On 22/02/11 22:06, Gary wrote:
for($i=1; $i=$_POST['counties']; $i++) {
if ( isset($_POST[county{$i}] ) ) {
You loop over $_POST['counties'] and look for
$_POST[county$i]
I suspect that there is no field 'counties' in your form, so
the
server
is
complaining about the missing index - seems likely that the
production
server
is not showing the error, but instead just giving up on the
page.
I think the IE7/Firefox browser difference is a red herring: it
wasn't
actually working in any browser, or the code changed between
tests.
Remember, PHP is server side and generates HTML (or whatever).
Unless
you
tell the PHP what browser you are using and write PHP code to
take
account
of that (not generally recommended) then the server output is
the
same
regardless of browser.
--
Peter Ford, Developer phone: 01580 89 fax:
01580
893399
Justcroft International Ltd.
www.justcroft.com
Justcroft House, High Street, Staplehurst, Kent TN12 0AH
United
Kingdom
Registered in England and Wales: 2297906
Registered office: Stag Gates House, 63/64 The Avenue,
Southampton
SO17
1XS
Pete
Once again, thank you for your help.
I was able to get it to work, I did not understand the browser
differences
either, but on my testing server on IE, it worked as I wanted
but
not
on
anything else. I chopped out most of the code and ended up with
this:
if ( isset($_POST['submit']) ) {} else {
print form action=\phpForm3.php\ method=\POST\\n;
if ($Recordset1) {
print table width=200 border=1\n;
print thnbsp; /th\n;
print th State /th\n; //2 fields in Counties table, State
and
County
print th County /th\n;
print /tr\n;
//create table
$i = 0;
while ( $row = mysql_fetch_array($Recordset1) ) {
$i++;
print tr\n;
print tdinput type=\checkbox\ name=\county$i\
value=\$row[name]\/td\n;
echo td{$row['state_id']}/td\n;
echo td{$row['name']}/td\n;
echo /tr\n;
}//end while
print /table\n;
} else {
echo(PError performing query: .
mysql_error() . /P);
}
print input type=\hidden\ name=\counties\
value=\$i\/\n;
print input type=\submit\ name=\submit\ value=\Go\/\n;
}
?
Again, thank you.
Gary
__ Information from ESET Smart Security, version of
virus
signature database 5899 (20110223) __
The message was checked by ESET Smart Security.
http://www.eset.com
If you would allow me to show you a little easier way of doing
this.
if ( !isset($_POST['submit']) ) {
echo 'form action=phpForm3.php method=POST';
if ($Recordset1) {
echo HEADER
table width=200 border=1
trth nbsp; /thth State /thth County /th/tr
HEADER;
while ( $row = mysql_fetch_array($Recordset1) ) {
echo ROW
tr
tdinput type=checkbox name=county[]
value={$row['name']}/td
td{$row['state_id']}/td
td{$row['name']}/td
/tr
ROW;
}//end while
echo '/table';
} else {
echo 'pError performing query: '.mysql_error().'/p';
}
echo 'input type=submit name=submit value=Go /';
}
Now, the main thing I want you to see is the line for the country
checkbox'es
The county[] turns the submitted values into an array. So, in
the
processing
script, you would do this.
?php
...
if ( !empty($_POST['county'])
foreach ( $_POST['county'] AS $id = $name )
echo {$id} {$name}\n;
...
?
Hope this clears things up for you a little.
Enjoy.
Jim
In a weird twist, I have/had the script working, but again it is
not
working in FF or Chrome. This is a link to the first page which
runs
you
through the pages.
I have the submit to trigger upon change so there is no submit
button.
This
works in IE7. If I add a submit button, it does not work in any of
them.
I am wondering if this is more a problem with the html.
http
Re: [PHP] Re: Dynamically Created Checkboxes
Steve Staples sstap...@mnsi.net wrote in message
news:1298570683.14826.2438.camel@webdev01...
On Thu, 2011-02-24 at 12:52 -0500, Gary wrote:
Steve Staples sstap...@mnsi.net wrote in message
news:1298568238.14826.2431.camel@webdev01...
On Thu, 2011-02-24 at 11:42 -0500, Gary wrote:
Steve Staples sstap...@mnsi.net wrote in message
news:1298492194.14826.2425.camel@webdev01...
On Wed, 2011-02-23 at 14:56 -0500, Gary wrote:
Steve Staples sstap...@mnsi.net wrote in message
news:1298490417.14826.2418.camel@webdev01...
On Wed, 2011-02-23 at 14:17 -0500, Gary wrote:
Jim Lucas li...@cmsws.com wrote in message
news:4d653673.7040...@cmsws.com...
On 2/23/2011 4:35 AM, Gary wrote:
Pete Ford p...@justcroft.com wrote in message
news:62.c1.32612.d49d4...@pb1.pair.com...
This bit?
On 22/02/11 22:06, Gary wrote:
for($i=1; $i=$_POST['counties']; $i++) {
if ( isset($_POST[county{$i}] ) ) {
You loop over $_POST['counties'] and look for
$_POST[county$i]
I suspect that there is no field 'counties' in your form, so
the
server
is
complaining about the missing index - seems likely that the
production
server
is not showing the error, but instead just giving up on the
page.
I think the IE7/Firefox browser difference is a red herring:
it
wasn't
actually working in any browser, or the code changed between
tests.
Remember, PHP is server side and generates HTML (or
whatever).
Unless
you
tell the PHP what browser you are using and write PHP code
to
take
account
of that (not generally recommended) then the server output
is
the
same
regardless of browser.
--
Peter Ford, Developer phone: 01580 89
fax:
01580
893399
Justcroft International Ltd.
www.justcroft.com
Justcroft House, High Street, Staplehurst, Kent TN12 0AH
United
Kingdom
Registered in England and Wales: 2297906
Registered office: Stag Gates House, 63/64 The Avenue,
Southampton
SO17
1XS
Pete
Once again, thank you for your help.
I was able to get it to work, I did not understand the
browser
differences
either, but on my testing server on IE, it worked as I wanted
but
not
on
anything else. I chopped out most of the code and ended up
with
this:
if ( isset($_POST['submit']) ) {} else {
print form action=\phpForm3.php\ method=\POST\\n;
if ($Recordset1) {
print table width=200 border=1\n;
print thnbsp; /th\n;
print th State /th\n; //2 fields in Counties table,
State
and
County
print th County /th\n;
print /tr\n;
//create table
$i = 0;
while ( $row = mysql_fetch_array($Recordset1) ) {
$i++;
print tr\n;
print tdinput type=\checkbox\ name=\county$i\
value=\$row[name]\/td\n;
echo td{$row['state_id']}/td\n;
echo td{$row['name']}/td\n;
echo /tr\n;
}//end while
print /table\n;
} else {
echo(PError performing query: .
mysql_error() . /P);
}
print input type=\hidden\ name=\counties\
value=\$i\/\n;
print input type=\submit\ name=\submit\
value=\Go\/\n;
}
?
Again, thank you.
Gary
__ Information from ESET Smart Security, version of
virus
signature database 5899 (20110223) __
The message was checked by ESET Smart Security.
http://www.eset.com
If you would allow me to show you a little easier way of doing
this.
if ( !isset($_POST['submit']) ) {
echo 'form action=phpForm3.php method=POST';
if ($Recordset1) {
echo HEADER
table width=200 border=1
trth nbsp; /thth State /thth County /th/tr
HEADER;
while ( $row = mysql_fetch_array($Recordset1) ) {
echo ROW
tr
tdinput type=checkbox name=county[]
value={$row['name']}/td
td{$row['state_id']}/td
td{$row['name']}/td
/tr
ROW;
}//end while
echo '/table';
} else {
echo 'pError performing query: '.mysql_error().'/p';
}
echo 'input type=submit name=submit value=Go /';
}
Now, the main thing I want you to see is the line for the
country
checkbox'es
The county[] turns the submitted values into an array. So, in
the
processing
script, you would do this.
?php
...
if ( !empty($_POST['county'])
foreach ( $_POST['county'] AS $id = $name )
echo {$id} {$name}\n;
...
?
Hope this clears things up for you a little.
Enjoy.
Jim
In a weird twist, I have/had the script working, but again it
is
not
working
[PHP] Re: Dynamically Created Checkboxes
Pete Ford p...@justcroft.com wrote in message
news:62.c1.32612.d49d4...@pb1.pair.com...
This bit?
On 22/02/11 22:06, Gary wrote:
for($i=1; $i=$_POST['counties']; $i++) {
if ( isset($_POST[county{$i}] ) ) {
You loop over $_POST['counties'] and look for $_POST[county$i]
I suspect that there is no field 'counties' in your form, so the server is
complaining about the missing index - seems likely that the production
server
is not showing the error, but instead just giving up on the page.
I think the IE7/Firefox browser difference is a red herring: it wasn't
actually working in any browser, or the code changed between tests.
Remember, PHP is server side and generates HTML (or whatever). Unless you
tell the PHP what browser you are using and write PHP code to take account
of that (not generally recommended) then the server output is the same
regardless of browser.
--
Peter Ford, Developer phone: 01580 89 fax: 01580
893399
Justcroft International Ltd.
www.justcroft.com
Justcroft House, High Street, Staplehurst, Kent TN12 0AH United
Kingdom
Registered in England and Wales: 2297906
Registered office: Stag Gates House, 63/64 The Avenue, Southampton SO17
1XS
Pete
Once again, thank you for your help.
I was able to get it to work, I did not understand the browser differences
either, but on my testing server on IE, it worked as I wanted but not on
anything else. I chopped out most of the code and ended up with this:
if ( isset($_POST['submit']) ) {} else {
print form action=\phpForm3.php\ method=\POST\\n;
if ($Recordset1) {
print table width=200 border=1\n;
print thnbsp; /th\n;
print th State /th\n; //2 fields in Counties table, State and County
print th County /th\n;
print /tr\n;
//create table
$i = 0;
while ( $row = mysql_fetch_array($Recordset1) ) {
$i++;
print tr\n;
print tdinput type=\checkbox\ name=\county$i\
value=\$row[name]\/td\n;
echo td{$row['state_id']}/td\n;
echo td{$row['name']}/td\n;
echo /tr\n;
}//end while
print /table\n;
} else {
echo(PError performing query: .
mysql_error() . /P);
}
print input type=\hidden\ name=\counties\ value=\$i\/\n;
print input type=\submit\ name=\submit\ value=\Go\/\n;
}
?
Again, thank you.
Gary
__ Information from ESET Smart Security, version of virus signature
database 5899 (20110223) __
The message was checked by ESET Smart Security.
http://www.eset.com
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Re: Dynamically Created Checkboxes
Jim Lucas li...@cmsws.com wrote in message
news:4d653673.7040...@cmsws.com...
On 2/23/2011 4:35 AM, Gary wrote:
Pete Ford p...@justcroft.com wrote in message
news:62.c1.32612.d49d4...@pb1.pair.com...
This bit?
On 22/02/11 22:06, Gary wrote:
for($i=1; $i=$_POST['counties']; $i++) {
if ( isset($_POST[county{$i}] ) ) {
You loop over $_POST['counties'] and look for $_POST[county$i]
I suspect that there is no field 'counties' in your form, so the server
is
complaining about the missing index - seems likely that the production
server
is not showing the error, but instead just giving up on the page.
I think the IE7/Firefox browser difference is a red herring: it wasn't
actually working in any browser, or the code changed between tests.
Remember, PHP is server side and generates HTML (or whatever). Unless
you
tell the PHP what browser you are using and write PHP code to take
account
of that (not generally recommended) then the server output is the same
regardless of browser.
--
Peter Ford, Developer phone: 01580 89 fax: 01580
893399
Justcroft International Ltd.
www.justcroft.com
Justcroft House, High Street, Staplehurst, Kent TN12 0AH United
Kingdom
Registered in England and Wales: 2297906
Registered office: Stag Gates House, 63/64 The Avenue, Southampton SO17
1XS
Pete
Once again, thank you for your help.
I was able to get it to work, I did not understand the browser
differences
either, but on my testing server on IE, it worked as I wanted but not on
anything else. I chopped out most of the code and ended up with this:
if ( isset($_POST['submit']) ) {} else {
print form action=\phpForm3.php\ method=\POST\\n;
if ($Recordset1) {
print table width=200 border=1\n;
print thnbsp; /th\n;
print th State /th\n; //2 fields in Counties table, State and
County
print th County /th\n;
print /tr\n;
//create table
$i = 0;
while ( $row = mysql_fetch_array($Recordset1) ) {
$i++;
print tr\n;
print tdinput type=\checkbox\ name=\county$i\
value=\$row[name]\/td\n;
echo td{$row['state_id']}/td\n;
echo td{$row['name']}/td\n;
echo /tr\n;
}//end while
print /table\n;
} else {
echo(PError performing query: .
mysql_error() . /P);
}
print input type=\hidden\ name=\counties\ value=\$i\/\n;
print input type=\submit\ name=\submit\ value=\Go\/\n;
}
?
Again, thank you.
Gary
__ Information from ESET Smart Security, version of virus
signature database 5899 (20110223) __
The message was checked by ESET Smart Security.
http://www.eset.com
If you would allow me to show you a little easier way of doing this.
if ( !isset($_POST['submit']) ) {
echo 'form action=phpForm3.php method=POST';
if ($Recordset1) {
echo HEADER
table width=200 border=1
trth nbsp; /thth State /thth County /th/tr
HEADER;
while ( $row = mysql_fetch_array($Recordset1) ) {
echo ROW
tr
tdinput type=checkbox name=county[] value={$row['name']}/td
td{$row['state_id']}/td
td{$row['name']}/td
/tr
ROW;
}//end while
echo '/table';
} else {
echo 'pError performing query: '.mysql_error().'/p';
}
echo 'input type=submit name=submit value=Go /';
}
Now, the main thing I want you to see is the line for the country
checkbox'es
The county[] turns the submitted values into an array. So, in the
processing
script, you would do this.
?php
...
if ( !empty($_POST['county'])
foreach ( $_POST['county'] AS $id = $name )
echo {$id} {$name}\n;
...
?
Hope this clears things up for you a little.
Enjoy.
Jim
Jim
Thank you very much, I have adapted your code and it is working. I
appriciate your help.
Gary
__ Information from ESET Smart Security, version of virus signature
database 5900 (20110223) __
The message was checked by ESET Smart Security.
http://www.eset.com
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Re: Dynamically Created Checkboxes
Jim Lucas li...@cmsws.com wrote in message
news:4d653673.7040...@cmsws.com...
On 2/23/2011 4:35 AM, Gary wrote:
Pete Ford p...@justcroft.com wrote in message
news:62.c1.32612.d49d4...@pb1.pair.com...
This bit?
On 22/02/11 22:06, Gary wrote:
for($i=1; $i=$_POST['counties']; $i++) {
if ( isset($_POST[county{$i}] ) ) {
You loop over $_POST['counties'] and look for $_POST[county$i]
I suspect that there is no field 'counties' in your form, so the server
is
complaining about the missing index - seems likely that the production
server
is not showing the error, but instead just giving up on the page.
I think the IE7/Firefox browser difference is a red herring: it wasn't
actually working in any browser, or the code changed between tests.
Remember, PHP is server side and generates HTML (or whatever). Unless
you
tell the PHP what browser you are using and write PHP code to take
account
of that (not generally recommended) then the server output is the same
regardless of browser.
--
Peter Ford, Developer phone: 01580 89 fax: 01580
893399
Justcroft International Ltd.
www.justcroft.com
Justcroft House, High Street, Staplehurst, Kent TN12 0AH United
Kingdom
Registered in England and Wales: 2297906
Registered office: Stag Gates House, 63/64 The Avenue, Southampton SO17
1XS
Pete
Once again, thank you for your help.
I was able to get it to work, I did not understand the browser
differences
either, but on my testing server on IE, it worked as I wanted but not on
anything else. I chopped out most of the code and ended up with this:
if ( isset($_POST['submit']) ) {} else {
print form action=\phpForm3.php\ method=\POST\\n;
if ($Recordset1) {
print table width=200 border=1\n;
print thnbsp; /th\n;
print th State /th\n; //2 fields in Counties table, State and
County
print th County /th\n;
print /tr\n;
//create table
$i = 0;
while ( $row = mysql_fetch_array($Recordset1) ) {
$i++;
print tr\n;
print tdinput type=\checkbox\ name=\county$i\
value=\$row[name]\/td\n;
echo td{$row['state_id']}/td\n;
echo td{$row['name']}/td\n;
echo /tr\n;
}//end while
print /table\n;
} else {
echo(PError performing query: .
mysql_error() . /P);
}
print input type=\hidden\ name=\counties\ value=\$i\/\n;
print input type=\submit\ name=\submit\ value=\Go\/\n;
}
?
Again, thank you.
Gary
__ Information from ESET Smart Security, version of virus
signature database 5899 (20110223) __
The message was checked by ESET Smart Security.
http://www.eset.com
If you would allow me to show you a little easier way of doing this.
if ( !isset($_POST['submit']) ) {
echo 'form action=phpForm3.php method=POST';
if ($Recordset1) {
echo HEADER
table width=200 border=1
trth nbsp; /thth State /thth County /th/tr
HEADER;
while ( $row = mysql_fetch_array($Recordset1) ) {
echo ROW
tr
tdinput type=checkbox name=county[] value={$row['name']}/td
td{$row['state_id']}/td
td{$row['name']}/td
/tr
ROW;
}//end while
echo '/table';
} else {
echo 'pError performing query: '.mysql_error().'/p';
}
echo 'input type=submit name=submit value=Go /';
}
Now, the main thing I want you to see is the line for the country
checkbox'es
The county[] turns the submitted values into an array. So, in the
processing
script, you would do this.
?php
...
if ( !empty($_POST['county'])
foreach ( $_POST['county'] AS $id = $name )
echo {$id} {$name}\n;
...
?
Hope this clears things up for you a little.
Enjoy.
Jim
In a weird twist, I have/had the script working, but again it is not
working in FF or Chrome. This is a link to the first page which runs you
through the pages.
I have the submit to trigger upon change so there is no submit button. This
works in IE7. If I add a submit button, it does not work in any of them.
I am wondering if this is more a problem with the html.
http://www.assessmentappeallawyer.com/forum_state.php
This is the code for the first processing page
if ( !isset($_POST['submit']) ) {
echo 'form action=forum_delete.php method=POST';
if ($Recordset1) {
echo HEADER
table width=200 border=0
trth nbsp; /thth State /thth County /th/tr
HEADER;
while ( $row = mysql_fetch_array($Recordset1) ) {
echo ROW
tr
tdinput type=checkbox name=county[] value={$row['name']}/td
td{$row['state_id']}/td
td{$row['name']}/td
/tr
ROW;
}//end while
echo '/table';
} else {
echo 'pError performing query: '.mysql_error().'/p';
}
echo 'input type=submit name=submit value=Select /';
}
?
This is the code for the second processing page.
?php
if ( !empty($_POST['county']))
foreach ( $_POST['county'] AS $id = $name )
echo 'You have selected '. {$name}.'br /';
$totalRows_county_result_net= ($totalRows_county_result) + (1);
echo [$totalRows_county_result_net];
echo $totalRows_county_result;
?
/div
!--eo center--
?php
$name=$_POST['name
Re: [PHP] Re: Dynamically Created Checkboxes
Steve Staples sstap...@mnsi.net wrote in message
news:1298490417.14826.2418.camel@webdev01...
On Wed, 2011-02-23 at 14:17 -0500, Gary wrote:
Jim Lucas li...@cmsws.com wrote in message
news:4d653673.7040...@cmsws.com...
On 2/23/2011 4:35 AM, Gary wrote:
Pete Ford p...@justcroft.com wrote in message
news:62.c1.32612.d49d4...@pb1.pair.com...
This bit?
On 22/02/11 22:06, Gary wrote:
for($i=1; $i=$_POST['counties']; $i++) {
if ( isset($_POST[county{$i}] ) ) {
You loop over $_POST['counties'] and look for $_POST[county$i]
I suspect that there is no field 'counties' in your form, so the
server
is
complaining about the missing index - seems likely that the
production
server
is not showing the error, but instead just giving up on the page.
I think the IE7/Firefox browser difference is a red herring: it
wasn't
actually working in any browser, or the code changed between tests.
Remember, PHP is server side and generates HTML (or whatever). Unless
you
tell the PHP what browser you are using and write PHP code to take
account
of that (not generally recommended) then the server output is the
same
regardless of browser.
--
Peter Ford, Developer phone: 01580 89 fax: 01580
893399
Justcroft International Ltd.
www.justcroft.com
Justcroft House, High Street, Staplehurst, Kent TN12 0AH United
Kingdom
Registered in England and Wales: 2297906
Registered office: Stag Gates House, 63/64 The Avenue, Southampton
SO17
1XS
Pete
Once again, thank you for your help.
I was able to get it to work, I did not understand the browser
differences
either, but on my testing server on IE, it worked as I wanted but not
on
anything else. I chopped out most of the code and ended up with this:
if ( isset($_POST['submit']) ) {} else {
print form action=\phpForm3.php\ method=\POST\\n;
if ($Recordset1) {
print table width=200 border=1\n;
print thnbsp; /th\n;
print th State /th\n; //2 fields in Counties table, State and
County
print th County /th\n;
print /tr\n;
//create table
$i = 0;
while ( $row = mysql_fetch_array($Recordset1) ) {
$i++;
print tr\n;
print tdinput type=\checkbox\ name=\county$i\
value=\$row[name]\/td\n;
echo td{$row['state_id']}/td\n;
echo td{$row['name']}/td\n;
echo /tr\n;
}//end while
print /table\n;
} else {
echo(PError performing query: .
mysql_error() . /P);
}
print input type=\hidden\ name=\counties\ value=\$i\/\n;
print input type=\submit\ name=\submit\ value=\Go\/\n;
}
?
Again, thank you.
Gary
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If you would allow me to show you a little easier way of doing this.
if ( !isset($_POST['submit']) ) {
echo 'form action=phpForm3.php method=POST';
if ($Recordset1) {
echo HEADER
table width=200 border=1
trth nbsp; /thth State /thth County /th/tr
HEADER;
while ( $row = mysql_fetch_array($Recordset1) ) {
echo ROW
tr
tdinput type=checkbox name=county[]
value={$row['name']}/td
td{$row['state_id']}/td
td{$row['name']}/td
/tr
ROW;
}//end while
echo '/table';
} else {
echo 'pError performing query: '.mysql_error().'/p';
}
echo 'input type=submit name=submit value=Go /';
}
Now, the main thing I want you to see is the line for the country
checkbox'es
The county[] turns the submitted values into an array. So, in the
processing
script, you would do this.
?php
...
if ( !empty($_POST['county'])
foreach ( $_POST['county'] AS $id = $name )
echo {$id} {$name}\n;
...
?
Hope this clears things up for you a little.
Enjoy.
Jim
In a weird twist, I have/had the script working, but again it is not
working in FF or Chrome. This is a link to the first page which runs you
through the pages.
I have the submit to trigger upon change so there is no submit button.
This
works in IE7. If I add a submit button, it does not work in any of them.
I am wondering if this is more a problem with the html.
http://www.assessmentappeallawyer.com/forum_state.php
This is the code for the first processing page
if ( !isset($_POST['submit']) ) {
echo 'form action=forum_delete.php method=POST';
if ($Recordset1) {
echo HEADER
table width=200 border=0
trth nbsp; /thth State /thth County /th/tr
HEADER;
while ( $row = mysql_fetch_array($Recordset1) ) {
echo ROW
tr
tdinput type=checkbox name=county[]
value={$row['name']}/td
td{$row['state_id']}/td
td{$row['name']}/td
/tr
ROW;
}//end while
echo '/table';
} else {
echo 'pError performing query: '.mysql_error().'/p';
}
echo 'input type=submit name=submit value=Select /';
}
?
This is the code for the second processing
Re: [PHP] mysql_num_rows()
Thanks for your reply
Deva devendra...@gmail.com wrote in message
news:aanlktimvycqsi5nejd1lxxaqbqsb+q_onhoc7omgj...@mail.gmail.com...
I tried your script locally. Its working for me.
On Tue, Feb 22, 2011 at 11:10 AM, Gary gp...@paulgdesigns.com wrote:
Can someone tell me why this is not working? I do not get an error
message,
the results are called and echo'd to screen, the count does not work, I
get
a 0 for a result...
$result = mysql_query(SELECT * FROM `counties` WHERE name = 'checked')
or
die(mysql_error());
if ( isset($_POST['submit']) ) {
for($i=1; $i=$_POST['counties']; $i++) {
if ( isset($_POST[county$i] ) ) {
echo You have chosen . $_POST[county$i]. br/;
}
}
}
$county_total=mysql_num_rows($result);
while($row = mysql_fetch_array($result))
{$i++;
echo $row['name'];
}
$county_total=mysql_num_rows($result);
echo $county_total;
echo 'You Have ' . $county_total . ' Counties ';
?
--
Gary
Thanks for your reply.
I'm curious, how did you try it locally without a database?
I am still getting either a 0 or a 1 for a result.
Gary
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Re: [PHP] mysql_num_rows()
Daniel Molina Wegener d...@coder.cl wrote in message
news:201102220747.11...@coder.cl...
On Tuesday 22 February 2011,
Gary gp...@paulgdesigns.com wrote:
Can someone tell me why this is not working? I do not get an error
message, the results are called and echo'd to screen, the count does not
work, I get a 0 for a result...
Are you sure that the table is called `counties` and not `countries`?
:)
$result = mysql_query(SELECT * FROM `counties` WHERE name = 'checked')
or die(mysql_error());
if ( isset($_POST['submit']) ) {
for($i=1; $i=$_POST['counties']; $i++) {
if ( isset($_POST[county$i] ) ) {
echo You have chosen . $_POST[county$i]. br/;
}
}
}
$county_total=mysql_num_rows($result);
while($row = mysql_fetch_array($result))
{$i++;
echo $row['name'];
}
$county_total=mysql_num_rows($result);
echo $county_total;
echo 'You Have ' . $county_total . ' Counties ';
?
Best regards,
--
Daniel Molina Wegener dmw [at] coder [dot] cl
System Programmer Web Developer
Phone: +56 (2) 979-0277 | Blog: http://coder.cl/
Daniel
Thank you for your reply, yes the database is named counties, working and
returning results.
Gary
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[PHP] Re: mysql_num_rows()
Pete Ford p...@justcroft.com wrote in message
news:a4.c0.39221.b3ca3...@pb1.pair.com...
On 22/02/11 05:40, Gary wrote:
Can someone tell me why this is not working? I do not get an error
message,
the results are called and echo'd to screen, the count does not work, I
get
a 0 for a result...
$result = mysql_query(SELECT * FROM `counties` WHERE name = 'checked')
or
die(mysql_error());
if ( isset($_POST['submit']) ) {
for($i=1; $i=$_POST['counties']; $i++) {
if ( isset($_POST[county$i] ) ) {
echo You have chosen . $_POST[county$i].br/;
}
}
}
$county_total=mysql_num_rows($result);
while($row = mysql_fetch_array($result))
{$i++;
echo $row['name'];
}
$county_total=mysql_num_rows($result);
echo $county_total;
echo 'You Have ' . $county_total . ' Counties ';
?
The first thing I see is that you do
$county_total=mysql_num_rows($result)
twice. Now I'm not to sure, but it is possible that looping over
mysql_fetch_array($result) moves an array pointer so the count is not
valid after the operation... that's a long shot.
But since you are looping over the database records anyway, why not just
count them as you go? Isn't $i the number of counties after all the
looping?
--
Peter Ford, Developer phone: 01580 89 fax: 01580
893399
Justcroft International Ltd.
www.justcroft.com
Justcroft House, High Street, Staplehurst, Kent TN12 0AH United
Kingdom
Registered in England and Wales: 2297906
Registered office: Stag Gates House, 63/64 The Avenue, Southampton SO17
1XS
Peter
Thank you for your reply.
I did notice that I had the $county_total=mysql_num_rows($result) twice. I
had moved and removed it, always getting either a 0 or 1 result.
I put up another test page with
$result = mysql_query(SELECT * FROM `counties` ) or die(mysql_error());
$tot=mysql_num_rows($result);
echo $tot;
?
And it worked fine.
I have tried count() as well as mysql_num_rows() but am getting the same
result.
Could you explain what you mean by count them as I go?
Again, Thank you for your reply.
Gary
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[PHP] Re: mysql_num_rows()
Pete Ford p...@justcroft.com wrote in message
news:76.48.39221.054c3...@pb1.pair.com...
On 22/02/11 13:59, Gary wrote:
Pete Fordp...@justcroft.com wrote in message
news:a4.c0.39221.b3ca3...@pb1.pair.com...
On 22/02/11 05:40, Gary wrote:
Can someone tell me why this is not working? I do not get an error
message,
the results are called and echo'd to screen, the count does not work, I
get
a 0 for a result...
$result = mysql_query(SELECT * FROM `counties` WHERE name =
'checked')
or
die(mysql_error());
if ( isset($_POST['submit']) ) {
for($i=1; $i=$_POST['counties']; $i++) {
if ( isset($_POST[county$i] ) ) {
echo You have chosen . $_POST[county$i].br/;
}
}
}
$county_total=mysql_num_rows($result);
while($row = mysql_fetch_array($result))
{$i++;
echo $row['name'];
}
$county_total=mysql_num_rows($result);
echo $county_total;
echo 'You Have ' . $county_total . ' Counties ';
?
The first thing I see is that you do
$county_total=mysql_num_rows($result)
twice. Now I'm not to sure, but it is possible that looping over
mysql_fetch_array($result) moves an array pointer so the count is not
valid after the operation... that's a long shot.
But since you are looping over the database records anyway, why not just
count them as you go? Isn't $i the number of counties after all the
looping?
--
Peter Ford, Developer phone: 01580 89 fax: 01580
893399
Justcroft International Ltd.
www.justcroft.com
Justcroft House, High Street, Staplehurst, Kent TN12 0AH United
Kingdom
Registered in England and Wales: 2297906
Registered office: Stag Gates House, 63/64 The Avenue, Southampton SO17
1XS
Peter
Thank you for your reply.
I did notice that I had the $county_total=mysql_num_rows($result) twice.
I
had moved and removed it, always getting either a 0 or 1 result.
I put up another test page with
$result = mysql_query(SELECT * FROM `counties` ) or die(mysql_error());
$tot=mysql_num_rows($result);
echo $tot;
?
And it worked fine.
I have tried count() as well as mysql_num_rows() but am getting the same
result.
Could you explain what you mean by count them as I go?
Again, Thank you for your reply.
Gary
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Well,
Lets go back to your original code and cut out the decoration:
$result = mysql_query(SELECT * FROM `counties` WHERE name = 'checked')
$county_total=mysql_num_rows($result);
while($row = mysql_fetch_array($result))
{
echo $row['name'];
}
echo $county_total;
That code should do pretty much what you want...
But, because you only show the number of records AFTER the loop, you could
do:
$result = mysql_query(SELECT * FROM `counties` WHERE name = 'checked')
$county_total = 0;
while($row = mysql_fetch_array($result))
{
echo $row['name'];
$county_total++;
}
echo $county_total;
Peter
Thank you for your suggestion...btw cut out the decoration LOL
I'm sorry to report it did not work. I tried various configurations of it
but to no avail.
Any other suggestions?
Again, thank you for your help.
Gary
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[PHP] Dynamically Created Checkboxes
I have a script that was working, however I discovered it was only working
in IE7, not in FF or Chrome. It also has stopped working in IE7 once I put
it on the remote server. It is to create checkboxes from data called from
the db. I have echo'd the mysql_num_row and am getting a result, so the
query seems to be working. At this point I get a blank page. Can someone
point me in the right direction to get this working (again).
$choice=$_POST['state'];
mysql_select_db($database_assess, $assess_remote);
$query_Recordset1 = (SELECT * FROM `counties` WHERE state_id =
'$choice')or die(mysql_error());
$Recordset1 = mysql_query($query_Recordset1, $assess_remote) or
die(mysql_error());
$row_Recordset1 = mysql_fetch_assoc($Recordset1);
$totalRows_Recordset1 = mysql_num_rows($Recordset1);
?php
echo $totalRows_Recordset1;
echo $row_Recordset1;
if ( isset($_POST['submit']) ) { // if form is submitted, process it
/I believe my issue is
below\\\
for($i=1; $i=$_POST['counties']; $i++) {
if ( isset($_POST[county{$i}] ) ) {
print $_POST[county{$i}]. is checked.br/;
}
}
} else { // if form isn't submitted, output the form
$county_choice=$_SESSION['county{$1}'];
print form action=\phpForm3.php\ method=\POST\\n;
if ($Recordset1) {
print table width=200 border=1\n;
print thnbsp; /th\n;
print th State /th\n; //2 fields in Counties table, State and County
print th County /th\n;
print /tr\n;
//create table
$i = 0;
while ( $row = mysql_fetch_array($Recordset1) ) {
$i++;
print tr\n;
print tdinput type=\checkbox\ name=\county$i\
value=\$row[name]\/td\n;
echo td{$row['state_id']}/td\n;
echo td{$row['name']}/td\n;
echo /tr\n;
}//end while
print /table\n;
} else {
echo(PError performing query: .
mysql_error() . /P);
}
print input type=\hidden\ name=\counties\ value=\$i\/\n;
print input type=\submit\ name=\submit\ value=\Go\/\n;
}
?
Would appriciate your help.
Thank you.
--
Gary
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[PHP] mysql_num_rows()
Can someone tell me why this is not working? I do not get an error message,
the results are called and echo'd to screen, the count does not work, I get
a 0 for a result...
$result = mysql_query(SELECT * FROM `counties` WHERE name = 'checked') or
die(mysql_error());
if ( isset($_POST['submit']) ) {
for($i=1; $i=$_POST['counties']; $i++) {
if ( isset($_POST[county$i] ) ) {
echo You have chosen . $_POST[county$i]. br/;
}
}
}
$county_total=mysql_num_rows($result);
while($row = mysql_fetch_array($result))
{$i++;
echo $row['name'];
}
$county_total=mysql_num_rows($result);
echo $county_total;
echo 'You Have ' . $county_total . ' Counties ';
?
--
Gary
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[PHP] All records not displaying...
I have an issue that the first record in a query is not being displayed. It
seems that the first row in alphabetical order is not being brought to the
screen.
I have run the query in the DB and it displays the correct result, so it has
to be in the php.
I have a MySQL DB that lists beers. I have a column for 'type' of beer
(imported, domestic, craft, light). The queries:
$result = MySQL_query(SELECT * FROM beer WHERE type = 'imported' AND stock
= 'YES' ORDER by beername );
When I run the query
if (mysql_num_rows($result) == !'0') {
$row = mysql_fetch_array($result);
echo 'h3Imported Beers/h3';
echo 'table width=100% border=0 cellspacing=1 cellpadding=1
id=tableone summary=
thBeer/th
thMaker/th
thType/th
thSingles/th
th6-Packs/th
thCans/th
thBottles/th
thDraft/th
thSize/th
thDescription/th';
while ($row = mysql_fetch_array($result)) {
echo 'tr td' . $row['beername'].'/td';
echo 'td' . $row['manu'] . '/td';
echo 'td' . $row['type'] . '/td';
echo 'td width=40' . $row['singles'] . '/td';
echo 'td width=20' . $row['six'] . '/td';
echo 'td width=40' . $row['can'] . '/td';
echo 'td width=20' . $row['bottles'] . '/td';
echo 'td width=40' . $row['tap'] . '/td';
echo 'td' . $row['size'] . '/td';
echo 'td' . $row['descrip'] . '/td';
'/tr';
}
echo '/tablebr /';
}
All but the first row in alphabetical order are displayed properly.
Can anyone tell me where I am going wrong?
--
Gary
BTW, I do have a bonus question that is about javascript in this same file,
so if anyone want to take a stab at that, I'll be happy to post it.
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Re: [PHP] All records not displaying...
Tamara Temple tamouse.li...@gmail.com wrote in message
news:c6993909-dd90-4f52-bf6b-ab888c281...@gmail.com...
On Dec 19, 2010, at 9:46 AM, Gary wrote:
I have an issue that the first record in a query is not being displayed.
It
seems that the first row in alphabetical order is not being brought to
the
screen.
I have run the query in the DB and it displays the correct result, so it
has
to be in the php.
I have a MySQL DB that lists beers. I have a column for 'type' of beer
(imported, domestic, craft, light). The queries:
$result = MySQL_query(SELECT * FROM beer WHERE type = 'imported' AND
stock
= 'YES' ORDER by beername );
When I run the query
if (mysql_num_rows($result) == !'0') {
$row = mysql_fetch_array($result);
echo 'h3Imported Beers/h3';
echo 'table width=100% border=0 cellspacing=1 cellpadding=1
id=tableone summary=
thBeer/th
thMaker/th
thType/th
thSingles/th
th6-Packs/th
thCans/th
thBottles/th
thDraft/th
thSize/th
thDescription/th';
while ($row = mysql_fetch_array($result)) {
echo 'tr td' . $row['beername'].'/td';
echo 'td' . $row['manu'] . '/td';
echo 'td' . $row['type'] . '/td';
echo 'td width=40' . $row['singles'] . '/td';
echo 'td width=20' . $row['six'] . '/td';
echo 'td width=40' . $row['can'] . '/td';
echo 'td width=20' . $row['bottles'] . '/td';
echo 'td width=40' . $row['tap'] . '/td';
echo 'td' . $row['size'] . '/td';
echo 'td' . $row['descrip'] . '/td';
'/tr';
}
echo '/tablebr /';
}
All but the first row in alphabetical order are displayed properly.
Can anyone tell me where I am going wrong?
--
Gary
BTW, I do have a bonus question that is about javascript in this same
file,
so if anyone want to take a stab at that, I'll be happy to post it.
This code will totally eliminate the first row of data.
if (mysql_num_rows($result) == !'0') {
$row = mysql_fetch_array($result);
Fetches the first row, but is not output. Because:
while ($row = mysql_fetch_array($result)) {
Fetches the second row before you do any output of the data.
Eliminate the first fetch_array and you're code should work fine.
BTW, if you put the td attributes 'width=n' in the preceding th
tags, you won't have to output them for each row. You should also put the
units those numbers are associated with.
Tamara
Thank you for your help and thank you for the explaination. I removed the
line and it works fine. I dont remember where or why I had that line in
there, it is code that I have recycled for a while now.
Gary
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[PHP] Error Querying Database
I cant seem to get this to connect. This is to my local testing server,
which is on, so we need not worry that I have posted the UN/PW.
This is a duplicate of a script I have used countless times and it worked.
The error message is 'Error querying database.'
Some one point out the error of my ways?
Gary
form action=?php echo $_SERVER[PHP_SELF]; ? method=post
tr
td
labelName of Beer/label/tdtdinput name=beername type=text /
/td
/tr
tr
td
labelMaker of Beer/label/tdtdinput name=manu type=text /
/td
/tr
tr
td
labelType of Beer/label/td
tdselect name=type size=1 id=type
optionImported/option
optionDomestic/option
optionCraft/option
optionLight/option
/select
!--select name=avail size=1 id=avail
optionAvailable/option
optionSold/option
/select--
/td
/tr
tr
tdlabelSold in/label
/tdtdinput type=checkbox name=singles value=Yes / Singlesbr /
input type=checkbox name=six value=Yes / Six Packs br /
input type=checkbox name=can value=Yes / Cansbr /
input type=checkbox name=bottles value=Yes / Bottles br /
input type=checkbox name=tap value=Yes / Draft br /
tr
td
labelSize/label/tdtdinput name=size type=text /
/td/tr
trtd
labelDescription/label/tdtdtextarea name=desc cols=40
rows=5/textarea
/td/tr
trtd
input name=submit type=submit value=Submit //td/tr
/form
/table
/div
div id=list
?php
$beername = $_POST['beername'];
$manu = $_POST['manu'];
$type = $_POST['type'];
$singles = $_POST['singles'];
$six = $_POST['six'];
$can = $_POST['can'];
$bottles = $_POST['bottles'];
$tap = $_POST['tap'];
$size = $_POST['size'];
$desc = $_POST['desc'];
$ip= $_SERVER['REMOTE_ADDR'];
$dbc = mysqli_connect('localhost','root','','rr')or die('Error connecting
with MySQL Database');
$query = INSERT INTO beer (beername, manu, type, singles, six, can,
bottles, tap, size, desc, ip ). VALUES ('$beername', '$manu', '$type',
'$singles', '$six', '$can', '$bottles', '$tap', '$size', '$desc', '$ip' );
$result = mysqli_query($dbc, $query)
or die('Error querying database.');
mysqli_close($dbc);
--
Gary
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Re: [PHP] Error Querying Database
Daniel P. Brown daniel.br...@parasane.net wrote in message
news:aanlkti=gemvzmpjg1uvb4_tdacqlbzyfjcgvd+har...@mail.gmail.com...
On Wed, Dec 15, 2010 at 13:42, Gary gp...@paulgdesigns.com wrote:
I cant seem to get this to connect. This is to my local testing server,
which is on, so we need not worry that I have posted the UN/PW.
This is a duplicate of a script I have used countless times and it worked.
The error message is 'Error querying database.'
Some one point out the error of my ways?
Replace:
or die('Error querying database.');
With:
or die('Error querying database. MySQL said: '.mysql_error());
--
/Daniel P. Brown
Dedicated Servers, Cloud and Cloud Hybrid Solutions, VPS, Hosting
(866-) 725-4321
http://www.parasane.net/
Daniel
Thank you for your reply, the new error message is
Error querying database. MySQL said:
Gary
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Re: [PHP] Error Querying Database
Steve Staples sstap...@mnsi.net wrote in message
news:1292440837.5460.8.ca...@webdev01...
On Wed, 2010-12-15 at 13:42 -0500, Gary wrote:
I cant seem to get this to connect. This is to my local testing server,
which is on, so we need not worry that I have posted the UN/PW.
This is a duplicate of a script I have used countless times and it
worked.
The error message is 'Error querying database.'
Some one point out the error of my ways?
Gary
form action=?php echo $_SERVER[PHP_SELF]; ? method=post
tr
td
labelName of Beer/label/tdtdinput name=beername type=text /
/td
/tr
tr
td
labelMaker of Beer/label/tdtdinput name=manu type=text /
/td
/tr
tr
td
labelType of Beer/label/td
tdselect name=type size=1 id=type
optionImported/option
optionDomestic/option
optionCraft/option
optionLight/option
/select
!--select name=avail size=1 id=avail
optionAvailable/option
optionSold/option
/select--
/td
/tr
tr
tdlabelSold in/label
/tdtdinput type=checkbox name=singles value=Yes / Singlesbr
/
input type=checkbox name=six value=Yes / Six Packs br /
input type=checkbox name=can value=Yes / Cansbr /
input type=checkbox name=bottles value=Yes / Bottles br /
input type=checkbox name=tap value=Yes / Draft br /
tr
td
labelSize/label/tdtdinput name=size type=text /
/td/tr
trtd
labelDescription/label/tdtdtextarea name=desc cols=40
rows=5/textarea
/td/tr
trtd
input name=submit type=submit value=Submit //td/tr
/form
/table
/div
div id=list
?php
$beername = $_POST['beername'];
$manu = $_POST['manu'];
$type = $_POST['type'];
$singles = $_POST['singles'];
$six = $_POST['six'];
$can = $_POST['can'];
$bottles = $_POST['bottles'];
$tap = $_POST['tap'];
$size = $_POST['size'];
$desc = $_POST['desc'];
$ip= $_SERVER['REMOTE_ADDR'];
$dbc = mysqli_connect('localhost','root','','rr')or die('Error connecting
with MySQL Database');
$query = INSERT INTO beer (beername, manu, type, singles, six, can,
bottles, tap, size, desc, ip ). VALUES ('$beername', '$manu', '$type',
'$singles', '$six', '$can', '$bottles', '$tap', '$size', '$desc',
'$ip' );
$result = mysqli_query($dbc, $query)
or die('Error querying database.');
mysqli_close($dbc);
--
Gary
Read Ash's reply... but basically, you're running the query with POST
variables, and inserting them on page display as well as on form submit.
can you ensure that you can connect from the command line?
if you may take some criticism, you should rethink your database design,
as well as the page flow/design... you should either post the form to a
new page, or if it is back to itself, you should check to see that you
have in fact posted it before just blindly inserting into the database
(as currently, every time you view the page, you will insert into the
database, even if completely empty values).
Steve
Thank you for your reply.
I did not see a reply from Ashley, but I would love to read it.
I always welcome criticism, however this form is for the owner of a bar
where he will inputing his list of beer that he sells. The rest of the code
that is not there is I will have the list then echo to screen below the
form. This is an internal list only, no customers will be seeing itif
that makes any difference to your suggestion.
On your one point
(as currently, every time you view the page, you will insert into the
database, even if completely empty values).
Is this always the case when you process a form onto itself? Or is there a
fix?
I did just create a new page, inserted the script onto it, and got the same
error message.
Again, thank you for your help.
Gary
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Re: [PHP] Error Querying Database
Sent: Wednesday, December 15, 2010 2:44 PM
Subject: Re: [PHP] Error Querying Database
On Wed, 2010-12-15 at 13:42 -0500, Gary wrote:
I cant seem to get this to connect. This is to my local testing
server,
which is on, so we need not worry that I have posted the UN/PW.
This is a duplicate of a script I have used countless times and it
worked.
The error message is 'Error querying database.'
Some one point out the error of my ways?
Gary
form action=?php echo $_SERVER[PHP_SELF]; ? method=post
tr
td
labelName of Beer/label/tdtdinput name=beername type=text
/
/td
/tr
tr
td
labelMaker of Beer/label/tdtdinput name=manu type=text /
/td
/tr
tr
td
labelType of Beer/label/td
tdselect name=type size=1 id=type
optionImported/option
optionDomestic/option
optionCraft/option
optionLight/option
/select
!--select name=avail size=1 id=avail
optionAvailable/option
optionSold/option
/select--
/td
/tr
tr
tdlabelSold in/label
/tdtdinput type=checkbox name=singles value=Yes /
Singlesbr
/
input type=checkbox name=six value=Yes / Six Packs br /
input type=checkbox name=can value=Yes / Cansbr /
input type=checkbox name=bottles value=Yes / Bottles br /
input type=checkbox name=tap value=Yes / Draft br /
tr
td
labelSize/label/tdtdinput name=size type=text /
/td/tr
trtd
labelDescription/label/tdtdtextarea name=desc cols=40
rows=5/textarea
/td/tr
trtd
input name=submit type=submit value=Submit //td/tr
/form
/table
/div
div id=list
?php
$beername = $_POST['beername'];
$manu = $_POST['manu'];
$type = $_POST['type'];
$singles = $_POST['singles'];
$six = $_POST['six'];
$can = $_POST['can'];
$bottles = $_POST['bottles'];
$tap = $_POST['tap'];
$size = $_POST['size'];
$desc = $_POST['desc'];
$ip= $_SERVER['REMOTE_ADDR'];
$dbc = mysqli_connect('localhost','root','','rr')or die('Error
connecting
with MySQL Database');
$query = INSERT INTO beer (beername, manu, type, singles, six, can,
bottles, tap, size, desc, ip ). VALUES ('$beername', '$manu',
'$type',
'$singles', '$six', '$can', '$bottles', '$tap', '$size', '$desc',
'$ip' );
$result = mysqli_query($dbc, $query)
or die('Error querying database.');
mysqli_close($dbc);
--
Gary
Read Ash's reply... but basically, you're running the query with POST
variables, and inserting them on page display as well as on form
submit.
can you ensure that you can connect from the command line?
if you may take some criticism, you should rethink your database
design,
as well as the page flow/design... you should either post the form to a
new page, or if it is back to itself, you should check to see that you
have in fact posted it before just blindly inserting into the database
(as currently, every time you view the page, you will insert into the
database, even if completely empty values).
Steve
Thank you for your reply.
I did not see a reply from Ashley, but I would love to read it.
I always welcome criticism, however this form is for the owner of a bar
where he will inputing his list of beer that he sells. The rest of the
code
that is not there is I will have the list then echo to screen below the
form. This is an internal list only, no customers will be seeing
itif
that makes any difference to your suggestion.
On your one point
(as currently, every time you view the page, you will insert into the
database, even if completely empty values).
Is this always the case when you process a form onto itself? Or is there
a
fix?
I did just create a new page, inserted the script onto it, and got the
same
error message.
Again, thank you for your help.
Gary
Gary
the line:
input name=submit type=submit value=Submit /
is the submit part... if you encapsulate the DB part of the code,
within:
if($_POST['submit'] == 'Submit')
{
# do db stuff in here and value sanitizing
}
basically, what that does is the submit button is name $_POST['submit']
and has the value Submit (the 'value' of the button) which will not be
set, until you submit the form. Personally, I do it another way, but
but is how most people check to see if a form is submitted (i think?)
but it seems as if your issue stems from the lack of being able to
connect to the database itself, so either your login credentials are
wrong, or you dont have the mysqli connector enabled with your php in
your development box.
check your phpinfo() to ensure it is enabled.
Steve
Steve
Again thank you, here is my phpini information, so it would seem mysqli is
enabled
MysqlI Support enabled
Client API library version 5.1.41
Active Persistent Links 0
Inactive Persistent Links 0
Active Links 13
Client API header version 5.1.41
MYSQLI_SOCKET MySQL
Directive Local Value Master Value
mysqli.allow_local_infile
[PHP] Re: Error Querying Database
Gary gp...@paulgdesigns.com wrote in message
news:81.d1.49824.33c09...@pb1.pair.com...
I cant seem to get this to connect. This is to my local testing server,
which is on, so we need not worry that I have posted the UN/PW.
This is a duplicate of a script I have used countless times and it worked.
The error message is 'Error querying database.'
Some one point out the error of my ways?
Gary
form action=?php echo $_SERVER[PHP_SELF]; ? method=post
tr
td
labelName of Beer/label/tdtdinput name=beername type=text /
/td
/tr
tr
td
labelMaker of Beer/label/tdtdinput name=manu type=text /
/td
/tr
tr
td
labelType of Beer/label/td
tdselect name=type size=1 id=type
optionImported/option
optionDomestic/option
optionCraft/option
optionLight/option
/select
!--select name=avail size=1 id=avail
optionAvailable/option
optionSold/option
/select--
/td
/tr
tr
tdlabelSold in/label
/tdtdinput type=checkbox name=singles value=Yes / Singlesbr
/
input type=checkbox name=six value=Yes / Six Packs br /
input type=checkbox name=can value=Yes / Cansbr /
input type=checkbox name=bottles value=Yes / Bottles br /
input type=checkbox name=tap value=Yes / Draft br /
tr
td
labelSize/label/tdtdinput name=size type=text /
/td/tr
trtd
labelDescription/label/tdtdtextarea name=desc cols=40
rows=5/textarea
/td/tr
trtd
input name=submit type=submit value=Submit //td/tr
/form
/table
/div
div id=list
?php
$beername = $_POST['beername'];
$manu = $_POST['manu'];
$type = $_POST['type'];
$singles = $_POST['singles'];
$six = $_POST['six'];
$can = $_POST['can'];
$bottles = $_POST['bottles'];
$tap = $_POST['tap'];
$size = $_POST['size'];
$desc = $_POST['desc'];
$ip= $_SERVER['REMOTE_ADDR'];
$dbc = mysqli_connect('localhost','root','','rr')or die('Error connecting
with MySQL Database');
$query = INSERT INTO beer (beername, manu, type, singles, six, can,
bottles, tap, size, desc, ip ). VALUES ('$beername', '$manu', '$type',
'$singles', '$six', '$can', '$bottles', '$tap', '$size', '$desc',
'$ip' );
$result = mysqli_query($dbc, $query)
or die('Error querying database.');
mysqli_close($dbc);
I have tried something completely different, changed to mysql instead of
mysqli and I put it up on my remote server/database, and it is still not
behaving...
mysql_connect('server','un','pw') or die(mysql_error());
mysql_select_db(db) or die(mysql_error());
mysql_query(INSERT INTO beer (beername, manu, type, singles, six, can,
bottles, tap, size, desc, ip).VALUES .('$beername', '$manu', '$type',
'$singles', '$six', '$can', '$bottles', '$tap', '$size', '$desc', '$ip' ))
or die(mysql_error());
echo Data Inserted!;
I am now getting this error message, but it does not make sense
You have an error in your SQL syntax; check the manual that corresponds to
your MySQL server version for the right syntax to use near 'desc, ip)VALUES
('', '', 'Imported', '', '', '', '', '', '', '', '68.80.24.11' at line 1
Does this mean I am getting closer to getting some actual work done today?
Thanks again
gary
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database 5706 (20101215) __
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Re: [PHP] Re: Error Querying Database
Bastien Koert phps...@gmail.com wrote in message
news:aanlktik8s8rgeatp_aq=qjbbuebq9y13zrnhoq+rb...@mail.gmail.com...
On Wed, Dec 15, 2010 at 3:41 PM, Gary gp...@paulgdesigns.com wrote:
Gary gp...@paulgdesigns.com wrote in message
news:81.d1.49824.33c09...@pb1.pair.com...
I cant seem to get this to connect. This is to my local testing server,
which is on, so we need not worry that I have posted the UN/PW.
This is a duplicate of a script I have used countless times and it
worked.
The error message is 'Error querying database.'
Some one point out the error of my ways?
Gary
form action=?php echo $_SERVER[PHP_SELF]; ? method=post
tr
td
labelName of Beer/label/tdtdinput name=beername type=text /
/td
/tr
tr
td
labelMaker of Beer/label/tdtdinput name=manu type=text /
/td
/tr
tr
td
labelType of Beer/label/td
tdselect name=type size=1 id=type
optionImported/option
optionDomestic/option
optionCraft/option
optionLight/option
/select
!--select name=avail size=1 id=avail
optionAvailable/option
optionSold/option
/select--
/td
/tr
tr
tdlabelSold in/label
/tdtdinput type=checkbox name=singles value=Yes / Singlesbr
/
input type=checkbox name=six value=Yes / Six Packs br /
input type=checkbox name=can value=Yes / Cansbr /
input type=checkbox name=bottles value=Yes / Bottles br /
input type=checkbox name=tap value=Yes / Draft br /
tr
td
labelSize/label/tdtdinput name=size type=text /
/td/tr
trtd
labelDescription/label/tdtdtextarea name=desc cols=40
rows=5/textarea
/td/tr
trtd
input name=submit type=submit value=Submit //td/tr
/form
/table
/div
div id=list
?php
$beername = $_POST['beername'];
$manu = $_POST['manu'];
$type = $_POST['type'];
$singles = $_POST['singles'];
$six = $_POST['six'];
$can = $_POST['can'];
$bottles = $_POST['bottles'];
$tap = $_POST['tap'];
$size = $_POST['size'];
$desc = $_POST['desc'];
$ip= $_SERVER['REMOTE_ADDR'];
$dbc = mysqli_connect('localhost','root','','rr')or die('Error connecting
with MySQL Database');
$query = INSERT INTO beer (beername, manu, type, singles, six, can,
bottles, tap, size, desc, ip ). VALUES ('$beername', '$manu', '$type',
'$singles', '$six', '$can', '$bottles', '$tap', '$size', '$desc',
'$ip' );
$result = mysqli_query($dbc, $query)
or die('Error querying database.');
mysqli_close($dbc);
I have tried something completely different, changed to mysql instead of
mysqli and I put it up on my remote server/database, and it is still not
behaving...
mysql_connect('server','un','pw') or die(mysql_error());
mysql_select_db(db) or die(mysql_error());
mysql_query(INSERT INTO beer (beername, manu, type, singles, six, can,
bottles, tap, size, desc, ip).VALUES .('$beername', '$manu', '$type',
'$singles', '$six', '$can', '$bottles', '$tap', '$size', '$desc',
'$ip' ))
or die(mysql_error());
echo Data Inserted!;
I am now getting this error message, but it does not make sense
You have an error in your SQL syntax; check the manual that corresponds to
your MySQL server version for the right syntax to use near 'desc,
ip)VALUES
('', '', 'Imported', '', '', '', '', '', '', '', '68.80.24.11' at line 1
Does this mean I am getting closer to getting some actual work done today?
Thanks again
gary
__ Information from ESET Smart Security, version of virus
signature database 5706 (20101215) __
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DESC is a reserved word in mysql (
http://dev.mysql.com/doc/refman/5.1/en/reserved-words.html )
Either change the field name or enclose the statement in backticks to
prevent this error. Changing the field name is my recommended action
--
Bastien
Cat, the other other white meat
And she appears.viola!
That was itthank you for your help...
I think this will require some onsite research at this site owners shop...
Thanks again
Gary
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Re: [PHP] Re: Error Querying Database
Steve Staples sstap...@mnsi.net wrote in message
news:1292447844.5460.16.ca...@webdev01...
On Wed, 2010-12-15 at 15:41 -0500, Gary wrote:
Gary gp...@paulgdesigns.com wrote in message
news:81.d1.49824.33c09...@pb1.pair.com...
I cant seem to get this to connect. This is to my local testing server,
which is on, so we need not worry that I have posted the UN/PW.
This is a duplicate of a script I have used countless times and it
worked.
The error message is 'Error querying database.'
Some one point out the error of my ways?
Gary
form action=?php echo $_SERVER[PHP_SELF]; ? method=post
tr
td
labelName of Beer/label/tdtdinput name=beername type=text
/
/td
/tr
tr
td
labelMaker of Beer/label/tdtdinput name=manu type=text /
/td
/tr
tr
td
labelType of Beer/label/td
tdselect name=type size=1 id=type
optionImported/option
optionDomestic/option
optionCraft/option
optionLight/option
/select
!--select name=avail size=1 id=avail
optionAvailable/option
optionSold/option
/select--
/td
/tr
tr
tdlabelSold in/label
/tdtdinput type=checkbox name=singles value=Yes /
Singlesbr
/
input type=checkbox name=six value=Yes / Six Packs br /
input type=checkbox name=can value=Yes / Cansbr /
input type=checkbox name=bottles value=Yes / Bottles br /
input type=checkbox name=tap value=Yes / Draft br /
tr
td
labelSize/label/tdtdinput name=size type=text /
/td/tr
trtd
labelDescription/label/tdtdtextarea name=desc cols=40
rows=5/textarea
/td/tr
trtd
input name=submit type=submit value=Submit //td/tr
/form
/table
/div
div id=list
?php
$beername = $_POST['beername'];
$manu = $_POST['manu'];
$type = $_POST['type'];
$singles = $_POST['singles'];
$six = $_POST['six'];
$can = $_POST['can'];
$bottles = $_POST['bottles'];
$tap = $_POST['tap'];
$size = $_POST['size'];
$desc = $_POST['desc'];
$ip= $_SERVER['REMOTE_ADDR'];
$dbc = mysqli_connect('localhost','root','','rr')or die('Error
connecting
with MySQL Database');
$query = INSERT INTO beer (beername, manu, type, singles, six, can,
bottles, tap, size, desc, ip ). VALUES ('$beername', '$manu',
'$type',
'$singles', '$six', '$can', '$bottles', '$tap', '$size', '$desc',
'$ip' );
$result = mysqli_query($dbc, $query)
or die('Error querying database.');
mysqli_close($dbc);
I have tried something completely different, changed to mysql instead of
mysqli and I put it up on my remote server/database, and it is still not
behaving...
mysql_connect('server','un','pw') or die(mysql_error());
mysql_select_db(db) or die(mysql_error());
mysql_query(INSERT INTO beer (beername, manu, type, singles, six, can,
bottles, tap, size, desc, ip).VALUES .('$beername', '$manu', '$type',
'$singles', '$six', '$can', '$bottles', '$tap', '$size', '$desc',
'$ip' ))
or die(mysql_error());
echo Data Inserted!;
I am now getting this error message, but it does not make sense
You have an error in your SQL syntax; check the manual that corresponds
to
your MySQL server version for the right syntax to use near 'desc,
ip)VALUES
('', '', 'Imported', '', '', '', '', '', '', '', '68.80.24.11' at line 1
Does this mean I am getting closer to getting some actual work done
today?
Thanks again
gary
not sure if this is the issue or not, but you have reserved words in
your field names... try escaping them all in backticks... and the way i
do things with variables inside , i put them in {}... here is what i
mean:
mysql_query(INSERT INTO `beer` (`beername`, `manu`, `type`, `singles`,
`six`, `can`, `bottles`, `tap`, `size`, `desc`, `ip`).VALUES
.('{$beername}', '{$manu}', '{$type}', '{$singles}', '{$six}',
'{$can}', '{$bottles}', '{$tap}', '{$size}', '{$desc', '{$ip}'));
Steve
Steve
Thanks again, that was the issue...let me throw that into my vault of
knowledge about mysql (seems there is plenty of open space in there anyway)
Thanks again
Gary
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[PHP] empty() in email message
I have an email message
$msg = 'Name: $fname ' . ' $lname\n'
. Phone: $phone\n
. Email: $email\n
and it works fine, however in this message there are about 30 variables that
are being called...as such
. Order: beefschnitzel $beefschnitzel\n
. Order: beefstrips $beefstrips\n
. Order: cheesesausage $cheesesausage\n
. Order: crumbedsausage $crumbedsausage\n
. Order: chucksteak $chucksteak\n
. Order: cornedbeef $cornedbeef\n
. Order: dicedsteak $dicedsteak\n
. Order: filletmignon $filletmignon\n
I want to only send the message if the submitter enters an amount in the
form for the corresponding variable, instead of having a bunch of empty
messages. So I have been trying to use the empty() function as such:
. if empty($beefolives){''} elseif (isset($beefolives)) { 'Order: beefolives
$beefolives\n'}
But I am getting the error
Parse error: syntax error, unexpected T_IF
Can someone point me in the right direction?
Thank you
--
Gary
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Re: [PHP] empty() in email message
Bastien Koert phps...@gmail.com wrote in message
news:aanlktimimt6h+g+zg4a_78wls4bu09zzsa3jvdujj...@mail.gmail.com...
On Mon, Dec 13, 2010 at 12:47 PM, Gary gp...@paulgdesigns.com wrote:
I have an email message
$msg = 'Name: $fname ' . ' $lname\n'
. Phone: $phone\n
. Email: $email\n
and it works fine, however in this message there are about 30 variables
that
are being called...as such
. Order: beefschnitzel $beefschnitzel\n
. Order: beefstrips $beefstrips\n
. Order: cheesesausage $cheesesausage\n
. Order: crumbedsausage $crumbedsausage\n
. Order: chucksteak $chucksteak\n
. Order: cornedbeef $cornedbeef\n
. Order: dicedsteak $dicedsteak\n
. Order: filletmignon $filletmignon\n
I want to only send the message if the submitter enters an amount in the
form for the corresponding variable, instead of having a bunch of empty
messages. So I have been trying to use the empty() function as such:
. if empty($beefolives){''} elseif (isset($beefolives)) { 'Order:
beefolives
$beefolives\n'}
But I am getting the error
Parse error: syntax error, unexpected T_IF
Can someone point me in the right direction?
Thank you
--
Gary
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Less complication is better, reduce the code to the below. The empty
portion needs to be inside the parentheses. Also double quotes are
needed to make the vairable parse correctly into the value you want to
see
if (!empty($beefolives)) { echo Order: beefolives $beefolives\n;}
--
Bastien
Bastien
Thank you for your response, and I agree, I like simpler. Your script
however did not work in the email message, it did work in another section
where I echo to screen the information to the screen.
Does the echo command work in emails?
Thank you again.
Gary
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Re: [PHP] empty() in email message
Daniel P. Brown daniel.br...@parasane.net wrote in message
news:aanlktikw+h1nvf6tcg5hgh-2xsubtuosvgkwaynb=...@mail.gmail.com...
On Mon, Dec 13, 2010 at 12:47, Gary gp...@paulgdesigns.com wrote:
I have an email message
$msg = 'Name: $fname ' . ' $lname\n'
. Phone: $phone\n
. Email: $email\n
and it works fine, however in this message there are about 30 variables
that
are being called...as such
. Order: beefschnitzel $beefschnitzel\n
. Order: beefstrips $beefstrips\n
. Order: cheesesausage $cheesesausage\n
. Order: crumbedsausage $crumbedsausage\n
. Order: chucksteak $chucksteak\n
. Order: cornedbeef $cornedbeef\n
. Order: dicedsteak $dicedsteak\n
. Order: filletmignon $filletmignon\n
I want to only send the message if the submitter enters an amount in the
form for the corresponding variable, instead of having a bunch of empty
messages. So I have been trying to use the empty() function as such:
. if empty($beefolives){''} elseif (isset($beefolives)) { 'Order:
beefolives
$beefolives\n'}
But I am getting the error
Parse error: syntax error, unexpected T_IF
Can someone point me in the right direction?
That's because you're concatenating. When appending to a
variable, you can't use constructs and conditions, unless using
ternaries:
. (!empty($beefolives) is_numeric($beefolives) ? 'Order: beefolives
'.$beefolives : null)
However, for readability and ease of management, you might want to
try something like this. It makes at least two assumptions: you're
using a form post, and you understand that it's untested (as I'm just
typing it here into the body of this email).
?php
// If we've posted our order
if (isset($_POST) is_array($_POST['order'])) {
// Instantiate the $order variable
$order = null;
// Iterate the order quantities
foreach ($_POST['order'] as $k = $v) {
// If the field is set and the amount is a number
if (isset($v) is_numeric($v)) {
$order .= 'Order: '.$k.': '.$v.PHP_EOL;
}
}
/**
* Note: in your example, you used single (literal)
* quotes and included the variables. That would
* literally print $fname and $lname.
*/
$msg = 'Name: '.$_POST['fname'].' '.$_POST['lname'].PHP_EOL;
$msg .= 'Phone: '.$_POST['phone'].PHP_EOL;
$msg .= 'Email: '.$_POST['email'].PHP_EOL;
$msg .= PHP_EOL;
$msg .= $order;
// And then handle the rest of your processing. Here, we just echo.
echo 'pre'.PHP_EOL.$msg.'/pre'.PHP_EOL;
}
?
form method=post action=?php echo $_SERVER['PHP_SELF']; ?
First Name: input type=text name=fname id=fnamebr/
Last Name: input type=text name=lname id=lnamebr/
Phone: input type=text name=phone id=phonebr/
Email: input type=text name=email id=emailbr/
br /
h3Order/h3
Beef Schnitzel: input type=text name=order[beefschnitzel]
id=order[beefschnitzel] size=2 maxlength=3br/
Beef Strips: input type=text name=order[beefstrips]
id=order[beefstrips] size=2 maxlength=3br/
Cheese Sausage: input type=text name=order[cheesesausage]
id=order[cheesesausage] size=2 maxlength=3br/
Crumbed Sausage: input type=text name=order[crumbedsausage]
id=order[crumbedsausage] size=2 maxlength=3br/
Chuck Steak: input type=text name=order[chucksteak]
id=order[chucksteak] size=2 maxlength=3br/
Corned Beef: input type=text name=order[cornedbeef]
id=order[cornedbeef] size=2 maxlength=3br/
Diced Steak: input type=text name=order[dicedsteak]
id=order[dicedsteak] size=2 maxlength=3br/
Filet Mignon: input type=text name=order[filletmignon]
id=order[filletmignon] size=2 maxlength=3br/
!-- NOTE: Filet as displayed on the form has one 'L', but the
variable is unchanged. --
br /
input type=submit value=Order Now
/form
--
/Daniel P. Brown
Dedicated Servers, Cloud and Cloud Hybrid Solutions, VPS, Hosting
(866-) 725-4321
http://www.parasane.net/
Daniel
As always, thank you for your input. I had tried your first script, and it
worked to a degree. What I was getting was a message that was correct, but,
well, this is what it looked like in an email
Order: beefschnitzel
10Order:
Which was supposed to look like
Order: beefschnitzel 10
Order:
I have not tried your second suggestion yet, I'll let you know how that
works.
Again, thank you for your help.
Gary
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Re: [PHP] empty() in email message
Paul M Foster pa...@quillandmouse.com wrote in message
news:20101213181124.gq21...@quillandmouse.com...
On Mon, Dec 13, 2010 at 12:47:49PM -0500, Gary wrote:
I have an email message
$msg = 'Name: $fname ' . ' $lname\n'
. Phone: $phone\n
. Email: $email\n
and it works fine, however in this message there are about 30 variables
that
are being called...as such
. Order: beefschnitzel $beefschnitzel\n
. Order: beefstrips $beefstrips\n
. Order: cheesesausage $cheesesausage\n
. Order: crumbedsausage $crumbedsausage\n
. Order: chucksteak $chucksteak\n
. Order: cornedbeef $cornedbeef\n
. Order: dicedsteak $dicedsteak\n
. Order: filletmignon $filletmignon\n
I want to only send the message if the submitter enters an amount in the
form for the corresponding variable, instead of having a bunch of empty
messages. So I have been trying to use the empty() function as such:
. if empty($beefolives){''} elseif (isset($beefolives)) { 'Order:
beefolives
$beefolives\n'}
But I am getting the error
Parse error: syntax error, unexpected T_IF
Can someone point me in the right direction?
It looks like you're trying to do something like:
$str = 'something' .
if ($somethingelse)
'another string'
else
'a different string';
For one thing, you can't put a conditional on the right side of a
concatenation mark (.). You'd have to do it this way:
$str = 'something';
if ($somethingelse) {
$str .= 'another string';
}
else {
$str .= 'a different string';
}
You also can't do:
elseif (isset($beefolives)) {
'Order: beefolives $beefolives\n';
}
For one thing, surrounding the 'Order...' line with single quotes will
cause the \n *not* to be interpreted as a newline. Second, your 'Order:
beefolives...' does nothing. You haven't assigned it to anything or
operated on that string in any way.
You might want to study up on single versus double quotes, the
concatenation operator (.), etc.
Paul
--
Paul M. Foster
Paul
Thank you for your reponse. I will admit I get a little fuzzy on the single
v double quotation marks, generally the rule I go by is if it is a string, I
use singles, if any processing I use doubles. I am not recalling why did it
that way, perhaps I had doubles and it did not work so I was whittling down
the options
Thank you again for your help.
Gary
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Re: [PHP] empty() in email message
Andre Polykanine an...@oire.org wrote in message
news:1110253876.20101214025...@oire.org...
Hello Gary,
Try using this:
if (!empty($beefolives)) $msg.=Order: beef olives;
or:
$msg.=empty($beefolives)? : Order: beef olives;
--
With best regards from Ukraine,
Andre
Skype: Francophile
Twitter: http://twitter.com/m_elensule
Facebook: http://facebook.com/menelion
- Original message -
From: Gary gp...@paulgdesigns.com
To: php-general@lists.php.net php-general@lists.php.net
Date: Monday, December 13, 2010, 7:47:49 PM
Subject: [PHP] empty() in email message
I have an email message
$msg = 'Name: $fname ' . ' $lname\n'
. Phone: $phone\n
. Email: $email\n
and it works fine, however in this message there are about 30 variables
that
are being called...as such
. Order: beefschnitzel $beefschnitzel\n
. Order: beefstrips $beefstrips\n
. Order: cheesesausage $cheesesausage\n
. Order: crumbedsausage $crumbedsausage\n
. Order: chucksteak $chucksteak\n
. Order: cornedbeef $cornedbeef\n
. Order: dicedsteak $dicedsteak\n
. Order: filletmignon $filletmignon\n
I want to only send the message if the submitter enters an amount in the
form for the corresponding variable, instead of having a bunch of empty
messages. So I have been trying to use the empty() function as such:
. if empty($beefolives){''} elseif (isset($beefolives)) { 'Order:
beefolives
$beefolives\n'}
But I am getting the error
Parse error: syntax error, unexpected T_IF
Can someone point me in the right direction?
Thank you
--
Gary
Andre
That looks good, I will give that a try in the morning, thank you for your
help.
Gary
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Re: [PHP] empty() in email message
Daevid Vincent dae...@daevid.com wrote in message
news:7d7c84d94dd24035a620e68b5b937...@mascorp.com...
- Original message -
From: Gary gp...@paulgdesigns.com
To: php-general@lists.php.net php-general@lists.php.net
Date: Monday, December 13, 2010, 7:47:49 PM
Subject: [PHP] empty() in email message
I have an email message
$msg = 'Name: $fname ' . ' $lname\n'
. Phone: $phone\n
. Email: $email\n
and it works fine, however in this message there are about 30
variables that
are being called...as such
. Order: beefschnitzel $beefschnitzel\n
. Order: beefstrips $beefstrips\n
. Order: cheesesausage $cheesesausage\n
. Order: crumbedsausage $crumbedsausage\n
. Order: chucksteak $chucksteak\n
. Order: cornedbeef $cornedbeef\n
. Order: dicedsteak $dicedsteak\n
. Order: filletmignon $filletmignon\n
I want to only send the message if the submitter enters an
amount in the
form for the corresponding variable, instead of having a
bunch of empty
messages. So I have been trying to use the empty() function as such:
. if empty($beefolives){''} elseif (isset($beefolives)) {
'Order: beefolives
$beefolives\n'}
You are setting this up fundamentally wrong.
You should be using an array and looping through it.
Something like:
$myorder['cowface'] = 1;
$myorder['beefenweiner'] = 2;
$myorder['chucksteak'] = 1;
foreach ($myorder as $item = $quantity)
{
echo Order: $item x $quantity\n;
}
Then your array only contains the items someone actually puchased and how
many.
d
Daevid
I knew someone was going to point out this was a convoluted method, and I
agree. This was sent to me by someone that needed to make the mail form
work. My suggestion was to look into a pre-made shopping cart, however that
was not going to work for them, so I made the mail() work for them.
I had thought about putting it into an array, but had not gotten that far
into it. I will look over the code to see how it works.
Thank you for your help.
gary
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[PHP] E-mail injection question
I have been testing various scripts to kill email injection attacks. I
adapted this script and it seems to work well. Does anyone see any issues
with this?
?php
$newlinecounter = 0;
foreach($_POST as $key = $val_newline){
if(stristr($val_newline, '\r')){$newlinecounter++;}
if(stristr($val_newline, '\n')){$newlinecounter++;}
if(stristr($val_newline, '\\r')){$newlinecounter++;}
if(stristr($val_newline, '\\n')){$newlinecounter++;}
if(stristr($val_newline, '\r\n')){$newlinecounter++;}
if(stristr($val_newline, '\\r\\n')){$newlinecounter++;}
if(stristr($val_newline, 'Bcc')){$newlinecounter++;}
}
if ($newlinecounter = 1){ die('die scum die');
}
?
Thank you,
Gary
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Re: [PHP] Stripslashes
Adam Richardson simples...@gmail.com wrote in message news:aanlktin_9_tfe9q+dc2hoynsavccoyuecudkqd919...@mail.gmail.com... On Tue, Nov 16, 2010 at 10:10 PM, Gary gp...@paulgdesigns.com wrote: I was doing a test of stripslashes on a $_POST, when I recieved the email, all of the slashes were still in the data posted. I used : $fname = stripslashes($_POST['fname']); I input G\\a//r\y\\, and was expecting, according to the manuel G\a//r*y\, but got the original spelling. In this case, you should get the original, if I'm understanding correctly. Think of it like a basic math problem: Step 1: Happens automatically when you submit the form and PHP receives the form variables input + slashes = slashed_input Step 2: This happens when you call stripslashes. slashed_input - slashes = input The goal of stripslashes is that it will undo what happened automatically using magic_quotes_gpc (which essentially calls addslashes on the GPC vars behind the scenes) so you'll end up with the original input. So, working through your example: 1. You inputted into a form G\\a//r\y\\ and submitted the form. 2. PHP received G\\a//r\y\\ and added slashes (Ga//r\\y). 3. You called stripslashes (G\\a//r\y\\). I added: echo stripslashes($fname); and did get the expected result on the page, but not in the email from the $_POST. Here, you called stripslashes on something already stripped once, so you now have a new value (G\a//ry\). I also tried $fname = (stripslashes($_POST['fname'])); This would be no different than your attempt without enclosing parentheses. Now, let me just say that I detest magic_quotes, and it's best to run with them disabled so you don't even have to worry about this kind of issue (they've been deprecated.) But, perhaps you were just trying to learn about some piece of legacy code. Hope the explanation helps, Gary. Adam -- Nephtali: PHP web framework that functions beautifully http://nephtaliproject.com Adam Thanks for your reply. So if I disable magic_quotes, and I assume I can do that a script, then the stripslashes would work as the manuel said it would, meaning G\\a//r\y\\ becomes G\a//r'y\ I also assume that until php 6 is out and or I upgrade to it, I will have to deal with magic_quotes? Thank you for your help. Gary __ Information from ESET Smart Security, version of virus signature database 5627 (20101117) __ The message was checked by ESET Smart Security. http://www.eset.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] Stripslashes
I was doing a test of stripslashes on a $_POST, when I recieved the email, all of the slashes were still in the data posted. I used : $fname = stripslashes($_POST['fname']); I input G\\a//r\y\\, and was expecting, according to the manuel G\a//r*y\, but got the original spelling. I added: echo stripslashes($fname); and did get the expected result on the page, but not in the email from the $_POST. I also tried $fname = (stripslashes($_POST['fname'])); But got the same result. Can anyone tell me what I am not understaning? Thank you Gary __ Information from ESET Smart Security, version of virus signature database 5625 (20101116) __ The message was checked by ESET Smart Security. http://www.eset.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Watermark with GD
Tamara Temple tamouse.li...@gmail.com wrote in message
news:94df613c-6b40-4897-b101-21cf7d83a...@gmail.com...
On Oct 30, 2010, at 9:31 AM, Gary wrote:
tedd tedd.sperl...@gmail.com wrote in message
news:p06240800c8f1d19b9...@[192.168.1.2]...
At 3:05 PM -0400 10/29/10, Gary wrote:
I am trying to get the watermark to work, however I am having a
problem in
that the image is being called from a database (image sits in images
file).
The script in question is this
$image = imagecreatefromjpeg($_GET['src']);
However it produces an error message of
Warning: imagecreatefromjpeg() [function.imagecreatefromjpeg]:
Filename
cannot be empty in /home/content/a/l/i/alinde52/html/ imagesDetail.php
on
line 233
I have tried various methods, for example: ($_GET ['images/'] or
($_GET
['images/$row_WADAimages[image_id]].
Can anyone shed some light on this for me.
Thank you
Gary
Gary:
Several things.
1. Getting an image from a database? You mean that you are getting the
path of the image in the file system, right?
Side note: You could place the image inside the database using a BLOB
and
do away with the path all together. That has the benefit of being
portable -- you simply move the database to where ever you want it. The
downside is that the database becomes very large, but no more so that
the
file system. There are pro's and con's in storing actual images in a
database.
2. Using a GET is not the way to get the path. Instead, you have to
retrieve the path from the database table where the path is stored --
and that requires a MySQL query similar to SELECT * FROM database
WHERE
id=whatever. There are lot's of examples of how to pull data from a
database.
3. After getting the path, then you can create the watermark like so:
http://webbytedd.com/b/watermark/
Hope this helps,
tedd
--
---
http://sperling.com/
tedd
Thank you for your reply.
I was under the impression that the image is stored in a folder called
images, in fact the images file do go in, however I have the DB set up
for
longblob, averaging about 20kb each, so now I am unsure. I exported the
sql
so perhaps you can tell me.
Table structure for table `images`
--
CREATE TABLE IF NOT EXISTS `images` (
`image_id` int(10) unsigned NOT NULL AUTO_INCREMENT,
`caption` varchar(50) NOT NULL,
`wheretaken` varchar(100) NOT NULL,
`description` text NOT NULL,
`file_name` varchar(25) NOT NULL,
`image_file` longblob NOT NULL,
`submitted` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP,
PRIMARY KEY (`image_id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=1447 ;
When I call the images, which works fine, I do need to specify the path
that
leads to the images folder. Am I being redundant in this structure.
This is the script that I use to call the images. I have pulled out some
of
the html that styles the data.
?php if ($totalRows_WADAimages 0) { // Show if recordset not empty ?
?php echo $row_WADAimages[caption]; ?
src=images/?php echo $row_WADAimages[image_file]; ? ?php echo
$row_WADAimages[description]; ?
?php echo $row_WADAimages[where_taken]; ?
?php echo $row_WADAimages[description]; ?
Thank you for your help.
Gary
is this the imageDetail.php script?
From what it appears in the code, `image_file` is holding the file name
rather than the actual image. Yet you have `image_file` defined as a
longblob, which would make sense if you were storing the actual image
data in the data base rather than on the file system. As Ashley noted,
you can do it either way. I notice you also have a field `file_name` --
what is this used for? It sounds like your design is a bit off -- neither
one way or the other. The php shown doesn't look complete enough,
though -- where is the img tag and such? Also, going in and out of php on
each line is kind of a waste and looks sloppy. If i read that code right,
it would emit something like this:
caption text
src=images/pathtoimagefile description text
where taken text
description text
Look at the html that's emitted from the script and see if that's what
you get.
Can you post or pastebin more of the .php script so I can see more of
what is going on?
Tamara
Thanks for the reply, here is a link to the code of the page.
http://www.paulgdesigns.com/detailcode.php
Again, thank you for your help.
Gary
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Re: [PHP] Watermark with GD
Tamara
I realized I forgot to answer your question about file_name. It is a text
field that the photographer enters the name of the image file for easy
reference. He does not have access to the DB and he is inserting the images
from a form I created.
I was sure that the images were being stored and called from the images
folder and not directly from the DB, that is until tedd brought it up. I
did have an issue of getting the images to show in the beginning, however I
solved it by inserting the path in the call from the DB. I am wondering now
if they are stored in both locations.
Thank again
Gary
Tamara Temple tamouse.li...@gmail.com wrote in message
news:94df613c-6b40-4897-b101-21cf7d83a...@gmail.com...
On Oct 30, 2010, at 9:31 AM, Gary wrote:
tedd tedd.sperl...@gmail.com wrote in message
news:p06240800c8f1d19b9...@[192.168.1.2]...
At 3:05 PM -0400 10/29/10, Gary wrote:
I am trying to get the watermark to work, however I am having a
problem in
that the image is being called from a database (image sits in images
file).
The script in question is this
$image = imagecreatefromjpeg($_GET['src']);
However it produces an error message of
Warning: imagecreatefromjpeg() [function.imagecreatefromjpeg]:
Filename
cannot be empty in /home/content/a/l/i/alinde52/html/ imagesDetail.php
on
line 233
I have tried various methods, for example: ($_GET ['images/'] or
($_GET
['images/$row_WADAimages[image_id]].
Can anyone shed some light on this for me.
Thank you
Gary
Gary:
Several things.
1. Getting an image from a database? You mean that you are getting the
path of the image in the file system, right?
Side note: You could place the image inside the database using a BLOB
and
do away with the path all together. That has the benefit of being
portable -- you simply move the database to where ever you want it. The
downside is that the database becomes very large, but no more so that
the
file system. There are pro's and con's in storing actual images in a
database.
2. Using a GET is not the way to get the path. Instead, you have to
retrieve the path from the database table where the path is stored --
and that requires a MySQL query similar to SELECT * FROM database
WHERE
id=whatever. There are lot's of examples of how to pull data from a
database.
3. After getting the path, then you can create the watermark like so:
http://webbytedd.com/b/watermark/
Hope this helps,
tedd
--
---
http://sperling.com/
tedd
Thank you for your reply.
I was under the impression that the image is stored in a folder called
images, in fact the images file do go in, however I have the DB set up
for
longblob, averaging about 20kb each, so now I am unsure. I exported the
sql
so perhaps you can tell me.
Table structure for table `images`
--
CREATE TABLE IF NOT EXISTS `images` (
`image_id` int(10) unsigned NOT NULL AUTO_INCREMENT,
`caption` varchar(50) NOT NULL,
`wheretaken` varchar(100) NOT NULL,
`description` text NOT NULL,
`file_name` varchar(25) NOT NULL,
`image_file` longblob NOT NULL,
`submitted` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP,
PRIMARY KEY (`image_id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=1447 ;
When I call the images, which works fine, I do need to specify the path
that
leads to the images folder. Am I being redundant in this structure.
This is the script that I use to call the images. I have pulled out some
of
the html that styles the data.
?php if ($totalRows_WADAimages 0) { // Show if recordset not empty ?
?php echo $row_WADAimages[caption]; ?
src=images/?php echo $row_WADAimages[image_file]; ? ?php echo
$row_WADAimages[description]; ?
?php echo $row_WADAimages[where_taken]; ?
?php echo $row_WADAimages[description]; ?
Thank you for your help.
Gary
is this the imageDetail.php script?
From what it appears in the code, `image_file` is holding the file name
rather than the actual image. Yet you have `image_file` defined as a
longblob, which would make sense if you were storing the actual image
data in the data base rather than on the file system. As Ashley noted,
you can do it either way. I notice you also have a field `file_name` --
what is this used for? It sounds like your design is a bit off -- neither
one way or the other. The php shown doesn't look complete enough,
though -- where is the img tag and such? Also, going in and out of php on
each line is kind of a waste and looks sloppy. If i read that code right,
it would emit something like this:
caption text
src=images/pathtoimagefile description text
where taken text
description text
Look at the html that's emitted from the script and see if that's what
you get.
Can you post or pastebin more of the .php script so I can see more of
what is going on?
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Re: [PHP] Watermark with GD
tedd tedd.sperl...@gmail.com wrote in message news:p06240800c8f339e20...@[192.168.1.2]... At 9:19 AM -0400 10/31/10, Gary wrote: I was sure that the images were being stored and called from the images folder and not directly from the DB, that is until tedd brought it up. I did have an issue of getting the images to show in the beginning, however I solved it by inserting the path in the call from the DB. I am wondering now if they are stored in both locations. Gary Gary: Both locations? I think you are confusing yourself.You either have the images stored in the database OR the file system, but not both. However, I think you have the images stored as files within the file directory, such as: http://yourdomain.com/images/picture1.jpg Now, in your database you should have only the path name stored (i.e., images/picture1.jpg) and not the image. However, I usually only store the file names and NOT the path. That way I can move the images to where ever I want and only change their reference in the HTML. To recap, to show any image you use the HTML img tag, such as: img src=images/picture1.jpg To tie this statement to a database, you need to pull out the image name (or file path) from the database and use that. Something like this: $query = SELECT file_name FROM pictures WHERE id = '$id' ; $result = mysql_query($query) or die(); $row = mysql_fetch_array($result); $file_name = $row['file_name']; And then in the HTML you would use: img src=images/?php echo($file_name);? If you can get to this part, then we can deal with placing a watermark on the image before showing it. Please let me know when you get this part to work. Cheers, tedd PS: There is no need to use a LONG BLOB in your database -- that's just a waste of space. -- --- http://sperling.com/ tedd Thank you for your reply. I have the images being displayed in the browser fine, my issue was that I was not able to get the watermark to show up. I believe the issue was here $img=imagecreatefromjpeg($_GET['pic']); in that 'pic' is asking for the image that I wanted the watermark displayed on, and what I put in did not work. Since the image file name is being stored in the DB, I tried $row_WADAimages[image_file] as well as a few others. What do you suggest I changed longblob to, and is it safe to do so? Again, thank you for your help. gary __ Information from ESET Smart Security, version of virus signature database 5579 (20101031) __ The message was checked by ESET Smart Security. http://www.eset.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Watermark with GD
tedd tedd.sperl...@gmail.com wrote in message
news:p06240800c8f1d19b9...@[192.168.1.2]...
At 3:05 PM -0400 10/29/10, Gary wrote:
I am trying to get the watermark to work, however I am having a problem in
that the image is being called from a database (image sits in images
file).
The script in question is this
$image = imagecreatefromjpeg($_GET['src']);
However it produces an error message of
Warning: imagecreatefromjpeg() [function.imagecreatefromjpeg]: Filename
cannot be empty in /home/content/a/l/i/alinde52/html/imagesDetail.php on
line 233
I have tried various methods, for example: ($_GET ['images/'] or ($_GET
['images/$row_WADAimages[image_id]].
Can anyone shed some light on this for me.
Thank you
Gary
Gary:
Several things.
1. Getting an image from a database? You mean that you are getting the
path of the image in the file system, right?
Side note: You could place the image inside the database using a BLOB and
do away with the path all together. That has the benefit of being
portable -- you simply move the database to where ever you want it. The
downside is that the database becomes very large, but no more so that the
file system. There are pro's and con's in storing actual images in a
database.
2. Using a GET is not the way to get the path. Instead, you have to
retrieve the path from the database table where the path is stored --
and that requires a MySQL query similar to SELECT * FROM database WHERE
id=whatever. There are lot's of examples of how to pull data from a
database.
3. After getting the path, then you can create the watermark like so:
http://webbytedd.com/b/watermark/
Hope this helps,
tedd
--
---
http://sperling.com/
tedd
Thank you for your reply.
I was under the impression that the image is stored in a folder called
images, in fact the images file do go in, however I have the DB set up for
longblob, averaging about 20kb each, so now I am unsure. I exported the sql
so perhaps you can tell me.
Table structure for table `images`
--
CREATE TABLE IF NOT EXISTS `images` (
`image_id` int(10) unsigned NOT NULL AUTO_INCREMENT,
`caption` varchar(50) NOT NULL,
`wheretaken` varchar(100) NOT NULL,
`description` text NOT NULL,
`file_name` varchar(25) NOT NULL,
`image_file` longblob NOT NULL,
`submitted` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP,
PRIMARY KEY (`image_id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=1447 ;
When I call the images, which works fine, I do need to specify the path that
leads to the images folder. Am I being redundant in this structure.
This is the script that I use to call the images. I have pulled out some of
the html that styles the data.
?php if ($totalRows_WADAimages 0) { // Show if recordset not empty ?
?php echo $row_WADAimages[caption]; ?
src=images/?php echo $row_WADAimages[image_file]; ? ?php echo
$row_WADAimages[description]; ?
?php echo $row_WADAimages[where_taken]; ?
?php echo $row_WADAimages[description]; ?
Thank you for your help.
Gary
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Re: [PHP] Watermark with GD
Tamara Temple tamouse.li...@gmail.com wrote in message
news:7f666311-4bc8-4064-8c70-4f2597e7b...@gmail.com...
On Oct 29, 2010, at 2:44 PM, Gary wrote:
Adam Richardson simples...@gmail.com wrote in message
news:aanlkti=kenxt7yewrztcm4+hyifrlqhozxse7ufmq...@mail.gmail.com...
On Fri, Oct 29, 2010 at 3:05 PM, Gary gp...@paulgdesigns.com wrote:
I am trying to get the watermark to work, however I am having a problem
in
that the image is being called from a database (image sits in images
file).
The script in question is this
$image = imagecreatefromjpeg($_GET['src']);
However it produces an error message of
Warning: imagecreatefromjpeg() [function.imagecreatefromjpeg]: Filename
cannot be empty in /home/content/a/l/i/alinde52/html/ imagesDetail.php
on
line 233
First things first. It looks like there's nothing in $_GET['src'].
From where are you getting the value 'src'?
Adam
**
Adam
Thanks for your reply, that is my question, what is to replace ['src']
if I
am calling the image from a database.
Gary
I'd really need to know more about this application to help. If you are
calling the image from a database (does this mean you have the image's
file spec saved in the database, or you actually storing the image data
in the database?), you need to use a query to do that. Is the
imagesDetail.php script being called by something else that already
queried the database and put the file spec in the src query string
argument? Before you dump the query string argument directly into the
imagecreatefromjpeg() funciton, you should verify that it exists:
if (isset($_GET['src'[) (!empty($_GET['src'[) {
$src = $_GET['src'];
if (fileexists($src)) {
$image = imageceatefromjpeg($src);
if ($image === FALSE) {
# process error from imagecreatefromjpeg
}
} else {
# process missing image
}
} else {
# process missing query string parameter
}
This is all prediated on the idea that something is calling your
imageDetail.php script with the source path for the image in question. If
that is not the case, and you need to in fact query the database for the
source path of the image, then you need to do a database query.
Not knowing anything about how your database is set up, I'll take a stab
at a generic method of doing it.
Somehow, imageDetail.php needs to know what image to get. Let's assume
you are calling it from a gallery that has several images displayed. Part
of the information needed is some what to identify the image's record in
the database. Let's assume you have images stored with an id, that is not
null and autoincrements when you store a new image. Here's a sample
schema:
CREATE TABLE `photos` (
`id` INT AUTO_INCREMENT NOT NULL,
`src` VARCHAR(255) NOT NULL,
`created` TIMESTAMP NOT NULL DEFAULT 0,
`updated` TIMESTAMP NOT NULL DEFAULT 0 ON UPDATE CURRENT_TIMESTAMP
PRIMARY KEY (`id`)
);
Of course, you would probably want to store a lot more info about an
image, but this will do for explanation.
Let's say you have a gallery shown at your application's index.php
program. In the html it generates, you may have something like this for
any particular thumbnail:
a href=imageDetail.php?id=25img src=thumbs/imageABC.jpg/a
This is making some assumptions:
* you have image thumbnails for each image stored in the subdirectory
thumbs/
* you've figured out somehow that the id for thumbs/imageABC.jpg is 25
When the user clicks on the image, they get taken to your imageDetail.php
script, with the query string paramter id set to 25.
In PHP you would then do:
if (isset($_GET['id'] (!empty($_GET['id'] is_numeric($_GET['id']) {
$id = $_GET['id'];
$sql = SELECT * FROM `photos` WHERE id=.$id. LIMIT 1;
$result = mysql_query($sql,$db);
if ($result) {
$imagedata = mysql_fetch_array($result,MYSQL_ASSOC);
$src = $imagedata['src'];
if (isset($src) !empty($src) fileexists($src)) {
$image = imagecreatefromjpeg($src);
# do stuff with image
} else {
# handle invalid src data
}
} else {
# handle error from query
}
} else {
# handle invalid id paramter on script
}
Note: some people like to handle error conditions before moving on to
working with the successful state. It's a matter of style. Either works.
Hope this helps.
Tamara
Tamara
Thanks for your reply. The image is already called, (and if you read the
response I posted to tedd, you'll see I am now not sure how it is). I also
posted the code that calls the image if that sheds any light.
I am digesting all the information to get it to work, if any of the code
changes your answer, I would love to hear it!
Thank you again.
Gary
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Re: [PHP] Watermark with GD
tedd tedd.sperl...@gmail.com wrote in message news:p06240804c8f1eaf38...@[192.168.1.2]... At 10:31 AM -0400 10/30/10, Gary wrote: I was under the impression that the image is stored in a folder called images, in fact the images file do go in, however I have the DB set up for longblob, averaging about 20kb each, so now I am unsure. I exported the sql so perhaps you can tell me. Gary: Impressions don't cut it -- you should *know* if the actual images are in the file system or in the database. Find that out before we can proceed. Cheers, tedd -- --- http://sperling.com/ tedd The images do go into the images folder, what I am now unclear about is where they are being called from. Gary __ Information from ESET Smart Security, version of virus signature database 5576 (20101029) __ The message was checked by ESET Smart Security. http://www.eset.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] Watermark with GD
I am trying to get the watermark to work, however I am having a problem in that the image is being called from a database (image sits in images file). The script in question is this $image = imagecreatefromjpeg($_GET['src']); However it produces an error message of Warning: imagecreatefromjpeg() [function.imagecreatefromjpeg]: Filename cannot be empty in /home/content/a/l/i/alinde52/html/imagesDetail.php on line 233 I have tried various methods, for example: ($_GET ['images/'] or ($_GET ['images/$row_WADAimages[image_id]]. Can anyone shed some light on this for me. Thank you Gary __ Information from ESET Smart Security, version of virus signature database 5575 (20101029) __ The message was checked by ESET Smart Security. http://www.eset.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Watermark with GD
Adam Richardson simples...@gmail.com wrote in message news:aanlkti=kenxt7yewrztcm4+hyifrlqhozxse7ufmq...@mail.gmail.com... On Fri, Oct 29, 2010 at 3:05 PM, Gary gp...@paulgdesigns.com wrote: I am trying to get the watermark to work, however I am having a problem in that the image is being called from a database (image sits in images file). The script in question is this $image = imagecreatefromjpeg($_GET['src']); However it produces an error message of Warning: imagecreatefromjpeg() [function.imagecreatefromjpeg]: Filename cannot be empty in /home/content/a/l/i/alinde52/html/imagesDetail.php on line 233 First things first. It looks like there's nothing in $_GET['src']. From where are you getting the value 'src'? Adam ** Adam Thanks for your reply, that is my question, what is to replace ['src'] if I am calling the image from a database. Gary __ Information from ESET Smart Security, version of virus signature database 5575 (20101029) __ The message was checked by ESET Smart Security. http://www.eset.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Unknown Table i field list
sueandant hollandsath...@tiscali.co.uk wrote in message
news:4c8c262adea3459d815fe3e70ae10...@sueandantpc...
Have you tried :
$query = INSERT INTO formcom(fname, lname, email, comment, ip) VALUES
('$fname','$lname','$email'.'$comment','$ip')
or die('error in query');
tholland
- Original Message -
From: Gary gp...@paulgdesigns.com
To: php-general@lists.php.net
Sent: Monday, October 11, 2010 8:09 PM
Subject: [PHP] Unknown Table i field list
I am getting an error of unkown talbe formcom in field list.
I dont understand this. I have a talbe called formcom in the database,
in fact I created this talbe to replace the first one named comments just
in case the table name comments was not allowed.
I have used this same script a number of time, this is the first time I
am getting this error
Here is the scritpt
$dbc = mysqli_connect('server','username',password','db name')
or die('Error connecting to MySQL server');
$query = INSERT INTO formcom(fname, lname, email, comment, ip) .
VALUES ('$fname','$lname','$email'.'$comment','$ip')
or die('error in query');
$result = mysqli_query($dbc, $query)
or die('Error in Result');
mysqli_close($dbc);
I have also tried formcom.fname, formcom.lname, formcomemail,
formcomcomment, formcom.ip
I have double triple quad checked to make sure I did not have comment and
comments mix up.
Anyone be able to help me?
Thank you
Gary
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Just tried it, still the same message...
Thank you for trying
Gary
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Re: [PHP] Unknown Table i field list
Ashley
I have used the concatenation before and I believe that since I closed the
INSERT INTO with a , I needed to add the . for the VAULES.
My post is laced with typo's, I just came back from an impromtu hosptial
stay and they had given me the good stuff,so I was a little cloudier than
usual.
Thank you again for trying to help.
Gary
a...@ashleysheridan.co.uk wrote in message
news:40.00.07410.a0783...@pb1.pair.com...
That's the same query.
One thing I did notice (and I wasn't sure if it was a typo as I saw a few
in the original post) but you've used a period instead of a comma to
separate the values in the query.
It shouldn't produce the error you're seeing, but sometimes you never
know!
Thanks,
Ash
http://www.ashleysheridan.co.uk
- Reply message -
From: sueandant hollandsath...@tiscali.co.uk
Date: Mon, Oct 11, 2010 22:26
Subject: [PHP] Unknown Table i field list
To: Gary gp...@paulgdesigns.com
Cc: PHP php-general@lists.php.net
Have you tried :
$query = INSERT INTO formcom(fname, lname, email, comment, ip) VALUES
('$fname','$lname','$email'.'$comment','$ip')
or die('error in query');
tholland
- Original Message -
From: Gary gp...@paulgdesigns.com
To: php-general@lists.php.net
Sent: Monday, October 11, 2010 8:09 PM
Subject: [PHP] Unknown Table i field list
I am getting an error of unkown talbe formcom in field list.
I dont understand this. I have a talbe called formcom in the database,
in
fact I created this talbe to replace the first one named comments just in
case the table name comments was not allowed.
I have used this same script a number of time, this is the first time I
am
getting this error
Here is the scritpt
$dbc = mysqli_connect('server','username',password','db name')
or die('Error connecting to MySQL server');
$query = INSERT INTO formcom(fname, lname, email, comment, ip) .
VALUES
('$fname','$lname','$email'.'$comment','$ip')
or die('error in query');
$result = mysqli_query($dbc, $query)
or die('Error in Result');
mysqli_close($dbc);
I have also tried formcom.fname, formcom.lname, formcomemail,
formcomcomment, formcom.ip
I have double triple quad checked to make sure I did not have comment and
comments mix up.
Anyone be able to help me?
Thank you
Gary
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[PHP] Re: Unknown Table i field list Solved
Gary gp...@paulgdesigns.com wrote in message
news:75.f4.27779.7f063...@pb1.pair.com...
I am getting an error of unkown talbe formcom in field list.
I dont understand this. I have a talbe called formcom in the database, in
fact I created this talbe to replace the first one named comments just in
case the table name comments was not allowed.
I have used this same script a number of time, this is the first time I am
getting this error
Here is the scritpt
$dbc = mysqli_connect('server','username',password','db name')
or die('Error connecting to MySQL server');
$query = INSERT INTO formcom(fname, lname, email, comment, ip) . VALUES
('$fname','$lname','$email'.'$comment','$ip')
or die('error in query');
$result = mysqli_query($dbc, $query)
or die('Error in Result');
mysqli_close($dbc);
I have also tried formcom.fname, formcom.lname, formcomemail,
formcomcomment, formcom.ip
I have double triple quad checked to make sure I did not have comment and
comments mix up.
Anyone be able to help me?
Thank you
Gary
I put the database into a serperate function and it seems to work nowgo
figure.
$dbc = mysql_connect(server,'username,'pw')
or die('Error connecting to MySQL server');
mysql_select_db(db,$dbc);
$query = INSERT INTO formcom(fname, lname, email, comment, ip) . VALUES
('$fname','$lname','$email','$comment','$ip')
or die('error in query');
$result = mysql_query($query, $dbc)
or die('Error in Result');
Thanks for all your help
Gary
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[PHP] Syntax Error
I have just created a registration page using Webassist, and I am getting a
syntax error that I am not understanding. Anyone be able to point me in the
right direction?
You have an error in your SQL syntax; check the manual that corresponds to
your MySQL server version for the right syntax to use near ' NULL, NULL)' at
line 1
This is the code (I have not modified it)
?php require_once('Connections/local.php'); ?
?php
if (!function_exists(GetSQLValueString)) {
function GetSQLValueString($theValue, $theType, $theDefinedValue = ,
$theNotDefinedValue = )
{
$theValue = get_magic_quotes_gpc() ? stripslashes($theValue) : $theValue;
$theValue = function_exists(mysql_real_escape_string) ?
mysql_real_escape_string($theValue) : mysql_escape_string($theValue);
switch ($theType) {
case text:
$theValue = ($theValue != ) ? ' . $theValue . ' : NULL;
break;
case long:
case int:
$theValue = ($theValue != ) ? intval($theValue) : NULL;
break;
case double:
$theValue = ($theValue != ) ? ' . doubleval($theValue) . ' : NULL;
break;
case date:
$theValue = ($theValue != ) ? ' . $theValue . ' : NULL;
break;
case defined:
$theValue = ($theValue != ) ? $theDefinedValue : $theNotDefinedValue;
break;
}
return $theValue;
}
}
?
?php
// *** Redirect if username exists
$MM_flag=MM_insert;
if (isset($_POST[$MM_flag])) {
$MM_dupKeyRedirect=;
$loginUsername = $_POST['id'];
$LoginRS__query = SELECT id FROM family WHERE id=' . $loginUsername . ';
mysql_select_db($database_local, $local);
$LoginRS=mysql_query($LoginRS__query, $local) or die(mysql_error());
$loginFoundUser = mysql_num_rows($LoginRS);
//if there is a row in the database, the username was found - can not add
the requested username
if($loginFoundUser){
$MM_qsChar = ?;
//append the username to the redirect page
if (substr_count($MM_dupKeyRedirect,?) =1) $MM_qsChar = ;
$MM_dupKeyRedirect = $MM_dupKeyRedirect . $MM_qsChar
.requsername=.$loginUsername;
header (Location: $MM_dupKeyRedirect);
exit;
}
}
?
?php
$editFormAction = $_SERVER['PHP_SELF'];
if (isset($_SERVER['QUERY_STRING'])) {
$editFormAction .= ? . htmlentities($_SERVER['QUERY_STRING']);
}
?
?php
if ((isset($_POST[MM_insert])) ($_POST[MM_insert] ==
WAATKRegistrationForm)) {
$insertSQL = sprintf(INSERT INTO family (firstname, lastname, email,
password, relationship, story, image, ip, submitted) VALUES (%s, %s, %s, %s,
%s, %s, %s, %s, %s),
GetSQLValueString($_POST['firstname'], text),
GetSQLValueString($_POST['lastname'], text),
GetSQLValueString($_POST['email'], text),
GetSQLValueString($_POST['password'], text),
GetSQLValueString($_POST['relationship'], text),
GetSQLValueString($_POST['story'], text),
GetSQLValueString($_POST['image'], ), GetSQLValueString($_POST['ip'],
text), GetSQLValueString($_POST['submitted'], date));
mysql_select_db($database_local, $local);
$Result1 = mysql_query($insertSQL, $local) or die(mysql_error());
$insertGoTo = family_LogIn.php;
if (isset($_SERVER['QUERY_STRING'])) {
$insertGoTo .= (strpos($insertGoTo, '?')) ? : ?;
$insertGoTo .= $_SERVER['QUERY_STRING'];
}
header(sprintf(Location: %s, $insertGoTo));
}
?
Thanks again for the help.
Gary
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[PHP] GD Watermark Question
Is there a way to insert a watermark on an image as it is being uploaded to the image file, then removed when it is called from a database to be viewed on a website? The rational behind this is I have a photographers site I am doing, and I am limiting the size of the images somewhat to reduce pilferage and I would like to be able to show the images a little larger, hence with a bit more clarity and detail. Thanks for your input. Gary __ Information from ESET Smart Security, version of virus signature database 5458 (20100917) __ The message was checked by ESET Smart Security. http://www.eset.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] Re: GD Watermark Question-
Gary gp...@paulgdesigns.com wrote in message news:1f.27.30333.1d5e3...@pb1.pair.com... Is there a way to insert a watermark on an image as it is being uploaded to the image file, then removed when it is called from a database to be viewed on a website? The rational behind this is I have a photographers site I am doing, and I am limiting the size of the images somewhat to reduce pilferage and I would like to be able to show the images a little larger, hence with a bit more clarity and detail. Thanks for your input. Gary More info. I was asked off board where the watermark would show, so I am sorry if I was less than clear. The watermark would show on an image that is being downloaded from the server. If this were to work, I could let viewers see an image with a size of 640px in width to show clarity, (they are only able to see an image now with a width of 250 px now) should they decide to help themselves to it, it would download with a watermark on it, but the watermark would not appear on the web page itself. Gary __ Information from ESET Smart Security, version of virus signature database 5458 (20100917) __ The message was checked by ESET Smart Security. http://www.eset.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Re: GD Watermark Question-
Tom No, I had never DD'd an image, however I just did. So I see your point. Thanks for your input. Gary TR Shaw ts...@oitc.com wrote in message news:cff72d3b-52cf-4baf-b60f-6b3709c98...@oitc.com... Gary you do realize that if you display the image in a browser without the watermark, simple drag and drop can copy the image as is (eg without the watermark) Tom On Sep 17, 2010, at 6:21 PM, Gary wrote: Gary gp...@paulgdesigns.com wrote in message news:1f.27.30333.1d5e3...@pb1.pair.com... Is there a way to insert a watermark on an image as it is being uploaded to the image file, then removed when it is called from a database to be viewed on a website? The rational behind this is I have a photographers site I am doing, and I am limiting the size of the images somewhat to reduce pilferage and I would like to be able to show the images a little larger, hence with a bit more clarity and detail. Thanks for your input. Gary More info. I was asked off board where the watermark would show, so I am sorry if I was less than clear. The watermark would show on an image that is being downloaded from the server. If this were to work, I could let viewers see an image with a size of 640px in width to show clarity, (they are only able to see an image now with a width of 250 px now) should they decide to help themselves to it, it would download with a watermark on it, but the watermark would not appear on the web page itself. Gary __ Information from ESET Smart Security, version of virus signature database 5458 (20100917) __ The message was checked by ESET Smart Security. http://www.eset.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php __ Information from ESET Smart Security, version of virus signature database 5458 (20100917) __ The message was checked by ESET Smart Security. http://www.eset.com __ Information from ESET Smart Security, version of virus signature database 5458 (20100917) __ The message was checked by ESET Smart Security. http://www.eset.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Re: GD Watermark Question-
Ashley Sheridan a...@ashleysheridan.co.uk wrote in message news:1284763747.12459.40.ca...@localhost... On Fri, 2010-09-17 at 18:41 -0400, TR Shaw wrote: Gary you do realize that if you display the image in a browser without the watermark, simple drag and drop can copy the image as is (eg without the watermark) Tom On Sep 17, 2010, at 6:21 PM, Gary wrote: Gary gp...@paulgdesigns.com wrote in message news:1f.27.30333.1d5e3...@pb1.pair.com... Is there a way to insert a watermark on an image as it is being uploaded to the image file, then removed when it is called from a database to be viewed on a website? The rational behind this is I have a photographers site I am doing, and I am limiting the size of the images somewhat to reduce pilferage and I would like to be able to show the images a little larger, hence with a bit more clarity and detail. Thanks for your input. Gary More info. I was asked off board where the watermark would show, so I am sorry if I was less than clear. The watermark would show on an image that is being downloaded from the server. If this were to work, I could let viewers see an image with a size of 640px in width to show clarity, (they are only able to see an image now with a width of 250 px now) should they decide to help themselves to it, it would download with a watermark on it, but the watermark would not appear on the web page itself. Gary __ Information from ESET Smart Security, version of virus signature database 5458 (20100917) __ The message was checked by ESET Smart Security. http://www.eset.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php There's no way to do this. Anything you see in your browser has already been downloaded in some form onto your computer, and once that happens it's out of your control. PHP has no way to detect if the image is being requested by the browser to view or download, so can't do what you want. Besides which, if an image is displayed in the browser, there are dozens of ways to get at it, from right clicking and saving it, using the media tab of the file info dialogue (firefox), using firebug to view it, saving it from the cache, saving the whole page, using wget to spider and save that page, etc. The only way to do what you want is to have your own custom browser app (possibly written in Java) but even then someone could simply do a print screen. At the end of the day, if you want to prevent people downloading your images, then just don't show them the image. Thanks, Ash http://www.ashleysheridan.co.uk Ashley When I right click on an image, I assumed it is being called from the server, not from the browser, which is why I thought this might work. Thanks for your help. Gary __ Information from ESET Smart Security, version of virus signature database 5458 (20100917) __ The message was checked by ESET Smart Security. http://www.eset.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] Re: GD Watermark Question
Thank you both for your input, my assumption was when an image file is gathered from a web page, through whatever method, the image was going to be served up from the server, thus if the image file on the server had the watermark, then so would the image that is being captured. Thanks again for your help. Gary Gary gp...@paulgdesigns.com wrote in message news:1f.27.30333.1d5e3...@pb1.pair.com... Is there a way to insert a watermark on an image as it is being uploaded to the image file, then removed when it is called from a database to be viewed on a website? The rational behind this is I have a photographers site I am doing, and I am limiting the size of the images somewhat to reduce pilferage and I would like to be able to show the images a little larger, hence with a bit more clarity and detail. Thanks for your input. Gary __ Information from ESET Smart Security, version of virus signature database 5458 (20100917) __ The message was checked by ESET Smart Security. http://www.eset.com __ Information from ESET Smart Security, version of virus signature database 5458 (20100917) __ The message was checked by ESET Smart Security. http://www.eset.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] How to set socket_read time out
How can I get calls to scoket_read to timeout if the server stops responding? It's not clear exactly what's happening, but occasionally the server simply stops sending data and my php code (client) then sits there waiting endlessly. I know about set_time_limit but I don't want to abandon the entire script, just the reading of data from the server (I can at least give a meaningful error message then, for example). -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] Re: How to set socket_read time out
On Mon, Jul 19, 2010 at 9:45 AM, Gary wrote:
How can I get calls to scoket_read to timeout if the server stops
responding?
Sorry. Found it:
socket_set_option($sock, SOL_SOCKET, SO_RCVTIMEO, array('sec' = 1,
'usec' = 0));
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[PHP] PHP - MOODLE - WORK IN AUSTRALIA
Hi We are looking for some one who knows Moodle well and can write for us, web service's so that we can integrate our application with Moodle. Expressions of interest are sought from persons or companies who can work on a project in Australia, pref Brisbane. Please email me on:- crouc...@gmail.com THX|GC Gary Crouch
[PHP] Serial Numbers
I'm sure it is possible, but I am unsure how to do this. I have created a Sale coupon that I was going to put up on a site that I manage, for visitors to print out and bring to the store. The coupon is currently a .png, however I was planning on converting to a pdf. I would like to put on the coupon a serial number that increases by 1 everytime the page is viewed. I dont really care if someone refreshes the page and skews the numbers. Is this possible and could someone give me some help? Thanks Gary __ Information from ESET Smart Security, version of virus signature database 5273 (20100712) __ The message was checked by ESET Smart Security. http://www.eset.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Serial Numbers
Floyd Resler fres...@adex-intl.com wrote in message news:e6321b00-590d-4294-9b81-81cc92883...@adex-intl.com... On Jul 12, 2010, at 2:52 PM, Gary wrote: I'm sure it is possible, but I am unsure how to do this. I have created a Sale coupon that I was going to put up on a site that I manage, for visitors to print out and bring to the store. The coupon is currently a .png, however I was planning on converting to a pdf. I would like to put on the coupon a serial number that increases by 1 everytime the page is viewed. I dont really care if someone refreshes the page and skews the numbers. Is this possible and could someone give me some help? Thanks Gary Is there any particular reason you need it to be a PDF? If not and the GD library is installed in your PHP, I would suggest using the GD library to draw your serial number on the coupon. As for keeping track of the counter I would do it either in a database table or save the number to file. Take care, Floyd Floyd Thanks for your reponse. No real reason that the file has to be a pdf, that was my original plan before I came up with the counter idea.. I also thought it might not be as difficult to put the script on a pdf. I have used the GD library in the past, so it is installed on the server, however I dont know how to apply to this situation.. After I posted the question, the thought came to me to simply create the coupon in html and include a simple counter, like a hit counter, but since I had the png already created, I was hoping there was a way to include the script on it instead of recreating in html/php. Gary __ Information from ESET Smart Security, version of virus signature database 5273 (20100712) __ The message was checked by ESET Smart Security. http://www.eset.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Serial Numbers
Ashley Sheridan a...@ashleysheridan.co.uk wrote in message news:1278962039.2202.5.ca...@localhost... On Mon, 2010-07-12 at 14:52 -0400, Gary wrote: I'm sure it is possible, but I am unsure how to do this. I have created a Sale coupon that I was going to put up on a site that I manage, for visitors to print out and bring to the store. The coupon is currently a .png, however I was planning on converting to a pdf. I would like to put on the coupon a serial number that increases by 1 everytime the page is viewed. I dont really care if someone refreshes the page and skews the numbers. Is this possible and could someone give me some help? Thanks Gary __ Information from ESET Smart Security, version of virus signature database 5273 (20100712) __ The message was checked by ESET Smart Security. http://www.eset.com You need some indicator on your server to keep track of the number. To me, the ideal solution would appear to be a MySQL database. You can set up a table with an auto_increment field and use the id generated from that. Two things to maybe note: 1. Don't use MAX(id) in a query to get the next auto value, use something like mysql_insert_id() instead. The MAX() method is just a race condition waiting to happen. 2. If you expect a lot of traffic, then consider setting this table to use the InnoDB engine instead of MyIsam which is usually the default. This allows MySQL to apply row-level locking instead of table-level, which can improve performance when PHP has to wait for MySQL. Thanks, Ash http://www.ashleysheridan.co.uk Ashley Thanks for your reply. I was not going to go to the trouble of calling the AI number from a database, although that is an interesting concept. The question really becomes How to insert into an image I could just write a simple hit counter and create a new php/html page, but since the image (coupon) was already created, I thought it would be easier, not to mention expand my abilities. It is a small shop and if they get 100 coupons, they will be extatic, so it will not kill us if it were too much trouble. I have gone to using InnoDB as my standard table, mainly for the foreign key ability, but I do keep reading that there are more and more advantage to using is over Myisam. Thanks again for your reply. Gary __ Information from ESET Smart Security, version of virus signature database 5273 (20100712) __ The message was checked by ESET Smart Security. http://www.eset.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Serial Numbers
Daevid Vincent dae...@daevid.com wrote in message news:a380a1570d4c4351b38e1cce2aabe...@mascorp.com... -Original Message- From: Gary [mailto:gp...@paulgdesigns.com] Sent: Monday, July 12, 2010 11:53 AM I'm sure it is possible, but I am unsure how to do this. I have created a Sale coupon that I was going to put up on a site that I manage, for visitors to print out and bring to the store. The coupon is currently a .png, however I was planning on converting to a pdf. I would like to put on the coupon a serial number that increases by 1 everytime the page is viewed. I dont really care if someone refreshes the page and skews the numbers. Is this possible and could someone give me some help? Is there a reason you need to increment by 1? I mean, are you going to validate the coupon someone brings in to see if it's in the database or is used already? Since you say you don't care if the number is skewed, it sounds as if you don't care about the code either really. Point being, why not just use md5(time()) or uniqid() or something and make some truly unique code (that you may or may not wish to store in a DB). This will be obscure enough the average person won't know that it isn't tracked and most likely won't try to make their own. If I saw 123457, I'm pretty sure I could make a coupon with 123500 on it. But if I saw 6ccd780c-baba-1026-9564-0040f4311e29 then I'm not really going to try and fudge one of those. Then I'd use the barcode library to print the code (again, doesn't have to actually work if you're not going to do a lookup) just to add more realizm to it. There are a billion different ways to make a unique ID. Even IF you are going to track them, you really shouldn't do a sequential code as you originally wanted to do. http://us.php.net/manual/en/function.uniqid.php http://php.net/manual/en/function.com-create-guid.php http://us.php.net/manual/en/function.mssql-guid-string.php http://us.php.net/manual/en/function.md5.php http://us.php.net/manual/en/function.sha1.php http://us.php.net/manual/en/function.hash.php http://us.php.net/manual/en/function.time.php http://dev.mysql.com/doc/refman/5.0/en/miscellaneous-functions.html#functio n_uuid Heck, you could even just use ip2long() on their IP address and then when someone redeems the coupon code, a simple long2ip() would tell you if it's a valid IP format rather than some hacked up string. http://us.php.net/manual/en/function.ip2long.php http://us.php.net/manual/en/function.long2ip.php Daevid. http://daevid.com Some people, when confronted with a problem, think 'I know, I'll use XML.' Now they have two problems. Daevid Thanks for your reply. The numbers are not for show, we are not trying to make it seem like we have thousands of coupons. We are going to use the numbers to guage how many times people see the coupon and to see how many it takes before we have someone show up to use it, but dont really care if someone gets a kick out of hitting the refresh button 10 or 20 times to watch the number go up. I was going to start the number at 72010001 (July 2010, 001) and if someone figured out what it was, we dont really care. As I mentioned, it is a small shop (one man show) that the site gets and average of 350 visits per month (which he is very happy with over what he had). Although I do like the idea of using there ip address... Thanks for your reply and all the links. Gary __ Information from ESET Smart Security, version of virus signature database 5273 (20100712) __ The message was checked by ESET Smart Security. http://www.eset.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] Login In script quesitons
I have a log in script that is not working. It is taken from a lesson book
that I read about a year or so ago so the lessons are not fresh in my mind.
My questions are.
1. Why is this not working?
2. Does is look secure?
3. In researching the issue, I was reading some older threads that it was a
better idea to use some commercial ready made scripts. Are there any
recommendations?
Here is the code to the script I am trying to get to work. I get no error
message, however if it will not log me in with a correct un/pw or if I leave
it blank, I get the same message and it allows the next page to be viewed
(which I do not want). DB is set up with data in it.
Login Page:
?php
// Start the session
session_start();
?
!DOCTYPE html PUBLIC -//W3C//DTD XHTML 1.0 Transitional//EN
http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd;
html xmlns=http://www.w3.org/1999/xhtml; xml:lang=en lang=en
head
meta http-equiv=Content-Type content=text/html; charset=utf-8 /
titlexxx- Log In/title
link rel=stylesheet type=text/css href=style.css /
/head
body
h3xxx- Log In/h3
?php
// If the session var is empty, show any error message and the log-in
form; otherwise confirm the log-in
if (empty($_SESSION['username'])) {
echo 'p class=error' . $error_msg . '/p';
?
form method=post action=story.php
fieldset
legendLog In/legend
label for=usernameUsername:/label
input type=text name=username value=?php if (!empty($username))
echo $username; ? /br /
label for=passwordPassword:/label
input type=password name=password /
/fieldset
input type=submit value=Log In name=submit /
/form
?php
}
else {
// Confirm the successful log-in
echo('p class=loginYou are logged in as ' . $_SESSION['username'] .
'./p');
}
?
/body
/html
Processing page:
?php
// Start the session
session_start();
// Clear the error message
$error_msg = ;
// If the user isn't logged in, try to log them in
if (!isset($_SESSION['username'])) {
if (isset($_POST['submit'])) {
// Connect to the database
$dbc = mysqli_connect(host, 'un', 'pw', 'db')//sanitized for board
or die('Error connecting with MySQL Database');
// Grab the user-entered log-in data
$username = mysqli_real_escape_string($dbc, trim($_POST['username']));
$password = mysqli_real_escape_string($dbc, trim($_POST['password']));
if (!empty($username) !empty($password)) {
// Look up the username and password in the database
$query = SELECT username, password FROM family WHERE username =
'$username' AND password = SHA('password');
$data = mysqli_query($dbc, $query)or die(mysqli_error($dbc));
if (mysqli_num_rows($data) == 1) {
// The log-in is OK so set the user ID and username session vars
(and cookies), and redirect to the home page
$row = mysqli_fetch_array($data);
$_SESSION['username'] = $row['username'];
setcookie('username', $row['username'], time() + (60 * 60 * 24 *
30)); // expires in 30 days
$home_url = 'http://' . $_SERVER['HTTP_HOST'] .
dirname($_SERVER['PHP_SELF']) . '/index.php';
header('Location: ' . $home_url);
}
else {
// The username/password are incorrect so set an error message
$error_msg = 'Sorry, you must enter a valid username and password
to log in.';
}
}
else {
// The username/password weren't entered so set an error message
$error_msg = 'Sorry, you must enter your username and password to
log in.';
}
}
}
?
!DOCTYPE html PUBLIC -//W3C//DTD XHTML 1.0 Transitional//EN
http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd;
html xmlns=http://www.w3.org/1999/xhtml;!-- InstanceBegin
template=/Templates/blindowl.dwt.php codeOutsideHTMLIsLocked=false --
head
meta http-equiv=Content-Type content=text/html; charset=utf-8 /
!-- InstanceBeginEditable name=doctitle --
titlexxx/title
!-- InstanceEndEditable --
style type=text/css
/style
link href=p7pmm/p7PMMh03.css rel=stylesheet type=text/css media=all
/
script type=text/javascript src=p7pmm/p7PMMscripts.js/script
script src=Scripts/AC_RunActiveContent.js
type=text/javascript/script
link href=xxx.css rel=stylesheet type=text/css /
!-- InstanceBeginEditable name=head --!-- InstanceEndEditable --
/head
body
div id=wrap
div id=header
div id=headercontent/div
/div
div id=menu
?php include('includes/menu.inc.php');?
/div
div id=mainwrap!-- InstanceBeginEditable name=main --
?php
// If the session var is empty, show any error message and the log-in
form; otherwise confirm the log-in
if (empty($_SESSION['username'])) {
echo 'p class=error' . $error_msg . '/p';
?
?php
}
else {
// Confirm the successful log-in
echo('p class=loginYou are logged in as ' . $_SESSION['username'] .
'./p');
}
?
Story//text to let me know if the page opened
!-- InstanceEndEditable --/div
div id=footer?php
Re: [PHP] Login In script quesitons
Richard Quadling rquadl...@gmail.com wrote in message
news:aanlktilbmyedd8paky9dwgn0q7t6kem4zzutu_49u...@mail.gmail.com...
On 9 July 2010 16:42, Gary gp...@paulgdesigns.com wrote:
[snip]
Take a look at https://code.google.com/p/loginsystem-rd/
Richard
Thank you for your quick reply and the link. Since I see you are one of the
creators, thank you for that as well.
I am getting the following error and I'm not sure how to correct it.
Warning: Cannot modify header information - headers already sent by (output
started at /home/content/45/6359745/html/login/include/loginGlobals.php:281)
in /home/content/45/6359745/html/login/include/form_token.php on line 15
Warning: Cannot modify header information - headers already sent by (output
started at /home/content/45/6359745/html/login/include/loginGlobals.php:281)
in /home/content/45/6359745/html/login/include/form_token.php on line 22
Line 15 is: setcookie(token, , time()-42000);
Line 22 is :if (setcookie(token, $_SESSION[token], time()+86400)) {
loginGlobals stops at line 278
Again, thank you for all your help.
gary
__ Information from ESET Smart Security, version of virus signature
database 5266 (20100709) __
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Re: [PHP] Login In script quesitons
Ashley Sheridan a...@ashleysheridan.co.uk wrote in message
news:1278705035.2295.2.ca...@localhost...
On Fri, 2010-07-09 at 15:43 -0400, Gary wrote:
Richard Quadling rquadl...@gmail.com wrote in message
news:aanlktilbmyedd8paky9dwgn0q7t6kem4zzutu_49u...@mail.gmail.com...
On 9 July 2010 16:42, Gary gp...@paulgdesigns.com wrote:
[snip]
Take a look at https://code.google.com/p/loginsystem-rd/
Richard
Thank you for your quick reply and the link. Since I see you are one of
the
creators, thank you for that as well.
I am getting the following error and I'm not sure how to correct it.
Warning: Cannot modify header information - headers already sent by
(output
started at
/home/content/45/6359745/html/login/include/loginGlobals.php:281)
in /home/content/45/6359745/html/login/include/form_token.php on line 15
Warning: Cannot modify header information - headers already sent by
(output
started at
/home/content/45/6359745/html/login/include/loginGlobals.php:281)
in /home/content/45/6359745/html/login/include/form_token.php on line 22
Line 15 is: setcookie(token, , time()-42000);
Line 22 is :if (setcookie(token, $_SESSION[token], time()+86400)) {
loginGlobals stops at line 278
Again, thank you for all your help.
gary
__ Information from ESET Smart Security, version of virus
signature database 5266 (20100709) __
The message was checked by ESET Smart Security.
http://www.eset.com
My guess is that you've put the login PHP code after some HTML in that
page? If you output any content at all (even a blank space) it will
output the default headers. What you need to do is have any logic that
includes a header() call before any output.
Thanks,
Ash
http://www.ashleysheridan.co.uk
Ashley
I am aware of that would cause a problem. I am not finding any html at all
on either page.
Thanks for your reply.
Gary
__ Information from ESET Smart Security, version of virus signature
database 5266 (20100709) __
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Re: [PHP] Login In script quesitons
Ashley Sheridan a...@ashleysheridan.co.uk wrote in message
news:1278705549.2295.4.ca...@localhost...
On Fri, 2010-07-09 at 15:58 -0400, Gary wrote:
Ashley Sheridan a...@ashleysheridan.co.uk wrote in message
news:1278705035.2295.2.ca...@localhost...
On Fri, 2010-07-09 at 15:43 -0400, Gary wrote:
Richard Quadling rquadl...@gmail.com wrote in message
news:aanlktilbmyedd8paky9dwgn0q7t6kem4zzutu_49u...@mail.gmail.com...
On 9 July 2010 16:42, Gary gp...@paulgdesigns.com wrote:
[snip]
Take a look at https://code.google.com/p/loginsystem-rd/
Richard
Thank you for your quick reply and the link. Since I see you are one
of
the
creators, thank you for that as well.
I am getting the following error and I'm not sure how to correct it.
Warning: Cannot modify header information - headers already sent by
(output
started at
/home/content/45/6359745/html/login/include/loginGlobals.php:281)
in /home/content/45/6359745/html/login/include/form_token.php on line
15
Warning: Cannot modify header information - headers already sent by
(output
started at
/home/content/45/6359745/html/login/include/loginGlobals.php:281)
in /home/content/45/6359745/html/login/include/form_token.php on line
22
Line 15 is: setcookie(token, , time()-42000);
Line 22 is :if (setcookie(token, $_SESSION[token], time()+86400))
{
loginGlobals stops at line 278
Again, thank you for all your help.
gary
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My guess is that you've put the login PHP code after some HTML in that
page? If you output any content at all (even a blank space) it will
output the default headers. What you need to do is have any logic that
includes a header() call before any output.
Thanks,
Ash
http://www.ashleysheridan.co.uk
Ashley
I am aware of that would cause a problem. I am not finding any html at
all
on either page.
Thanks for your reply.
Gary
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What does your code look like now?
Thanks,
Ash
http://www.ashleysheridan.co.uk
Ashley
This is form_token.php
?php
if (!isset($_SESSION)) {
session_start();
}
?
?php
function generateToken(){
/*
* Create and set a new token for CSRF protection * on initial entry or after
form errors and we are going to redisplay theform.
**/
$salt=; $tokenStr=; $salt = sha1($_SERVER[HTTP_HOST]);
setcookie(token, , time()-42000); $_SESSION[salt]=$salt;
$_SESSION[guid] = getGUID(); $_SESSION[ip] = $_SERVER[REMOTE_ADDR];
$_SESSION[time] = time(); $tokenStr = IP: . $_SESSION[ip] . ,SESSIONID:
. session_id() .,GUID: . $_SESSION[guid];
$_SESSION[token]=sha1(($tokenStr.$_SESSION[salt]).$_SESSION[salt]); if
(setcookie(token, $_SESSION[token], time()+86400)) {
$_SESSION[usecookie]=True; }}function checkToken() {
/*
* Check the posted token for correctness
**/
$oldToken=; $testToken=; $tokenStr=;
$page=basename($_SERVER['PHP_SELF']); $oldToken=$_POST[token]; $tokenStr =
IP: . $_SESSION[ip] . ,SESSIONID: . session_id() .,GUID: .
$_SESSION[guid];
$testToken=sha1(($tokenStr.$_SESSION[salt]).$_SESSION[salt]);
$checkToken=False; If ($oldToken===$testToken) { $diff = time() -
$_SESSION[time]; If ($diff=300) { // Five minutes max If
($_SESSION[usecookie]) { If ($_COOKIE[token]===$oldToken) {
/*
* Destroy the old form token, then * generate a new token for the
form, which may or may not be needed. Wewant to do this * before headers
are written. When writeToken() or writeTokenH() iscalled we are only *
writing the pre-generated token to the form. The cookie will havealready been
written.
**/
setcookie(token, '', time()-42000); generateToken(); return
true;}else{ $_SESSION = array(); if
(isset($_COOKIE[session_name()])) { setcookie(session_name(), '',
time()-42000); } session_destroy(); header(Location: http://;.
lg_domain . lg_form_error .?p= . $page .t=ec);} }else{ return
True; }}else{ $_SESSION = array(); if
(isset($_COOKIE[session_name()])) { setcookie
Re: [PHP] Login In script quesitons
Ashley Sheridan a...@ashleysheridan.co.uk wrote in message news:1278706121.2295.5.ca...@localhost... On Fri, 2010-07-09 at 16:04 -0400, Gary wrote: ? ?php That bit of the code has a newline in it, which counts as output :p I've not looked over the rest yet, but see if that helps. Thanks, Ash http://www.ashleysheridan.co.uk Ashley Actually I had added that closing and opening tag in trying to solve the problem, I put it back and still have the issue. I looked at the code as it looks on the board, would it be easier if I sent you the files? Thanks again. Gary __ Information from ESET Smart Security, version of virus signature database 5266 (20100709) __ The message was checked by ESET Smart Security. http://www.eset.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Login In script quesitons
Ashley Richard I think I found the issue. In loginGlobals.php, the error was pointing to line 281, when the code stopped and 278. (I know most of the time this just means there is a missing bracket or semi-colon in the code), however, what I did is put my curser on link 281, backspaced to eliminate white space, and it seems to be working. I will let you know if this is just premature exhuberation. Thanks again for all your help. Gary Ashley Sheridan a...@ashleysheridan.co.uk wrote in message news:1278706121.2295.5.ca...@localhost... On Fri, 2010-07-09 at 16:04 -0400, Gary wrote: ? ?php That bit of the code has a newline in it, which counts as output :p I've not looked over the rest yet, but see if that helps. Thanks, Ash http://www.ashleysheridan.co.uk __ Information from ESET Smart Security, version of virus signature database 5266 (20100709) __ The message was checked by ESET Smart Security. http://www.eset.com __ Information from ESET Smart Security, version of virus signature database 5266 (20100709) __ The message was checked by ESET Smart Security. http://www.eset.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] php -l - does it find *anything*?
On 7/6/10, Per Jessen wrote: It really is _only_ the syntax. Same goes for e.g. the C lint - Sorry, but that's not what I remember from my C days, nor what wikipedia says lint was the name originally given to a particular program that flagged some suspicious and non-portable constructs (likely to be bugs) in C language source code. In fact, what would be the point of a C lint that does that? It's already done by the parsing/syntactical analysis part of the compiler, there'd be no point writing a separate program (lint) to do that. Anyway, yeah, I accept that that (syntax checking only) is what php -l does, even if I think it's wrong, or at least incorrectly described :) -- Per Jessen, Zürich (24.2°C) It feels warmer than that today. Maybe it's because my code isn't working X( -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] Simple XML - problem with errors
Why am I still getting an exception when I do this: libxml_use_internal_errors(true); $this-xml = new SimpleXMLElement($this-htmlString); or this $this-xml = new SimpleXMLElement($this-htmlString, LIBXML_NOERROR|LIBXML_NOWARNING); ? The exception says Exception: String could not be parsed as XML. Not a hint of why not, of course. I thought the point of those things was to just stuff the content in, and let user code handle errors? I mean, I *know* the provided HTML is broken. I also know there's not a chance in hell of it ever being fixed (completely out of my control). And yes, I'd rather use DOM, but I can't. -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Simple XML - problem with errors
Richard Quadling writes: On 8 July 2010 08:07, Gary wrote: Why am I still getting an exception when I do this: libxml_use_internal_errors(true); $this-xml = new SimpleXMLElement($this-htmlString); or this $this-xml = new SimpleXMLElement($this-htmlString, LIBXML_NOERROR|LIBXML_NOWARNING); ? The exception says Exception: String could not be parsed as XML. ... The XML needs to be well formed [1]. I thought so, thanks. What does libxml_use_internal_errors do then, if it doesn't allow me to handle those problems in my own code? So, if it is junk, you can't read it using SimpleXML as the XML is not well formed. I'm trying to just use xml_parse and so on now. This problem really should be *so* easy. In fact I've already solved it once X( Try putting it through Tidy first - that is, tidy the file first. Ha ha! Sorry. It's almost certainly not available. I don't want to talk about it *cries* -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Simple XML - problem with errors
Marc Guay writes: libxml_use_internal_errors(true); $this-xml = new SimpleXMLElement($this-htmlString); I have code that looks like this: libxml_use_internal_errors(true); $xml = simplexml_load_string($val); Yeah. I tried simplexml_load_string and found that worked (in that it didn't cause an exception - there are errors which caused the conversion not to work). I wonder what the difference is between doing new SimpleXMLElement and calling simplexml_load_string which results in the libxml_use_internal_errors call being ineffective. Odd. -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] Re: Simple XML - problem with errors
Okay. At least one of the problems with this so called HTML seems to be that the body tag looks like BODY vlink=#ff ... and xml_parse complains that required on that line (i.e. it is claiming it can't find the end of the tag!). I'm guessing that those attributes must be quoted in XML and should be in HTML (but patently aren't)? Is there any way to get xml_parse to ignore that? My element_handler functions never even get a chance to see that line. Regex to insert quotes or remove the attributes entirely, perhaps? *gulp* I hope there's a better way than that. -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Re: Simple XML - problem with errors
On 7/8/10, Richard Quadling wrote: On 8 July 2010 16:15, Gary wrote: Okay. At least one of the problems with this so called HTML seems to be that the body tag looks like BODY vlink=#ff ... and xml_parse complains that required on that line (i.e. it is claiming it can't find the end of the tag!). I'm guessing that those attributes must be quoted in XML and should be in HTML (but patently aren't)? Is there any way to get xml_parse to ignore that? My element_handler functions never even get a chance to see that line. Regex to insert quotes or remove the attributes entirely, perhaps? *gulp* I hope there's a better way than that. So. Essentially, you want to parse some plain text which may or may not be well formed XML. No. I don't *want* to And it isn't plain text, it's just sh*t html (no doctype, missing closing tags in some cases, etc. It's an absolute mess). Browsers are pretty good at handling it. XML parsers... less so. How badly formed is the file going to be? It's not a file. It comes from an embedded web server on a device. I could ask them to change it. I can hear the laughter already. If it is things like missing , then this could be managed with regex. Essentially you are going to have to do the clean up that Tidy could do for you. Yeah :( -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Simple XML - problem with errors
On 7/8/10, Marc Guay wrote: And yes, I'd rather use DOM, but I can't. Could you use this: http://simplehtmldom.sourceforge.net/? Interesting. Although I can't use DOM or Tidy (because they're normally built in, but TPTB decided to recompile PHP and exclude them, and I am not allowed to recompile it with them in), that's external so might be a possibility. Thanks. -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Re: Simple XML - problem with errors
On 7/8/10, Nisse Engström wrote: On Thu, 8 Jul 2010 17:15:02 +0200, Gary . wrote: I'm guessing that those attributes must be quoted in XML and should be in HTML (but patently aren't)? For that attribute value, it's a must in both cases. Okay. Please tell L**! :) -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] php -l - does it find *anything*?
Or, alternatively put, is there any way to find the kind of problems in
foo2 foo3 (below), at *compile* time?
,[ lint-test.php ]
| ?php
|
| error_reporting(E_ALL | E_STRICT);
|
| function foo1()
| {
| $bar = 'cheese';
| echo $bar;
| }
|
| function foo2()
| {
| $bar = 'cheese';
| echo 'cheese';
| }
|
| function foo3()
| {
| //$bar = 'cheese';
| echo $bar;
| }
|
| foo1();
| foo2();
| foo3();
|
| ?
`
I only get errors displayed when code happens to pass down the code
path, i.e. at runtime:
,
| /home/jg/work $ php -l lint-test.php
| No syntax errors detected in lint-test.php
| /home/jg/work $ php lint-test.php
| cheesecheese
| Notice: Undefined variable: bar in
| /home/jg/work/lint-test.php on line 20
|
| Call Stack:
| 1. 61488 1. {main}()
| /home/jg/work/lint-test.php:0
| 1. 61680 2. foo3()
| /home/jg/work/lint-test.php:25
`
If foo3 never happens to be called when I am doing my testing (for
example if the call is in some if branch that is never exercised) then
it only gets found in production, so I would like to find this kind of
thing using a static analyser. The kind of problem in foo2 I could live
with, but would like to find as well, if possible. (Obviously I am using
these two example problems as indicative of the type of things I want to
find, it isn't an exhaustive list!)
BTW, what problems *does* php -l pick up? I can't find a description
anywhere.
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Re: [PHP] Question about the Board
On 7/5/10, Gary[1] wrote: The last few times I have posted to the board, I recieved an email, which has the subject line of ??? ??? and is written in what appears to be greek Ukrainian or Russian I would guess, looking at the headers. Anyone else have this happen? Yes. Either some spammer harvesting addresses, or some setting up filters on his email and getting it wrong. [1] Another Gary, not this one :) -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] php -l - does it find *anything*?
Ashley Sheridan writes: On Tue, 2010-07-06 at 10:54 +0200, Gary . wrote: If foo3 never happens to be called when I am doing my testing (for example if the call is in some if branch that is never exercised) then it only gets found in production, so I would like to find this kind of thing using a static analyser. The kind of problem in foo2 I could live with, but would like to find as well, if possible. (Obviously I am using these two example problems as indicative of the type of things I want to find, it isn't an exhaustive list!) BTW, what problems *does* php -l pick up? I can't find a description anywhere. According to the man page for php, the -l flag only checks the syntax, so a warning wouldn't be displayed, as technically it's not a show-stopper. Well, I think foo3 is, but yeah, I know what you mean. Maybe some sort of unit testing would help pick out these sorts of issues? I do. Actually I posted this by mistake to the phpunit mailing list first :) As PHP isn't a compiled language, I guess it's harder for it to pick up on things like this. Yeah. There are static checkers out there, even some FOSS ones. I guess I'm just a bit frustrated that (as you say) the man page says that -l checks syntax but doesn't really detail what kind of things that covers. Actually, I can't even find a decent description of what E_STRICT covers :( just that Enabling E_STRICT during development has some benefits. STRICT messages will help you to use the latest and greatest suggested method of coding, for example warn you about using deprecated functions. Some benefits? For example? Telling people exactly what is covered seems more useful, to me. Meh. Anyway, that's my whine over for the day :) -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Question about the Board
On 7/6/10, Ashley Sheridan wrote: I have had a couple such emails, requesting me to log in somewhere in order to have my address 'validated' as non-spam, but I ignored it and nothing bad has happened to me yet! Not yet. *hides cattle prod behind back* -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Inner join woes... and sweet tea!
Richard Quadling rquadl...@gmail.com wrote in message news:aanlktinredvfb5cjran9ati-ildvzctlkat-i7amb...@mail.gmail.com... On 5 July 2010 14:48, Pete Ford p...@justcroft.com wrote: P.S. I don't have an emu. Clearly, or you'd know that they can't fly either... :) GIGO!!! Does that mean you do have a deathwatch beetle? Gary __ Information from ESET Smart Security, version of virus signature database 5252 (20100705) __ The message was checked by ESET Smart Security. http://www.eset.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] Question about the Board
The last few times I have posted to the board, I recieved an email, which has the subject line of ??? ??? and is written in what appears to be greek, then english then seemingly some arabic language. It tells me that it has recieved an email from me that it suspects is spam and it has held until I confirm the e-mail (which I DO NOT), then my post appears on the board. I dont open the email, I view it (if there is a difference) in outlook express. I dont want to post the email address or content, but it does not reference the php.net board. Anyone else have this happen? Gary __ Information from ESET Smart Security, version of virus signature database 5252 (20100705) __ The message was checked by ESET Smart Security. http://www.eset.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Attachment to email from form.
Richard
Thank you again for your kind help.
Gary
Richard Quadling rquadl...@gmail.com wrote in message
news:aanlktikwujmpbqijd_ee9mbfpxa92ssexjddng2fg...@mail.gmail.com...
On 26 June 2010 13:14, Gary gp...@paulgdesigns.com wrote:
Richard
Please accept my apology in that my response was sent directly to you and
not the group, I hit 'Reply instead of 'Reply to Group'.
Gary
No problem.
Here is an example of my code using the RMail script from phpguru.org.
It was previously known as html_mime_mail5.
?php
// Create a new message.
$o_Mail = new RMail();
// As we are going to use SMTP to send the email, we need to know
where to send it.
// On my setup, the SMTP ini setting is what I'll be using.
// You can also supply any SMTP AUTH settings needed here.
$o_Mail-setSMTPParams
(
ini_get('SMTP'),
25,
Null,
registrySettings_V2::read(registrySettings_V2::REG_SMTP_AUTH_LOGIN_REQUIRED),
registrySettings_V2::read(registrySettings_V2::REG_SMTP_AUTH_LOGIN),
registrySettings_V2::read(registrySettings_V2::REG_SMTP_AUTH_PASSWORD)
);
// Set the From header
$o_Mail-setHeader('From', 'Richard Quadling r...@nowhere.co.uk');
// Some other useful headers.
// $o_Mail-setHeader('Return-Path', 'Return to me
returnt...@nowhere.co.uk');
// $o_Mail-setHeader('Return-Receipt-To', 'Delivery Receipt
deliveryconfirmati...@nowhere.co.uk');
// $o_Mail-setHeader('Disposition-Notification-To', 'Read or Delete
Notification disposit...@nowhere.co.uk');
// Add the subject.
$o_Mail-setSubject($s_Subject);
// Add the HTML with a location of the images to be added to the email
automatically, rather than relying on internet to provide images
// which is blocked by various security levels of Outlook.
$o_Mail-setHTML($s_HTMLMessage, 'D:/Global Web Documents/images/');
// Let's add an attachment.
// The filename, the mime type, the encoding and the file name to show
the attachment as.
$o_Mail-addAttachment(new fileAttachment('D:/Uploaded/content.zip',
'application/zip', new Base64Encoding(), 'YourContent.zip'));
// If you want to keep copies of your emails, then setCc is the method to
use.
$o_Mail-setCc('r...@nowhere.co.uk');
// Finally, send the email to the recipients (held in an array to
allow for multiple addresses) using SMTP.
$o_Mail-send(array('Your recipient\'s name
your_recipient_em...@domain.com'), 'smtp');
?
--
-
Richard Quadling
Standing on the shoulders of some very clever giants!
EE : http://www.experts-exchange.com/M_248814.html
EE4Free : http://www.experts-exchange.com/becomeAnExpert.jsp
Zend Certified Engineer : http://zend.com/zce.php?c=ZEND002498r=213474731
ZOPA : http://uk.zopa.com/member/RQuadling
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__ Information from ESET Smart Security, version of virus signature
database 5233 (20100628) __
The message was checked by ESET Smart Security.
http://www.eset.com
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Re: [PHP] Attachment to email from form.
Ashley Sheridan a...@ashleysheridan.co.uk wrote in message
news:1277474393.2787.82.ca...@localhost...
On Fri, 2010-06-25 at 09:51 -0400, Gary wrote:
I am trying to have an attachment to an email from a form. Email is
working
fine, am unable to get attachment. The attachment will be a word.doc.
I am getting error message
Warning: file_get_contents(attachment.zip) [function.file-get-contents]:
failed to open stream: No such file or directory in
/home/oneonel1/public_html/emailreminderresult.inc.php on line 24
Mail failed
Line 24 reads:
$attachment =
chunk_split(base64_encode(file_get_contents('attachment.zip')));
here is the all of the code that I have removed the email addresses
such.
Can someone point me in the right direction?
Thank you
Gary
?php
$fname=stripslashes($_POST['fname']);
$lname=stripslashes($_POST['lname']);
$email=stripslashes($_POST['email']);
$comments=stripslashes($_POST['comments']);
$ip= $_SERVER['REMOTE_ADDR'];
$attachment = $_POST['attachment'];
$attachment = $_FILES['attachment']['name'];
$attachment_type = $_FILES['attachment']['type'];
$attachment_size = $_FILES['attachment']['size'];
//create a boundary string. It must be unique
//so we use the MD5 algorithm to generate a random hash
$random_hash = md5(date('r', time()));
//define the headers we want passed. Note that they are separated with
\r\n
$headers = From: myemail\r\nReply-To: myemail.com;
//add boundary string and mime type specification
$headers .= \r\nContent-Type: multipart/mixed;
boundary=\PHP-mixed-.$random_hash.\;
//read the atachment file contents into a string,
//encode it with MIME base64,
//and split it into smaller chunks
$attachment =
chunk_split(base64_encode(file_get_contents('attachment.zip'))); //line
24
//define the body of the message.
ob_start();
//Turn on output buffering
//--PHP-mixed-
echo $random_hash;
//Content-Type: multipart/alternative; boundary=PHP-alt-
echo $random_hash;
//--PHP-alt-
echo $random_hash;
/* Content-Type: text/plain; charset=iso-8859-1
Content-Transfer-Encoding: 7bit */
//--PHP-alt-
echo $random_hash;
/* Content-Type: text/html; charset=iso-8859-1
Content-Transfer-Encoding: 7bit */
//--PHP-alt-
echo $random_hash;
//--PHP-mixed-
echo $random_hash;
/* Content-Type: application/zip; name=attachment.zip
Content-Transfer-Encoding: base64
Content-Disposition: attachment */
echo $attachment;
//--PHP-mixed-
echo $random_hash;
//copy current buffer contents into $message variable and delete current
output buffer
$message = ob_get_clean();
//send the email
$mail_sent = @mail( $to, $subject, $message, $headers );
//if the message is sent successfully print Mail sent. Otherwise print
Mail failed
echo $mail_sent ? Mail sent : Mail failed;
echo Thank you for contacting b888!/bbr /br /;
echo You have submitted the following information:br /br /;
echo Name: $fname $lnamebr /;
echo E-Mail Address: $emailbr /;
echo Your comments or request: $commentsbr /br /br /;
echo We have also sent you an e-mail to $email with the submitted
information as well as our contact information for your convienience. br
/br /
Thank you for the opportunity to serve you!;
/*This is the email message to submitter*/
$contact=888\n 888\n 888;
$from_d=$email;
$to_d=$email;
$subject_d='Thank you from 888';
$msg_d=Thank you $fname for your submission, find our contact
information
listed for your convenience.\n\n
.$contact\n\n
. You have submitted the following information\n\n
. Name: $fname $lname \n
. E-Mail Address: $email\n
. Comments: $comments\n
;
mail($to_d, $subject_d, $msg_d, 'From:' . $from_d);
/*this is to form owner, */
$from=$email;
$to=myemail;
$subject=Submission from 888;
$msg= This is a submission from 888com. \n\n
. Clients Name: $fname . $lname \n
. Email Address: $email\n
. Comments: $comments\n
;
mail($to, $subject, $msg, 'From:' .$from);
?
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Your script can't find the attachment.zip file. As you're using a
relative path to it, it should be in the same directory as your php
script, or somewhere directly in the path environment variable.
Also, make sure that the file has read properties set to allow Apache to
read it.
Thanks,
Ash
http://www.ashleysheridan.co.uk
Ashley
Thank you for your response and I'm sorry for the delay. I'm not sure I
understand your answer. I tried changing the 'attahcment.zip' to
$attachment, even creating a temp folder with reference to the path, however
it does not seem to work.
Perhaps you could explain a little further?
Again, thank you for your response.
Gary
__ Information from
Re: [PHP] Attachment to email from form.
Bastien Koert phps...@gmail.com wrote in message
news:aanlktilfxsbs83zaaa6ix32yhurqdg9azd6twq_hn...@mail.gmail.com...
On Fri, Jun 25, 2010 at 9:59 AM, Ashley Sheridan
a...@ashleysheridan.co.uk wrote:
On Fri, 2010-06-25 at 09:51 -0400, Gary wrote:
I am trying to have an attachment to an email from a form. Email is
working
fine, am unable to get attachment. The attachment will be a word.doc.
I am getting error message
Warning: file_get_contents(attachment.zip) [function.file-get-contents]:
failed to open stream: No such file or directory in
/home/oneonel1/public_html/emailreminderresult.inc.php on line 24
Mail failed
Line 24 reads:
$attachment =
chunk_split(base64_encode(file_get_contents('attachment.zip')));
here is the all of the code that I have removed the email addresses
such.
Can someone point me in the right direction?
Thank you
Gary
?php
$fname=stripslashes($_POST['fname']);
$lname=stripslashes($_POST['lname']);
$email=stripslashes($_POST['email']);
$comments=stripslashes($_POST['comments']);
$ip= $_SERVER['REMOTE_ADDR'];
$attachment = $_POST['attachment'];
$attachment = $_FILES['attachment']['name'];
$attachment_type = $_FILES['attachment']['type'];
$attachment_size = $_FILES['attachment']['size'];
//create a boundary string. It must be unique
//so we use the MD5 algorithm to generate a random hash
$random_hash = md5(date('r', time()));
//define the headers we want passed. Note that they are separated with
\r\n
$headers = From: myemail\r\nReply-To: myemail.com;
//add boundary string and mime type specification
$headers .= \r\nContent-Type: multipart/mixed;
boundary=\PHP-mixed-.$random_hash.\;
//read the atachment file contents into a string,
//encode it with MIME base64,
//and split it into smaller chunks
$attachment =
chunk_split(base64_encode(file_get_contents('attachment.zip'))); //line
24
//define the body of the message.
ob_start();
//Turn on output buffering
//--PHP-mixed-
echo $random_hash;
//Content-Type: multipart/alternative; boundary=PHP-alt-
echo $random_hash;
//--PHP-alt-
echo $random_hash;
/* Content-Type: text/plain; charset=iso-8859-1
Content-Transfer-Encoding: 7bit */
//--PHP-alt-
echo $random_hash;
/* Content-Type: text/html; charset=iso-8859-1
Content-Transfer-Encoding: 7bit */
//--PHP-alt-
echo $random_hash;
//--PHP-mixed-
echo $random_hash;
/* Content-Type: application/zip; name=attachment.zip
Content-Transfer-Encoding: base64
Content-Disposition: attachment */
echo $attachment;
//--PHP-mixed-
echo $random_hash;
//copy current buffer contents into $message variable and delete current
output buffer
$message = ob_get_clean();
//send the email
$mail_sent = @mail( $to, $subject, $message, $headers );
//if the message is sent successfully print Mail sent. Otherwise print
Mail failed
echo $mail_sent ? Mail sent : Mail failed;
echo Thank you for contacting b888!/bbr /br /;
echo You have submitted the following information:br /br /;
echo Name: $fname $lnamebr /;
echo E-Mail Address: $emailbr /;
echo Your comments or request: $commentsbr /br /br /;
echo We have also sent you an e-mail to $email with the submitted
information as well as our contact information for your convienience. br
/br /
Thank you for the opportunity to serve you!;
/*This is the email message to submitter*/
$contact=888\n 888\n 888;
$from_d=$email;
$to_d=$email;
$subject_d='Thank you from 888';
$msg_d=Thank you $fname for your submission, find our contact
information
listed for your convenience.\n\n
.$contact\n\n
. You have submitted the following information\n\n
. Name: $fname $lname \n
. E-Mail Address: $email\n
. Comments: $comments\n
;
mail($to_d, $subject_d, $msg_d, 'From:' . $from_d);
/*this is to form owner, */
$from=$email;
$to=myemail;
$subject=Submission from 888;
$msg= This is a submission from 888com. \n\n
. Clients Name: $fname . $lname \n
. Email Address: $email\n
. Comments: $comments\n
;
mail($to, $subject, $msg, 'From:' .$from);
?
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Your script can't find the attachment.zip file. As you're using a
relative path to it, it should be in the same directory as your php
script, or somewhere directly in the path environment variable.
Also, make sure that the file has read properties set to allow Apache to
read it.
Thanks,
Ash
http://www.ashleysheridan.co.uk
Consider using something like phpmailer to handle the emails. Makes
attaching things really simple.
http://phpmailer.worxware.com/
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Cat, the other other white meat
Bastien
How could anyone not love your advice when you seem to enjoy cats so much!
I did download
[PHP] in_array - what the...
If I have an array that looks like
array(1) {
[mac_address]=
string(2) td
}
and I call
if (in_array($name, self::$aboveArray))
with $name as
string(11) mac_address
what should be the result?
Because it is *false* and it is driving me nuts! This despite the fact
that if I do
$foo = self::$aboveArray[$name];
$foo then has the value
string(2) td
Am I not understanding what in_array does? Is there some bug in php that
has been present since 5.2.12 and still is? *bangs head against desk*
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Re: [PHP] in_array - what the...
Ford, Mike writes:
-Original Message-
If I have an array that looks like
array(1) {
[mac_address]=
string(2) td
}
and I call
if (in_array($name, self::$aboveArray))
with $name as
string(11) mac_address
what should be the result?
FALSE -- in_array checks the *values*, not the keys, so would be
looking at the td for this element.
Agh! So it does.
You know what's worse? I even looked at the documentation of the
function this morning wondering if that's what the problem was and
*still* didn't see it!
*slinks away in embarrassment*
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[PHP] Attachment to email from form.
I am trying to have an attachment to an email from a form. Email is working
fine, am unable to get attachment. The attachment will be a word.doc.
I am getting error message
Warning: file_get_contents(attachment.zip) [function.file-get-contents]:
failed to open stream: No such file or directory in
/home/oneonel1/public_html/emailreminderresult.inc.php on line 24
Mail failed
Line 24 reads:
$attachment =
chunk_split(base64_encode(file_get_contents('attachment.zip')));
here is the all of the code that I have removed the email addresses such.
Can someone point me in the right direction?
Thank you
Gary
?php
$fname=stripslashes($_POST['fname']);
$lname=stripslashes($_POST['lname']);
$email=stripslashes($_POST['email']);
$comments=stripslashes($_POST['comments']);
$ip= $_SERVER['REMOTE_ADDR'];
$attachment = $_POST['attachment'];
$attachment = $_FILES['attachment']['name'];
$attachment_type = $_FILES['attachment']['type'];
$attachment_size = $_FILES['attachment']['size'];
//create a boundary string. It must be unique
//so we use the MD5 algorithm to generate a random hash
$random_hash = md5(date('r', time()));
//define the headers we want passed. Note that they are separated with \r\n
$headers = From: myemail\r\nReply-To: myemail.com;
//add boundary string and mime type specification
$headers .= \r\nContent-Type: multipart/mixed;
boundary=\PHP-mixed-.$random_hash.\;
//read the atachment file contents into a string,
//encode it with MIME base64,
//and split it into smaller chunks
$attachment =
chunk_split(base64_encode(file_get_contents('attachment.zip'))); //line 24
//define the body of the message.
ob_start();
//Turn on output buffering
//--PHP-mixed-
echo $random_hash;
//Content-Type: multipart/alternative; boundary=PHP-alt-
echo $random_hash;
//--PHP-alt-
echo $random_hash;
/* Content-Type: text/plain; charset=iso-8859-1
Content-Transfer-Encoding: 7bit */
//--PHP-alt-
echo $random_hash;
/* Content-Type: text/html; charset=iso-8859-1
Content-Transfer-Encoding: 7bit */
//--PHP-alt-
echo $random_hash;
//--PHP-mixed-
echo $random_hash;
/* Content-Type: application/zip; name=attachment.zip
Content-Transfer-Encoding: base64
Content-Disposition: attachment */
echo $attachment;
//--PHP-mixed-
echo $random_hash;
//copy current buffer contents into $message variable and delete current
output buffer
$message = ob_get_clean();
//send the email
$mail_sent = @mail( $to, $subject, $message, $headers );
//if the message is sent successfully print Mail sent. Otherwise print
Mail failed
echo $mail_sent ? Mail sent : Mail failed;
echo Thank you for contacting b888!/bbr /br /;
echo You have submitted the following information:br /br /;
echo Name: $fname $lnamebr /;
echo E-Mail Address: $emailbr /;
echo Your comments or request: $commentsbr /br /br /;
echo We have also sent you an e-mail to $email with the submitted
information as well as our contact information for your convienience. br
/br /
Thank you for the opportunity to serve you!;
/*This is the email message to submitter*/
$contact=888\n 888\n 888;
$from_d=$email;
$to_d=$email;
$subject_d='Thank you from 888';
$msg_d=Thank you $fname for your submission, find our contact information
listed for your convenience.\n\n
.$contact\n\n
. You have submitted the following information\n\n
. Name: $fname $lname \n
. E-Mail Address: $email\n
. Comments: $comments\n
;
mail($to_d, $subject_d, $msg_d, 'From:' . $from_d);
/*this is to form owner, */
$from=$email;
$to=myemail;
$subject=Submission from 888;
$msg= This is a submission from 888com. \n\n
. Clients Name: $fname . $lname \n
. Email Address: $email\n
. Comments: $comments\n
;
mail($to, $subject, $msg, 'From:' .$from);
?
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Re: [PHP] Re: function within a class function
On Mon, Jun 21, 2010 at 10:49 AM, Pete Ford wrote: On 21/06/10 00:45, Rick Pasotto wrote: I think that's right - doesn't the manual describe this sort of thing about using callback functions that are class members? Yes. http://www.php.net/manual/en/language.pseudo-types.php#language.types.callback -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] PHP Udate MySQL command
I am trying to get an update command to work in PHP. I am able to update
records going directly to phpmyadmin command line. I have even let it
produce the php code to insert, but have not been able to get it to work.
I have it stripped down to one command hoping to get it to work then
replicate entire forms for clients to use direct.I get no error codes, I
only get my message It did not enter into DB;
Anyone see where I am going wrong?
Gary
!DOCTYPE html PUBLIC -//W3C//DTD XHTML 1.0 Transitional//EN
http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd;
html xmlns=http://www.w3.org/1999/xhtml;
head
meta http-equiv=Content-Type content=text/html; charset=utf-8 /
titleUntitled Document/title
/head
body
form action=?php echo $_SERVER['PHP_SELF']; ? method=post
testinput name=test type=text /
input name=submit type=submit value=submit /
/form
?php
$batchconnetion = mysql_connect(host, 'un', 'pw', 'db')//sanatized for board
or die('Error connecting with MySQL Database');
$test=$_POST['test'];
//$sql=update contact set type = \'$test\' where item_id = \'164\'; //this
is the code created by phpmyadmin
$sql = INSERT INTO contact comments, VALUES = $test WHERE contact_id =
33;
mysql_query($sql,$batchconnetion);
$result = mysql_query($sql,$batchconnetion);
if($result == true) {
echo Successfully Inserted Records;
} else {
echo It did not enter into DB;
}
mysql_close($batchconnetion);
?
/body
/html
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[PHP] Re: PHP Udate MySQL command
Sorry, just noticed that I had this line in the post. (should not post
before coffee).
$sql = INSERT INTO contact comments, VALUES = $test WHERE contact_id =
33;
It is supposed to be
$sql = UPDATE contact comments = '$test' WHERE contact_id = '33'
Gary gwp...@ptd.net wrote in message
news:49.e6.07323.599d8...@pb1.pair.com...
I am trying to get an update command to work in PHP. I am able to update
records going directly to phpmyadmin command line. I have even let it
produce the php code to insert, but have not been able to get it to work.
I have it stripped down to one command hoping to get it to work then
replicate entire forms for clients to use direct.I get no error codes, I
only get my message It did not enter into DB;
Anyone see where I am going wrong?
Gary
!DOCTYPE html PUBLIC -//W3C//DTD XHTML 1.0 Transitional//EN
http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd;
html xmlns=http://www.w3.org/1999/xhtml;
head
meta http-equiv=Content-Type content=text/html; charset=utf-8 /
titleUntitled Document/title
/head
body
form action=?php echo $_SERVER['PHP_SELF']; ? method=post
testinput name=test type=text /
input name=submit type=submit value=submit /
/form
?php
$batchconnetion = mysql_connect(host, 'un', 'pw', 'db')//sanatized for
board
or die('Error connecting with MySQL Database');
$test=$_POST['test'];
//$sql=update contact set type = \'$test\' where item_id = \'164\';
//this is the code created by phpmyadmin
$sql = INSERT INTO contact comments, VALUES = $test WHERE contact_id =
33;
mysql_query($sql,$batchconnetion);
$result = mysql_query($sql,$batchconnetion);
if($result == true) {
echo Successfully Inserted Records;
} else {
echo It did not enter into DB;
}
mysql_close($batchconnetion);
?
/body
/html
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[PHP] xml_set_element_handler and equivalents
I need to do essentially do what can be found in the XML Tag Mapping example (http://www.php.net/manual/en/example.xml-map-tags.php) - transform the names of some XML elements. Normally I'd use XSLT for this kind of thing, but due to environmental considerations I can't. I would prefer to do this in a somewhat OO manner so I can hide this nastiness from the rest of the code. Obviously I can't pass any of my XML handler's to xml_set_element_handler as handlers. Nor can I find a way of encapsulating the map_array (at least not without duplicating the map). Any ideas? Is there an equivalent for DOM or SimpleXML or something? -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] PHP Udate MySQL command
Ashley Sheridan a...@ashleysheridan.co.uk wrote in message
news:1275649100.2217.45.ca...@localhost...
On Fri, 2010-06-04 at 06:46 -0400, Gary wrote:
I am trying to get an update command to work in PHP. I am able to update
records going directly to phpmyadmin command line. I have even let it
produce the php code to insert, but have not been able to get it to work.
I have it stripped down to one command hoping to get it to work then
replicate entire forms for clients to use direct.I get no error codes, I
only get my message It did not enter into DB;
Anyone see where I am going wrong?
Gary
!DOCTYPE html PUBLIC -//W3C//DTD XHTML 1.0 Transitional//EN
http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd;
html xmlns=http://www.w3.org/1999/xhtml;
head
meta http-equiv=Content-Type content=text/html; charset=utf-8 /
titleUntitled Document/title
/head
body
form action=?php echo $_SERVER['PHP_SELF']; ? method=post
testinput name=test type=text /
input name=submit type=submit value=submit /
/form
?php
$batchconnetion = mysql_connect(host, 'un', 'pw', 'db')//sanatized for
board
or die('Error connecting with MySQL Database');
$test=$_POST['test'];
//$sql=update contact set type = \'$test\' where item_id = \'164\';
//this
is the code created by phpmyadmin
$sql = INSERT INTO contact comments, VALUES = $test WHERE contact_id =
33;
mysql_query($sql,$batchconnetion);
$result = mysql_query($sql,$batchconnetion);
if($result == true) {
echo Successfully Inserted Records;
} else {
echo It did not enter into DB;
}
mysql_close($batchconnetion);
?
/body
/html
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Your problem is the way you're trying to connect to the DB:
$batchconnetion = mysql_connect(host, 'un', 'pw', 'db')//sanatized for
board
or die('Error connecting with MySQL Database');
The 4th argument to this function isn't a database name, it's a boolean
value for whether PHP should create a new link or reuse the old one. You
have to use the mysql_select_db() function to select the database to use
on that connection in order for your queries to run the way you have
them.
There does seem to be a bit of an inconsistency with the way you're
using quotation marks on your query strings as well. Your first
commented out query is escaping the single quotes within a double quoted
string (which isn't necessary), you've omitted the single quotes on your
insert line, and then on your amended email you've omitted the double
quotes. If this is your actual code and not a mistake made when copying
it into an email, then it would also be a second reason why things
aren't working as expected even once you get PHP to connect properly.
Thanks,
Ash
http://www.ashleysheridan.co.uk
__ Information from ESET Smart Security, version of virus
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Ashley
Thank you for your quick reply.
The $batchconnection works fine in numerous other files (the insert files)
and is a c/p from one of those files. The db noted in my post was just to
specify that the name of the database at this point, I thought it was a
standard practice. I will try your suggestion.
I will admit that I dont have the full grasp of the quotes, single or
double. Typically in a mysql command I start with a , however if I get a
script that is not working, I will try to resolve the issue with switching
up the singles with the doubles as well and adding parenthsis.
The mistake in the copying the code was I was trying to see if the insert
command would work in place of the update.it did not..
Aside from that, do you see a problem with my update commands?
Thanks for your help.
gary
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Re: [PHP] PHP Udate MySQL command Success
Gary gwp...@ptd.net wrote in message
news:a9.ea.07323.381e8...@pb1.pair.com...
Ashley Sheridan a...@ashleysheridan.co.uk wrote in message
news:1275649100.2217.45.ca...@localhost...
On Fri, 2010-06-04 at 06:46 -0400, Gary wrote:
I am trying to get an update command to work in PHP. I am able to
update
records going directly to phpmyadmin command line. I have even let it
produce the php code to insert, but have not been able to get it to
work.
I have it stripped down to one command hoping to get it to work then
replicate entire forms for clients to use direct.I get no error codes, I
only get my message It did not enter into DB;
Anyone see where I am going wrong?
Gary
!DOCTYPE html PUBLIC -//W3C//DTD XHTML 1.0 Transitional//EN
http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd;
html xmlns=http://www.w3.org/1999/xhtml;
head
meta http-equiv=Content-Type content=text/html; charset=utf-8 /
titleUntitled Document/title
/head
body
form action=?php echo $_SERVER['PHP_SELF']; ? method=post
testinput name=test type=text /
input name=submit type=submit value=submit /
/form
?php
$batchconnetion = mysql_connect(host, 'un', 'pw', 'db')//sanatized for
board
or die('Error connecting with MySQL Database');
$test=$_POST['test'];
//$sql=update contact set type = \'$test\' where item_id = \'164\';
//this
is the code created by phpmyadmin
$sql = INSERT INTO contact comments, VALUES = $test WHERE contact_id
=
33;
mysql_query($sql,$batchconnetion);
$result = mysql_query($sql,$batchconnetion);
if($result == true) {
echo Successfully Inserted Records;
} else {
echo It did not enter into DB;
}
mysql_close($batchconnetion);
?
/body
/html
__ Information from ESET Smart Security, version of virus
signature database 5171 (20100604) __
The message was checked by ESET Smart Security.
http://www.eset.com
Your problem is the way you're trying to connect to the DB:
$batchconnetion = mysql_connect(host, 'un', 'pw', 'db')//sanatized for
board
or die('Error connecting with MySQL Database');
The 4th argument to this function isn't a database name, it's a boolean
value for whether PHP should create a new link or reuse the old one. You
have to use the mysql_select_db() function to select the database to use
on that connection in order for your queries to run the way you have
them.
There does seem to be a bit of an inconsistency with the way you're
using quotation marks on your query strings as well. Your first
commented out query is escaping the single quotes within a double quoted
string (which isn't necessary), you've omitted the single quotes on your
insert line, and then on your amended email you've omitted the double
quotes. If this is your actual code and not a mistake made when copying
it into an email, then it would also be a second reason why things
aren't working as expected even once you get PHP to connect properly.
Thanks,
Ash
http://www.ashleysheridan.co.uk
__ Information from ESET Smart Security, version of virus
signature database 5171 (20100604) __
The message was checked by ESET Smart Security.
http://www.eset.com
Ashley
Thank you for your quick reply.
The $batchconnection works fine in numerous other files (the insert files)
and is a c/p from one of those files. The db noted in my post was just to
specify that the name of the database at this point, I thought it was a
standard practice. I will try your suggestion.
I will admit that I dont have the full grasp of the quotes, single or
double. Typically in a mysql command I start with a , however if I get a
script that is not working, I will try to resolve the issue with switching
up the singles with the doubles as well and adding parenthsis.
The mistake in the copying the code was I was trying to see if the insert
command would work in place of the update.it did not..
Aside from that, do you see a problem with my update commands?
Thanks for your help.
gary
__ Information from ESET Smart Security, version of virus
signature database 5171 (20100604) __
The message was checked by ESET Smart Security.
http://www.eset.com
Ashley
Thanks for your input, I think what corrected the problem was I added single
qoutes to the $test variable.
I did also change to your suggestion.
The working code is
body
form action=?php echo $_SERVER['PHP_SELF']; ? method=post
testinput name=test type=text /
input name=submit type=submit value=submit /
/form
?php
$batchconnetion = mysql_connect(sanatized)
or die('Error connecting with MySQL Database');
mysql_select_db('sanatized', $batchconnetion)or die('Error connecting with
MySQL Database2');;
$test=$_POST['test'];
//$sql=update contact set type = \'$test\' where item_id = \'164\';
$sql = UPDATE contact SET comments = '$test' WHERE contact_id = 33;
mysql_query($sql,$batchconnetion);
$result
[PHP] Re: xml_set_element_handler and equivalents
On 6/4/10, Gary wrote:
I need to do essentially do what can be found in the XML Tag Mapping
example (http://www.php.net/manual/en/example.xml-map-tags.php)
...
I would prefer to do this in a somewhat OO manner so I can hide this
nastiness from the rest of the code.
Obviously I can't pass any of my XML handler's to
xml_set_element_handler as handlers.
Oh. Hum. Of course, I can (where //... == elided code):
class Foo {
//...
public function bar() {
//...
xml_set_element_handler($parser,
array($this, startElement),
array($this, endElement));
//...
}
public function startElement($parser, $name, $attrs) {
//...
}
}
So just ignore me, I'm having a bad day.
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Re: [PHP] Remove blank lines from a file
Anton Heuschen writes: Hi Im trying do something like this, have a function which uploads my file and returns file pointer ... but at same time ... I want to remove all Blank lines in a file and update it sed -i '/^$/d' yourFile -- Gary -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Email from php
Ferdi writes: I have met with little success sending mail from PHP. I have used mainly the mail() function but have also tried imap_mail() which the documentation says is just a wrapper around mail(). IMAP is only incoming (at least, the protocol is - the clue is in the name Internet Message Access Protocol - I can't say anything about imap_mail). I agree with others, the easiest solution is a local MTA such as sendmail, or, preferably, something more simple. Here is my understanding of the situation: On Windows (WampServer 2.0i ) I manage to send email after setting SMTP = smtp.someserver.com. I guess this works because the server I use relays all mail received; it does not check if the user has been registered or not. Oof. Yes, most remote/ISP MTAs these days will require some kind of authentication, which is why it is best (IMO) to use a local MTA. It won't mean you don't have to authenticate to the remote MTA, but it will push the responsibility of configuring the MTA onto someone who knows what it's all about. On Linux (Centos 5.XX, XAMPP 1.7.2), the above function [mail()] does not work. On Linux, what I have understood, is that the smtp server settings in php.ini do not matter. The sendmail / mail (something like that) What happens if you type 'man mail'? utility is called which sends the mail. I have put in the correct sendmail_path setting in php.ini, but, I guess this utility is not configured since I don't receive the mail and running the code does not throw up errors or warnings. You can try sending email from the commandline using 'mail' or even sendmail, to see if it works. You could try installing something simple locally (i.e. as your user), perhaps msmtp, which is a piece of cake to setup, especially compared with stuff like sendmail (which it is, anyway, compatible with, to a large extent). What I need to achieve is the ability to send attachments in an email from PHP. I would like one of the following options (in order of preference): 1. Create an email account (specifically Google Apps Mail) and send email as that user. This will not be trivial. I will need to get a fix on authenticating, logging in etc. A simple config to send mail through Google Apps for Domains via a local msmtp installation follows: #-- defaults host smtp.googlemail.com tls on tls_starttls on port 587 tls_trust_file /home/YOUR_USER/.certs/equifax.pem tls_cert_file tls_key_file auth on logfile ~/.logfiles/msmtp.log account foo from f...@yourdomain user f...@yourdomain password YOUR_PASSWORD_HERE #-- You'll need to get the referenced tls_trust_file but I forget where from. It will be different anyway, depending on the system - for example from looking at the msmtp docs it seems you can simply install a package on Debian. 3. Configuring the sendmail utility on Linux From what I remember, I wouldn't recommend sendmail :) It's really complicated, and probably overkill for what you want. -- Gary -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] Generating PHP from WSDL
Let's if anyone can come up with a non-brittle solution to this. In order to create the framework code for a webservice client I have generated it from the server's wsdl. Unfortunately the wsdl (which is out of my control) tries to use extension base=... to say that 'B' extends/inherits from 'A', and the tool I use doesn't understand it, with the result that the generated code doesn't include that relatonship, and so sending a 'B' to the webservice doesn't work because it doesn't also include the expected properties of 'A'. Possible solutions I have thought of are: * create 'C', inheriting from 'A' and 'B' and send one of those instead, but of course php doesn't support multiple-inheritance; * manually change 'B' so that it does inherit from 'A'. I don't like this because if the wsdl changes then that client-side code needs to again be changed manually. * use composition instead. Doesn't seem it will work because the marshalling of the objects seems to want to use the properties directly (i.e. instead of using getters as it should do). Does anyone have any other ideas? Are there any generators out there that handle something like: s:complexType name=B s:complexContent mixed=false s:extension base=tns:A ... /s:complexType and produce nusoap client code? -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] xpath help
On 5/9/10, Peter Lind wrote: On 9 May 2010 10:31, Gary wrote: If I have a document containing, say: html ... How do I get at bar2? I tried everything, based on an xpath from Firebug (Firefox plugin), but kept getting NULL. try //table//font - that should give you all the font elements in table elements. Given your layout, you're then looking for $list-item(3) I found out what my problem was. Something, somewhere - the plugin, Firefox itself, something - decided to help me by inserting TBODY elements into the xpath which just plain did not exist in the original HTML. -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] xpath help
If I have a document containing, say: html body table tbody tr tdfont ...foo1/font/td tdfont ...bar1/font/td /tr tr td font ...foo2/font /td td font ...bar2/font /td /tr ... How do I get at bar2? I tried everything, based on an xpath from Firebug (Firefox plugin), but kept getting NULL. -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Any One See where this is going wrong?
Adam Richardson simples...@gmail.com wrote in message
news:aanlktikuqehnrjcannzq8ezvmx5pneg_bjvezglj0...@mail.gmail.com...
On Sat, May 1, 2010 at 6:54 PM, Gary gwp...@ptd.net wrote:
Adam Richardson simples...@gmail.com wrote in message
news:aanlktinqixxb9oipu4op2xztrze_vwpbymaywziwb...@mail.gmail.com...
$sqlStatements = INSERT INTO guns (manufacturer, type, model,
caliber,
condition, price, description, image_file_name, available) VALUES
('$manufacturer',
'$type',
'$model','$caliber','$condition','$price','$description','$image_file_name','$available');
INSERT INTO images(id, image_file) VALUES ('', '$image_file');
Gary, the second insert has what appears to be a troublesome bit of
code:
VALUES ('', '$image_file');
Shouldn't it read:
INSERT INTO images(image_file) VALUES ('$image_file');
That's assuming you have id set to auto-increment, so no id in the
first
or
second set of parentheses. Additionally, the , before the $image_file
var
would cause issues, too.
Adam
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Adam
The second INSERT is part of a multi_query, and yes there is a
auto-increment field in front named id, it is the foreign key that links
the tables.
I had removed, and just tried again, but the same result.
I have tried it so many ways my fingers and eyes are about to bleed. It
makes not sense how it works in one but not the other.
I cant see to get the error called out.
Thanks again.
Gary
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And you also removed the double quote and the comma next to it (sorry,
just
making sure?) That is to say, you copied the line of code I sent?
$sqlStatements = INSERT INTO guns( *id,*manufacturer, type, model, //
*id,*should be removed
caliber, condition, price, description, image_file_name,submitted
,available) VALUES ('*',*'$manufacturer', '$type', '$model', '$caliber',
// the *,* combo should be removed
'$condition', '$price', '$description','$image_file_name', ' ',
'$available');
INSERT INTO images (*id,* image_file) VALUES('','$image_file');
//
the *id,* should be removed AND the *,* combo should be removed
Also, have you tried directly echoing out the error (e.g., *echo
mysql_error();* if it's mysql you're using, or the comparable method in
the
library you're using.)
If that didn't work, I'd break up the multi-query into 2 separate queries
to
better troubleshoot the issue. I like using PDO and setting it up so it
throws exceptions when something goes wrong (additionally, I like using
the
prepared statements, too.) Have you been trying that?
I'm still hopeful you'll get this figured out before there's any
blood-loss
;)
Feel free to send back the new code after the edits if you still have
issues.
Adam
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Adam
What I ended up doing was I rebuilt the entire thing from start. I created
one table with one column, made sure it worked, then added a few more, then
added the child table.
The only thing that I can think of is that I had one of the columns named
id, and maybe that is not allowed.
I also just had the form processed on itself, but I cant really imagine that
would have anything to do with it.
I'll post the code later if interested, I am still adding columns, but the
child is working so that would seem to be the only real challenge.
Thanks for all your input.
Gary
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Re: [PHP] Any One See where this is going wrong?
Jim
Thanks for your reponse. With the variable defined. it is supposed to
reduce the size of the image before upload.
Thanks again.
Gary
Jim Lucas li...@cmsws.com wrote in message
news:4bdc311b.4060...@cmsws.com...
Gary wrote:
I have this duplicate code on another site and it works fine. The image
is uploaded to the images folder, the information is not submitted to the
database. I get the error
Some Error Occured While Inserting Records
This is only on a local machine so I have not yet included and safegaurds
like stripslashes or my_real_escape_string.
Thanks for your help
Gary
?php
if (isset($_POST['submit'])) {
$manufacturer=($_POST['manufacturer']);
$type=($_POST['type']);
$model=($_POST['model']);
$caliber=($_POST['caliber']);
$condition=($_POST['condition']);
$price=($_POST['price']);
$description=($_POST['description']);
$image_file_name=($_POST['image_file_name']);
$image_file=($_FILES['image_file']);
$available=($_POST['available']);
$image_file = $_FILES['image_file']['name'];
$image_type = $_FILES['image_file']['type'];
$image_size = $_FILES['image_file']['size'];
include ('includes/connect_local.inc.php');
Is the following line suppose to be working with a constant or the
variable that you defined/extracted just above here?
if(image_size 300) {
class ImgResizer {
private $originalFile = 'image_file';
public function __construct($originalFile = 'image_file') {
$this - originalFile = $originalFile;
}
public function resize($newWidth, $targetFile) {
if (empty($newWidth) || empty($targetFile)) {
return false;
}
$src = imagecreatefromjpeg($this - originalFile);
list($width, $height) = getimagesize($this - originalFile);
$newHeight = ($height / $width) * $newWidth;
$tmp = imagecreatetruecolor($newWidth, $newHeight);
imagecopyresampled($tmp, $src, 0, 0, 0, 0, $newWidth, $newHeight,
$width, $height);
if (file_exists($targetFile)) {
unlink($targetFile);
}
imagejpeg($tmp, $targetFile, 85); // 85 is my choice, make it between
0 - 100 for output image quality with 100 being the most luxurious
}
}
}
if (!empty($type) !empty($image_file)) {
if (($image_type == 'image/gif') || ($image_type == 'image/jpeg')
|| ($image_type == 'image/pjpeg') || ($image_type == 'image/png')
By the use of a variable below, I'm guessing it was suppose to be a
variable.
($image_size 300)) {
if ($_FILES['image_file']['error'] == 0) {
// Move the file to the target upload folder
$target = 'images/' . $image_file;
if (move_uploaded_file($_FILES['image_file']['tmp_name'],
$target)){
$batchconnection;
$sqlStatements = INSERT INTO guns( id,manufacturer, type, model,
caliber, condition, price, description, image_file_name,submitted
,available) VALUES ('','$manufacturer', '$type', '$model', '$caliber',
'$condition', '$price', '$description','$image_file_name', ' ',
'$available');
INSERT INTO images (id, image_file) VALUES('','$image_file');
$sqlResult = $batchconnection-multi_query($sqlStatements);
if($sqlResult == true) {
echo Successfully Inserted Records;
} else {
echo Some Error Occured While Inserting Records;
}
}
}
}
}
mysqli_close($batchconnection);
}
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Jim Lucas
A: Maybe because some people are too annoyed by top-posting.
Q: Why do I not get an answer to my question(s)?
A: Because it messes up the order in which people normally read text.
Q: Why is top-posting such a bad thing?
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