Re: [PHP-DB] Passing variables between html forms and updating sql table
On Wednesday 17 September 2003 22:34, David wrote: I am putting together some pages where I will be able to update mysql database using php pages. I am having a problem with updating. I put a record into textboxes and when I press submit it passes the values to the next page. The problem I am having is updating the values. Here is my attempt [snip] And can you describe what this problem is? -- Jason Wong - Gremlins Associates - www.gremlins.biz Open Source Software Systems Integrators * Web Design Hosting * Internet Intranet Applications Development * -- Search the list archives before you post http://marc.theaimsgroup.com/?l=php-db -- /* I learned to play guitar just to get the girls, and anyone who says they didn't is just lyin'! -- Willie Nelson */ -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Passing variables between html forms and updating sql table
Dear Jason
Here is my original problem below as I am treated as spam. When it
reaches the next page, when the update query is run, it does not update
the database.
Regards
David
Anagram Systems
http://www.anagram-sys.co.uk/ http://www.web-planets.com/davec/techsitedb/
Dear all
I am putting together some pages where I will be able to update mysql
database using php pages. I am having a problem with updating. I put a
record into textboxes and when I press submit it passes the values to the
next page. The problem I am having is updating the values. Here is my
attempt
?php
$MySQLLink = mysql_pconnect (davecp4, root, )
or die(Could not attach to database. Please try later or contact
[EMAIL PROTECTED]);
mysql_select_db($database_name, $MySQLLink) or die(ERROR--CAN'T CONNECT
TO DB);
$result = mysql_query(SELECT * FROM .$table_name) or die(Error: .
mysql_error());
for($i = 1; $i mysql_num_fields($result); $i++)
{
$fieldname = mysql_field_name($result, $i);
$fieldname2 = $.$fieldname;
mysql_query(UPDATE $table_name SET $fieldname=$fieldname2 WHERE id
=$id);
echo $fieldname.br.$fieldname2.br;
}
?
-- Kind Regards David Anagram Systems http://www.anagram-sys.co.uk/
http://www.web-planets.com/davec/techsitedb/
Jason Wong wrote:
On Wednesday 17 September 2003 22:34, David wrote:
I am putting together some pages where I will be able to update mysql
database using php pages. I am having a problem with updating. I put a
record into textboxes and when I press submit it passes the values to the
next page. The problem I am having is updating the values. Here is my
attempt
[snip]
And can you describe what this problem is?
--
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Re: [PHP-DB] Passing variables between html forms and updating sql table
On Wednesday 17 September 2003 23:42, David wrote: Here is my original problem below as I am treated as spam. When it reaches the next page, when the update query is run, it does not update the database. *sigh* Could you please describe *exactly* what happens. Eg does your error messages kick-in? Does the php error logs show anything? etc -- Jason Wong - Gremlins Associates - www.gremlins.biz Open Source Software Systems Integrators * Web Design Hosting * Internet Intranet Applications Development * -- Search the list archives before you post http://marc.theaimsgroup.com/?l=php-db -- /* I am two fools, I know, for loving, and for saying so. -- John Donne */ -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Passing variables between html forms and updating sql table
Dear Jason
I need a secure way of updating mysql on the website, phpadmin is not
secure enough
So I have created these pages in order:
A page to choose the databases
Next a page to choose the table
Next a list of the records in the table
Next to edit the page using textboxes and textareas
I pass the database name, table name, and the records to change to the
page with the problem.
The passed values have the same field name as the record in the database
e.g. id
The problem I can see is that I cannot access the passed variables, that
is why the update is failing , although I do not get any errors.
I am trying to access the passed variables with this line $fieldname2 =
$.$fieldname;
Does this make sense
David
?php
$MySQLLink = mysql_pconnect (davecp4, root, )
or die(Could not attach to database. Please try later or contact
[EMAIL PROTECTED]);
mysql_select_db($database_name, $MySQLLink) or die(ERROR--CAN'T CONNECT
TO DB);
$result = mysql_query(SELECT * FROM .$table_name) or die(Error: .
mysql_error());
for($i = 1; $i mysql_num_fields($result); $i++)
{
$fieldname = mysql_field_name($result, $i);
$fieldname2 = $.$fieldname;
mysql_query(UPDATE $table_name SET $fieldname=$fieldname2 WHERE id
=$id);
echo $fieldname.br.$fieldname2.br;
}
?
Jason Wong wrote:
On Wednesday 17 September 2003 23:42, David wrote:
Here is my original problem below as I am treated as spam. When it
reaches the next page, when the update query is run, it does not update
the database.
*sigh*
Could you please describe *exactly* what happens. Eg does your error messages
kick-in? Does the php error logs show anything? etc
--
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Re: [PHP-DB] Passing variables between html forms and updating sql table
On Thursday 18 September 2003 01:03, David wrote:
I need a secure way of updating mysql on the website, phpadmin is not
secure enough
Why is phpadmin not secure enough? Or to put it another way, what are you
doing differently that makes it more secure than php admin?
So I have created these pages in order:
A page to choose the databases
Next a page to choose the table
Next a list of the records in the table
Next to edit the page using textboxes and textareas
I pass the database name, table name, and the records to change to the
page with the problem.
The passed values have the same field name as the record in the database
e.g. id
The problem I can see is that I cannot access the passed variables, that
is why the update is failing ,
Most likely register_globals (google or search archive or RTFM).
although I do not get any errors.
Have you set the highest level of error reporting? And set it to display
errors (or log to file)?
I am trying to access the passed variables with this line $fieldname2 =
$.$fieldname;
Not sure what you're trying to do here but ...
?php
$MySQLLink = mysql_pconnect (davecp4, root, )
or die(Could not attach to database. Please try later or contact
[EMAIL PROTECTED]);
mysql_select_db($database_name, $MySQLLink) or die(ERROR--CAN'T CONNECT
TO DB);
$result = mysql_query(SELECT * FROM .$table_name) or die(Error: .
mysql_error());
for($i = 1; $i mysql_num_fields($result); $i++)
{
$fieldname = mysql_field_name($result, $i);
$fieldname2 = $.$fieldname;
mysql_query(UPDATE $table_name SET $fieldname=$fieldname2 WHERE id
=$id);
echo $fieldname.br.$fieldname2.br;
... does the above echo display what you expected it to?
--
Jason Wong - Gremlins Associates - www.gremlins.biz
Open Source Software Systems Integrators
* Web Design Hosting * Internet Intranet Applications Development *
--
Search the list archives before you post
http://marc.theaimsgroup.com/?l=php-db
--
/*
Every time I think I know where it's at, they move it.
*/
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RE: [PHP-DB] Passing variables between html forms and updating sql table
I need a secure way of updating mysql on the website, phpadmin is not secure enough [snip] $MySQLLink = mysql_pconnect (davecp4, root, ) or die(Could not attach to database. Please try later or contact [EMAIL PROTECTED]); Please tell me you just took the password out of that for posting purposes and root isn't really set up without a password. If root really doesn't need a password, I'd say phpMyadmin is the least of your security concerns. -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] passing variables
That's just it, for some reason the variable isn't being set.
Here is the code:
When the page loads, the form loads under
if (!$login_clients and !$register_clients){
When I click the minus.gif image, I want the page to reload with the other
code but it still loads the form under
if (!$login_clients and !$register_clients){
If I use isset it does the same thing.
Thanks!
Eddie
?
require(template_1.inc);
?
!-- If this page is opened from the template, then show the log-in
screen --
A HREF=login_clients.phpimg src=images\minus.gif
name=register_clients/a
? if (!$login_clients and !$register_clients){
?
center
form name=login method=post action=login_clients.php
TABLE width=480 bgColor=black
TBODY
TR bgColor=#363636
TDFONT face=arial color=#fffafa size=2BUser
Name:/B/font
TDinput name=uname type=text id=uname/td/tr
TR bgColor=#363636
TDFONT face=arial color=#fffafa
size=2BPassword:/B/font
TDinput name=pword type=password id=pword/td
/tr
/table
INPUT type=image src=images/bidsubmit.gif border=0
/form
/center
? }
?
!-- If the user clicks the 'Register' button, show this --
? if ($register_clients){
?
center
form name=nclient method=post action=newclient.php
TABLE width=480 bgColor=black
TR bgColor=#363636
tdFONT face=arial color=#fffafa size=2strongFirst
Name:/strong/font/td
tdinput name=fname type=text id=fname/td
/tr
tr
td FONT face=arial color=#fffafa size=2strongLast Name:
/strong/font/td
tdinput name=lname type=text id=lname/td
/tr
TR bgColor=#363636
td FONT face=arial color=#fffafa
size=2strongAddress:/strong/font/td
tdinput name=address type=text id=address/td
/tr
TR bgColor=#363636
td FONT face=arial color=#fffafa size=2strongCity:
/strong/font/td
tdinput name=city type=text id=city/td
/tr
TR bgColor=#363636
TR bgColor=#363636
td FONT face=arial color=#fffafa size=2strongState:
/strong/font/td
tdinput name=state type=text id=state/td
/tr
TR bgColor=#363636
TR bgColor=#363636
td FONT face=arial color=#fffafa size=2strongZip
/strong/font/td
tdinput name=zip type=text id=zip/td
/tr
TR bgColor=#363636
td FONT face=arial color=#fffafa size=2strongTelephone
Number:/strong/font/td
tdinput name=phone type=text id=phone/td
/tr
TR bgColor=#363636
td FONT face=arial color=#fffafa size=2strongFax
Number:/strong/font/td
tdinput name=fax type=text id=fax/td
/tr
TR bgColor=#363636
td FONT face=arial color=#fffafa size=2stronge-mail
address:/strong/font/td
tdinput name=email type=text id=email/td
/tr
TR bgColor=#363636
td FONT face=arial color=#fffafa size=2strongConfirm
E-mail:/strong/font/td
tdinput name=cemail type=text id=cemail/td
/tr
TR bgColor=#363636
td FONT face=arial color=#fffafa
size=2strongUsername:/strong/font/td
tdinput name=uname type=text id=uname/td
/tr
TR bgColor=#363636
td FONT face=arial color=#fffafa
size=2strongPassword:/strong/font/td
tdinput name=pword type=text id=pword/td
/tr
/table
INPUT type=image src=images/bidsubmit.gif border=0
/form
/center
?
}
?
?
require(template_2.inc);
?
-Original Message-
From: Ryan Jameson (USA) [mailto:[EMAIL PROTECTED]]
Sent: Monday, December 09, 2002 10:09 AM
To: [EMAIL PROTECTED]
Subject: RE: [PHP-DB] passing variables
I think you want:
if (isset($variable)){}
Ryan
-Original Message-
From: Edward Peloke [mailto:[EMAIL PROTECTED]]
Sent: Monday, December 09, 2002 8:33 AM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] passing variables
Hello all,
I have a login/register screen for my php/mysql db. When the page opens,
the user sees the log in screen but I have a button and reloads the same
page, hopefully with the register screen. I know I can simply check to see
if a variable is set with
if ($variable){} . This does not seem to work when I give an image a name
and link it to the same page. The image is named register and it links back
to the same page, so the page loads, I click on the register image , the
page reloads but the variable is not set.
Any ideas?
Thanks,
Eddie
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RE: [PHP-DB] passing variables
You may need to look in the global array. The more recent versions of PHP default to
register_globals off ... which means that form variables do not turn into individual
variables, but they are still in their respective arrays $_SERVER,$_GET,$_POST ...
depending on the method your using it will be in either get or post.
Ryan
-Original Message-
From: Edward Peloke [mailto:[EMAIL PROTECTED]]
Sent: Monday, December 09, 2002 9:03 AM
To: [EMAIL PROTECTED]
Subject: RE: [PHP-DB] passing variables
That's just it, for some reason the variable isn't being set.
Here is the code:
When the page loads, the form loads under
if (!$login_clients and !$register_clients){
When I click the minus.gif image, I want the page to reload with the other
code but it still loads the form under
if (!$login_clients and !$register_clients){
If I use isset it does the same thing.
Thanks!
Eddie
?
require(template_1.inc);
?
!-- If this page is opened from the template, then show the log-in
screen --
A HREF=login_clients.phpimg src=images\minus.gif
name=register_clients/a
? if (!$login_clients and !$register_clients){
?
center
form name=login method=post action=login_clients.php
TABLE width=480 bgColor=black
TBODY
TR bgColor=#363636
TDFONT face=arial color=#fffafa size=2BUser
Name:/B/font
TDinput name=uname type=text id=uname/td/tr
TR bgColor=#363636
TDFONT face=arial color=#fffafa
size=2BPassword:/B/font
TDinput name=pword type=password id=pword/td
/tr
/table
INPUT type=image src=images/bidsubmit.gif border=0
/form
/center
? }
?
!-- If the user clicks the 'Register' button, show this --
? if ($register_clients){
?
center
form name=nclient method=post action=newclient.php
TABLE width=480 bgColor=black
TR bgColor=#363636
tdFONT face=arial color=#fffafa size=2strongFirst
Name:/strong/font/td
tdinput name=fname type=text id=fname/td
/tr
tr
td FONT face=arial color=#fffafa size=2strongLast Name:
/strong/font/td
tdinput name=lname type=text id=lname/td
/tr
TR bgColor=#363636
td FONT face=arial color=#fffafa
size=2strongAddress:/strong/font/td
tdinput name=address type=text id=address/td
/tr
TR bgColor=#363636
td FONT face=arial color=#fffafa size=2strongCity:
/strong/font/td
tdinput name=city type=text id=city/td
/tr
TR bgColor=#363636
TR bgColor=#363636
td FONT face=arial color=#fffafa size=2strongState:
/strong/font/td
tdinput name=state type=text id=state/td
/tr
TR bgColor=#363636
TR bgColor=#363636
td FONT face=arial color=#fffafa size=2strongZip
/strong/font/td
tdinput name=zip type=text id=zip/td
/tr
TR bgColor=#363636
td FONT face=arial color=#fffafa size=2strongTelephone
Number:/strong/font/td
tdinput name=phone type=text id=phone/td
/tr
TR bgColor=#363636
td FONT face=arial color=#fffafa size=2strongFax
Number:/strong/font/td
tdinput name=fax type=text id=fax/td
/tr
TR bgColor=#363636
td FONT face=arial color=#fffafa size=2stronge-mail
address:/strong/font/td
tdinput name=email type=text id=email/td
/tr
TR bgColor=#363636
td FONT face=arial color=#fffafa size=2strongConfirm
E-mail:/strong/font/td
tdinput name=cemail type=text id=cemail/td
/tr
TR bgColor=#363636
td FONT face=arial color=#fffafa
size=2strongUsername:/strong/font/td
tdinput name=uname type=text id=uname/td
/tr
TR bgColor=#363636
td FONT face=arial color=#fffafa
size=2strongPassword:/strong/font/td
tdinput name=pword type=text id=pword/td
/tr
/table
INPUT type=image src=images/bidsubmit.gif border=0
/form
/center
?
}
?
?
require(template_2.inc);
?
-Original Message-
From: Ryan Jameson (USA) [mailto:[EMAIL PROTECTED]]
Sent: Monday, December 09, 2002 10:09 AM
To: [EMAIL PROTECTED]
Subject: RE: [PHP-DB] passing variables
I think you want:
if (isset($variable)){}
Ryan
-Original Message-
From: Edward Peloke [mailto:[EMAIL PROTECTED]]
Sent: Monday, December 09, 2002 8:33 AM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] passing variables
Hello all,
I have a login/register screen for my php/mysql db. When the page opens,
the user sees the log in screen but I have a button and reloads the same
page, hopefully with the register screen. I know I can simply check to see
if a variable is set with
if ($variable){} . This does not seem to work when I give an image a name
and link it to the same page. The image is named register and it links back
to the same page, so the page loads, I click on the register image , the
page reloads but the variable is not set.
Any ideas?
Thanks,
Eddie
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http
RE: [PHP-DB] passing variables
One of many possible things.
Dependent on the version of PHP you are using, you might have
register_globals set to off in your php.ini file, which is preferable.
In which case you'll have to use
If($_REQUEST['variable'])
Or
If(isset($_REQUEST['variable']))
Second possible is:
I've never really used an image within a link to submit a form. What I
usually do is make the image a submit button, e.g.,input type=image
src=path to your image height=XX width=XX border=0 You could
name it or create a hidden field input type=hidden name=register
value=register
Then upon form submission (form action=? Echo $_SERVER['PHP_SELF']
?)
You would look for register
If(isset($_REQUEST['register']))
{
//whatever
}
hope one of them works.
-Original Message-
From: Edward Peloke [mailto:[EMAIL PROTECTED]]
Sent: Monday, December 09, 2002 9:33 AM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] passing variables
Hello all,
I have a login/register screen for my php/mysql db. When the page
opens,
the user sees the log in screen but I have a button and reloads the same
page, hopefully with the register screen. I know I can simply check to
see
if a variable is set with
if ($variable){} . This does not seem to work when I give an image a
name
and link it to the same page. The image is named register and it links
back
to the same page, so the page loads, I click on the register image , the
page reloads but the variable is not set.
Any ideas?
Thanks,
Eddie
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] passing variables
I missed the part where he was using an image. Without a value property, I don't see
how it could pass anything at all
A note on my recent post, to emulate register_globals do this:
if (!empty($_SERVER))
extract($_SERVER);
if (!empty($_GET)) {
extract($_GET);
} else if (!empty($HTTP_GET_VARS)) {
extract($HTTP_GET_VARS);
}
if (!empty($_POST)) {
extract($_POST);
} else if (!empty($HTTP_POST_VARS)) {
extract($HTTP_POST_VARS);
}
This registers all of the different arrays.
Ryan
-Original Message-
From: Jonathan [mailto:[EMAIL PROTECTED]]
Sent: Monday, December 09, 2002 8:40 AM
To: [EMAIL PROTECTED]
Subject: RE: [PHP-DB] passing variables
One of many possible things.
Dependent on the version of PHP you are using, you might have
register_globals set to off in your php.ini file, which is preferable.
In which case you'll have to use
If($_REQUEST['variable'])
Or
If(isset($_REQUEST['variable']))
Second possible is:
I've never really used an image within a link to submit a form. What I
usually do is make the image a submit button, e.g.,input type=image
src=path to your image height=XX width=XX border=0 You could
name it or create a hidden field input type=hidden name=register
value=register
Then upon form submission (form action=? Echo $_SERVER['PHP_SELF']
?)
You would look for register
If(isset($_REQUEST['register']))
{
//whatever
}
hope one of them works.
-Original Message-
From: Edward Peloke [mailto:[EMAIL PROTECTED]]
Sent: Monday, December 09, 2002 9:33 AM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] passing variables
Hello all,
I have a login/register screen for my php/mysql db. When the page
opens,
the user sees the log in screen but I have a button and reloads the same
page, hopefully with the register screen. I know I can simply check to
see
if a variable is set with
if ($variable){} . This does not seem to work when I give an image a
name
and link it to the same page. The image is named register and it links
back
to the same page, so the page loads, I click on the register image , the
page reloads but the variable is not set.
Any ideas?
Thanks,
Eddie
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
--
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To unsubscribe, visit: http://www.php.net/unsub.php
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Re: [PHP-DB] passing variables
On Tuesday 10 December 2002 00:03, Edward Peloke wrote:
That's just it, for some reason the variable isn't being set.
Here is the code:
When the page loads, the form loads under
if (!$login_clients and !$register_clients){
When I click the minus.gif image, I want the page to reload with the other
code but it still loads the form under
if (!$login_clients and !$register_clients){
If I use isset it does the same thing.
Quickest solution, stick a phpinfo() in the first line your code to see
exactly what is and isn't being set.
--
Jason Wong - Gremlins Associates - www.gremlins.biz
Open Source Software Systems Integrators
* Web Design Hosting * Internet Intranet Applications Development *
/*
Did you hear that there's a group of South American Indians that worship
the number zero?
Is nothing sacred?
*/
--
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RE: [PHP-DB] passing variables
Eddie, whenever I start having problems with variables, I drop this code in:
begin code snip...
?
// these lines format the output as HTML comments
// and call dump_array repeatedly
echo \n!-- BEGIN VARIABLE DUMP --\n\n;
echo !-- BEGIN GET VARS --\n;
echo !-- .dump_array($HTTP_GET_VARS). --\n;
echo !-- BEGIN POST VARS --\n;
echo !-- .dump_array($HTTP_POST_VARS). --\n;
echo !-- BEGIN SESSION VARS --\n;
echo !-- .dump_array($HTTP_SESSION_VARS). --\n;
echo !-- BEGIN COOKIE VARS --\n;
echo !-- .dump_array($HTTP_COOKIE_VARS). --\n;
echo \n!-- END VARIABLE DUMP --\n;
// dump_array() takes one array as a parameter
// It iterates through that array, creating a string
// to represent the array as a set
function dump_array($array)
{
if(is_array($array))
{
$size = count($array);
$string = ;
if($size)
{
$count = 0;
$string .= { ;
// add each element's key and value to the string
foreach($array as $var = $value)
{
$string .= $var = '$value';
if($count++ ($size-1))
{
$string .= , ;
}
}
$string .= };
}
return $string;
}
else
{
// if it is not an array, just return it
return $array;
}
}
?
...end code snip
Actually, I have it as a separate file that I include when I need to.
Anyway, it'll list every single variable and array passed to the page along
with its value(s).
It is entirely possible that your variable exists, but does not have a
value. In which case, your
if ($register_clients){
//your HTML stuff here
}
code should evaluate to true because it exists, but doesn't have a value. As
Ryan suggested, the proper thing to do would be to use
if(isset($register_clients)){
//your HTML stuff here
}
That's all assuming you have register_globals set to on. Otherwise, you'll
have to switch to the $_REQUEST['register_clients'] notation (or $_POST or
$_GET, or etc.) if register_globals is set to off.
Either way, try using the code snippet and look at what your values are
actually set to (you'll have to view the source of the page into which you
place the snippet). That may give you some further insight into what's
happening with your script.
Hope this helps and that I haven't misinterpreted your post.
-Original Message-
From: Ryan Jameson (USA) [mailto:[EMAIL PROTECTED]]
Sent: Monday, December 09, 2002 10:44 AM
To: [EMAIL PROTECTED]
Subject: RE: [PHP-DB] passing variables
I missed the part where he was using an image. Without a
value property, I don't see how it could pass anything at all
A note on my recent post, to emulate register_globals do this:
if (!empty($_SERVER))
extract($_SERVER);
if (!empty($_GET)) {
extract($_GET);
} else if (!empty($HTTP_GET_VARS)) {
extract($HTTP_GET_VARS);
}
if (!empty($_POST)) {
extract($_POST);
} else if (!empty($HTTP_POST_VARS)) {
extract($HTTP_POST_VARS);
}
This registers all of the different arrays.
Ryan
-Original Message-
From: Jonathan [mailto:[EMAIL PROTECTED]]
Sent: Monday, December 09, 2002 8:40 AM
To: [EMAIL PROTECTED]
Subject: RE: [PHP-DB] passing variables
One of many possible things.
Dependent on the version of PHP you are using, you might have
register_globals set to off in your php.ini file, which is preferable.
In which case you'll have to use
If($_REQUEST['variable'])
Or
If(isset($_REQUEST['variable']))
Second possible is:
I've never really used an image within a link to submit a
form. What I
usually do is make the image a submit button, e.g.,input type=image
src=path to your image height=XX width=XX border=0 You could
name it or create a hidden field input type=hidden name=register
value=register
Then upon form submission (form action=? Echo $_SERVER['PHP_SELF']
?)
You would look for register
If(isset($_REQUEST['register']))
{
//whatever
}
hope one of them works.
-Original Message-
From: Edward Peloke [mailto:[EMAIL PROTECTED]]
Sent: Monday, December 09, 2002 9:33 AM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] passing variables
Hello all,
I have a login/register screen for my php/mysql db. When the page
opens,
the user sees the log in screen but I have a button and
reloads the same
page, hopefully with the register screen. I know I can
simply check to
see
if a variable is set with
if ($variable){} . This does not seem to work when I give an image a
name
and link it to the same page. The image is named register
and it links
back
to the same page, so the page loads, I click on the register
image , the
page reloads but the variable is not set.
Any ideas?
Thanks,
Eddie
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RE: [PHP-DB] passing variables
Thanks for all the help everyone
-Original Message-
From: Hutchins, Richard [mailto:[EMAIL PROTECTED]]
Sent: Monday, December 09, 2002 11:14 AM
To: [EMAIL PROTECTED]
Cc: '[EMAIL PROTECTED]'
Subject: RE: [PHP-DB] passing variables
Eddie, whenever I start having problems with variables, I drop this code in:
begin code snip...
?
// these lines format the output as HTML comments
// and call dump_array repeatedly
echo \n!-- BEGIN VARIABLE DUMP --\n\n;
echo !-- BEGIN GET VARS --\n;
echo !-- .dump_array($HTTP_GET_VARS). --\n;
echo !-- BEGIN POST VARS --\n;
echo !-- .dump_array($HTTP_POST_VARS). --\n;
echo !-- BEGIN SESSION VARS --\n;
echo !-- .dump_array($HTTP_SESSION_VARS). --\n;
echo !-- BEGIN COOKIE VARS --\n;
echo !-- .dump_array($HTTP_COOKIE_VARS). --\n;
echo \n!-- END VARIABLE DUMP --\n;
// dump_array() takes one array as a parameter
// It iterates through that array, creating a string
// to represent the array as a set
function dump_array($array)
{
if(is_array($array))
{
$size = count($array);
$string = ;
if($size)
{
$count = 0;
$string .= { ;
// add each element's key and value to the string
foreach($array as $var = $value)
{
$string .= $var = '$value';
if($count++ ($size-1))
{
$string .= , ;
}
}
$string .= };
}
return $string;
}
else
{
// if it is not an array, just return it
return $array;
}
}
?
...end code snip
Actually, I have it as a separate file that I include when I need to.
Anyway, it'll list every single variable and array passed to the page along
with its value(s).
It is entirely possible that your variable exists, but does not have a
value. In which case, your
if ($register_clients){
//your HTML stuff here
}
code should evaluate to true because it exists, but doesn't have a value. As
Ryan suggested, the proper thing to do would be to use
if(isset($register_clients)){
//your HTML stuff here
}
That's all assuming you have register_globals set to on. Otherwise, you'll
have to switch to the $_REQUEST['register_clients'] notation (or $_POST or
$_GET, or etc.) if register_globals is set to off.
Either way, try using the code snippet and look at what your values are
actually set to (you'll have to view the source of the page into which you
place the snippet). That may give you some further insight into what's
happening with your script.
Hope this helps and that I haven't misinterpreted your post.
-Original Message-
From: Ryan Jameson (USA) [mailto:[EMAIL PROTECTED]]
Sent: Monday, December 09, 2002 10:44 AM
To: [EMAIL PROTECTED]
Subject: RE: [PHP-DB] passing variables
I missed the part where he was using an image. Without a
value property, I don't see how it could pass anything at all
A note on my recent post, to emulate register_globals do this:
if (!empty($_SERVER))
extract($_SERVER);
if (!empty($_GET)) {
extract($_GET);
} else if (!empty($HTTP_GET_VARS)) {
extract($HTTP_GET_VARS);
}
if (!empty($_POST)) {
extract($_POST);
} else if (!empty($HTTP_POST_VARS)) {
extract($HTTP_POST_VARS);
}
This registers all of the different arrays.
Ryan
-Original Message-
From: Jonathan [mailto:[EMAIL PROTECTED]]
Sent: Monday, December 09, 2002 8:40 AM
To: [EMAIL PROTECTED]
Subject: RE: [PHP-DB] passing variables
One of many possible things.
Dependent on the version of PHP you are using, you might have
register_globals set to off in your php.ini file, which is preferable.
In which case you'll have to use
If($_REQUEST['variable'])
Or
If(isset($_REQUEST['variable']))
Second possible is:
I've never really used an image within a link to submit a
form. What I
usually do is make the image a submit button, e.g.,input type=image
src=path to your image height=XX width=XX border=0 You could
name it or create a hidden field input type=hidden name=register
value=register
Then upon form submission (form action=? Echo $_SERVER['PHP_SELF']
?)
You would look for register
If(isset($_REQUEST['register']))
{
//whatever
}
hope one of them works.
-Original Message-
From: Edward Peloke [mailto:[EMAIL PROTECTED]]
Sent: Monday, December 09, 2002 9:33 AM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] passing variables
Hello all,
I have a login/register screen for my php/mysql db. When the page
opens,
the user sees the log in screen but I have a button and
reloads the same
page, hopefully with the register screen. I know I can
simply check to
see
if a variable is set with
if ($variable){} . This does not seem to work when I give an image a
name
and link it to the same page. The image is named register
and it links
back
to the same page, so the page loads, I click on the register
image , the
page reloads but the variable is not set.
Any ideas?
Thanks,
Eddie
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Re: [PHP-DB] passing variables
www.php.net/isset
if !isset($var)
{
echo var is not set to anything;
}
else
{
echo var is set to $var;
}
On Mon, 9 Dec 2002, Edward Peloke wrote:
Hello all,
I have a login/register screen for my php/mysql db. When the page opens,
the user sees the log in screen but I have a button and reloads the same
page, hopefully with the register screen. I know I can simply check to see
if a variable is set with
if ($variable){} . This does not seem to work when I give an image a name
and link it to the same page. The image is named register and it links back
to the same page, so the page loads, I click on the register image , the
page reloads but the variable is not set.
Any ideas?
Thanks,
Eddie
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RE: [PHP-DB] passing variables
--- Ryan Jameson (USA) [EMAIL PROTECTED] wrote:
You are absolutely correct. However, the only value that security
feature has is in the case that the program really cares which
method the variable was received. I never have, and I doubt I ever
will. I have had cases where I thought I'd be concerned with this
but a rethink of the logic proved there was a better way. Even if I
am concerned with such an issue I can always check the post array
to make sure it is there.
The decision to default to 'off' was a good one, but it only
protects certain types of programmers from accidentally creating
holes. I am a bit more deliberate, and see no security value in it
for myself. Therefore my installations remain
register_globals=on...
If you'd like to pass your username and password on a query string
be my guest, it'll work just fine. I don't recommend it though.
Actually, it's more than just checking for variables you expect.
Let's say you have a variable in your script that is set before user
data is parsed. Then you use extract() to pull all the variables out
of the superglobals. If someone attaches a variable to a GET with the
same name as a variable you've already set, then it will overwrite
your variable if you have register_globals on or simply extract() all
superglobals.
A **trivial** example:
?
$var1=Ryan;
// some set of PHP code
extract($HTTP_GET_VARS); // or $_GET
// more PHP code
echo Hello .$var1;
?
If I add ?var1=Mark to the URL for this page, It will respond with
Hello Mark, not Hello Ryan as expected. Obviously I've
oversimplified this. But I believe a big part of the the point of the
superglobals was to eliminate the ability for a malicious user to
overwrite values that the programmer didn't want to let them set.
I don't mean to get into a debate over the value of register_globals
being turned off. I do agree that for some it has value and for
others it doesn't. I don't think being deliberate is the only test
for whether there's use for it.
Mark
Ryan
-Original Message-
From: Mark [mailto:[EMAIL PROTECTED]]
Sent: Monday, December 09, 2002 10:51 AM
To: Ryan Jameson (USA); [EMAIL PROTECTED]
Subject: RE: [PHP-DB] passing variables
--- Ryan Jameson (USA) [EMAIL PROTECTED] wrote:
I missed the part where he was using an image. Without a value
property, I don't see how it could pass anything at all
A note on my recent post, to emulate register_globals do this:
if (!empty($_SERVER))
extract($_SERVER);
if (!empty($_GET)) {
extract($_GET);
} else if (!empty($HTTP_GET_VARS)) {
extract($HTTP_GET_VARS);
}
if (!empty($_POST)) {
extract($_POST);
} else if (!empty($HTTP_POST_VARS)) {
extract($HTTP_POST_VARS);
}
This registers all of the different arrays.
And completely nullifies the security value of having
register_globals turned off. But I guess if you don't have access
to
the php.ini file this is as good...
Ryan
=
Mark Weinstock
[EMAIL PROTECTED]
***
You can't demand something as a right unless you are willing to
fight to death to defend everyone else's right to the same thing.
-Stolen from the now-defunct Randy's Random mailing list.
***
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=
Mark Weinstock
[EMAIL PROTECTED]
***
You can't demand something as a right unless you are willing to fight to death to
defend everyone else's right to the same thing.
-Stolen from the now-defunct Randy's Random mailing list.
***
__
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Re: [PHP-DB] passing variables from one page to another
If that is your exact code, then you need to have an ending quote after the query is
written. Like this:
$query = Select * from course, disc, instit, prof where course.CourseID =
$CourseID and course.DiscID = disc.DiscID and course.InstitID =
instit.InstitID and course.ProfID = prof.ProfID;
Martin
Matthew K. Gold [EMAIL PROTECTED] 07/08/02 03:40PM
Hi Everybody,
Please forgive the basic nature of this question--I have looked at the
manual, but I found that much of it went over my head.
I'm trying to create a second level of a website that displays course
listings. The first level lists a bunch of courses. I'd like users to be
able to click on the title of a course to go to a page that contains details
about that course.
From what I understand, I should do this by making the title of the course
on the first page (courses.php) a link to the second page (courseinfo.php)
with a query string attached--so that the link would look like
a href=courseinfo.php?CourseID=12Accounting 101/a
What I'm having trouble doing is coding the second page. Here's what I've
done. When I run this, I get a blank page in return...if anyone can help, I
would greatly appreciate it. Thanks in advance.
Matt
?php
$db = @mysql_connect(host,user,pword);
mysql_select_db(dln, $db);
if ( !$CourseID ) {
print (no course id included)
exit;
};
$query = Select * from course, disc, instit, prof where course.CourseID =
$CourseID and course.DiscID = disc.DiscID and course.InstitID =
instit.InstitID and course.ProfID = prof.ProfID;
$result = mysql_query ($query);
echo MySQL error number . mysql_errno() . : . mysql_error();
print (table border=\0\ cellpadding=\5\ class=\default\\n);
print (tr
bgcolor=\#ff9900\th$CourseID/th/trtrtd$CourseTitle/td/tr/t
able);
?
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Re: [PHP-DB] passing variables from one page to another
You're right--I did leave out that end quotation mark. Unfortunately,
though, I'm still getting the same result after adding it.
best,
Matt
- Original Message -
From: Martin Clifford [EMAIL PROTECTED]
To: [EMAIL PROTECTED]; [EMAIL PROTECTED]
Sent: Monday, July 08, 2002 3:47 PM
Subject: Re: [PHP-DB] passing variables from one page to another
If that is your exact code, then you need to have an ending quote after the
query is written. Like this:
$query = Select * from course, disc, instit, prof where course.CourseID =
$CourseID and course.DiscID = disc.DiscID and course.InstitID =
instit.InstitID and course.ProfID = prof.ProfID;
Martin
Matthew K. Gold [EMAIL PROTECTED] 07/08/02 03:40PM
Hi Everybody,
Please forgive the basic nature of this question--I have looked at the
manual, but I found that much of it went over my head.
I'm trying to create a second level of a website that displays course
listings. The first level lists a bunch of courses. I'd like users to be
able to click on the title of a course to go to a page that contains details
about that course.
From what I understand, I should do this by making the title of the course
on the first page (courses.php) a link to the second page (courseinfo.php)
with a query string attached--so that the link would look like
a href=courseinfo.php?CourseID=12Accounting 101/a
What I'm having trouble doing is coding the second page. Here's what I've
done. When I run this, I get a blank page in return...if anyone can help, I
would greatly appreciate it. Thanks in advance.
Matt
?php
$db = @mysql_connect(host,user,pword);
mysql_select_db(dln, $db);
if ( !$CourseID ) {
print (no course id included)
exit;
};
$query = Select * from course, disc, instit, prof where course.CourseID =
$CourseID and course.DiscID = disc.DiscID and course.InstitID =
instit.InstitID and course.ProfID = prof.ProfID;
$result = mysql_query ($query);
echo MySQL error number . mysql_errno() . : . mysql_error();
print (table border=\0\ cellpadding=\5\ class=\default\\n);
print (tr
bgcolor=\#ff9900\th$CourseID/th/trtrtd$CourseTitle/td/tr/t
able);
?
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Re: [PHP-DB] passing variables from one page to another
Hello!!
How are you fetching the rows???
mysql_fetch_rows?? or mysql_fetch_object???
Dan
On Monday, July 8, 2002, at 02:52 PM, [EMAIL PROTECTED] wrote:
You're right--I did leave out that end quotation mark. Unfortunately,
though, I'm still getting the same result after adding it.
best,
Matt
- Original Message -
From: Martin Clifford [EMAIL PROTECTED]
To: [EMAIL PROTECTED]; [EMAIL PROTECTED]
Sent: Monday, July 08, 2002 3:47 PM
Subject: Re: [PHP-DB] passing variables from one page to another
If that is your exact code, then you need to have an ending quote after
the
query is written. Like this:
$query = Select * from course, disc, instit, prof where
course.CourseID =
$CourseID and course.DiscID = disc.DiscID and course.InstitID =
instit.InstitID and course.ProfID = prof.ProfID;
Martin
Matthew K. Gold [EMAIL PROTECTED] 07/08/02 03:40PM
Hi Everybody,
Please forgive the basic nature of this question--I have looked at the
manual, but I found that much of it went over my head.
I'm trying to create a second level of a website that displays course
listings. The first level lists a bunch of courses. I'd like users to
be
able to click on the title of a course to go to a page that contains
details
about that course.
From what I understand, I should do this by making the title of the
course
on the first page (courses.php) a link to the second page
(courseinfo.php)
with a query string attached--so that the link would look like
a href=courseinfo.php?CourseID=12Accounting 101/a
What I'm having trouble doing is coding the second page. Here's what
I've
done. When I run this, I get a blank page in return...if anyone can
help, I
would greatly appreciate it. Thanks in advance.
Matt
?php
$db = @mysql_connect(host,user,pword);
mysql_select_db(dln, $db);
if ( !$CourseID ) {
print (no course id included)
exit;
};
$query = Select * from course, disc, instit, prof where
course.CourseID =
$CourseID and course.DiscID = disc.DiscID and course.InstitID =
instit.InstitID and course.ProfID = prof.ProfID;
$result = mysql_query ($query);
echo MySQL error number . mysql_errno() . : . mysql_error();
print (table border=\0\ cellpadding=\5\ class=\default\\n);
print (tr
bgcolor=\#ff9900\th$CourseID/th/trtrtd$CourseTitle/td/tr
/t
able);
?
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Re: [PHP-DB] passing variables from one page to another
ummm:).I'm now using fetch_rows, though I wasn't using anything
before. Still, it's not working. Here's the new script (thanks again for
your help):
?php
$db = @mysql_connect(host,user,pword);
mysql_select_db(dln, $db);
if ( !$CourseID ) {
print (no course id included)
exit
};
$query = Select * from course, disc, instit, prof where course.CourseID =
$CourseID and course.DiscID = disc.DiscID and course.InstitID =
instit.InstitID and course.ProfID = prof.ProfID;
$result = mysql_query ($query);
echo MySQL error number . mysql_errno() . : . mysql_error();
print (table border=\0\ cellpadding=\5\ class=\default\\n);
while ($row = mysql_fetch_row ($result))
{
print (tr
bgcolor=\#ff9900\th$CourseID/th/trtrtd$CourseTitle/td/tr);
}
print (/table);
?
- Original Message -
From: Daniel Brunner [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED]
Sent: Monday, July 08, 2002 4:04 PM
Subject: Re: [PHP-DB] passing variables from one page to another
Hello!!
How are you fetching the rows???
mysql_fetch_rows?? or mysql_fetch_object???
Dan
On Monday, July 8, 2002, at 02:52 PM, [EMAIL PROTECTED] wrote:
You're right--I did leave out that end quotation mark. Unfortunately,
though, I'm still getting the same result after adding it.
best,
Matt
- Original Message -
From: Martin Clifford [EMAIL PROTECTED]
To: [EMAIL PROTECTED]; [EMAIL PROTECTED]
Sent: Monday, July 08, 2002 3:47 PM
Subject: Re: [PHP-DB] passing variables from one page to another
If that is your exact code, then you need to have an ending quote after
the
query is written. Like this:
$query = Select * from course, disc, instit, prof where
course.CourseID =
$CourseID and course.DiscID = disc.DiscID and course.InstitID =
instit.InstitID and course.ProfID = prof.ProfID;
Martin
Matthew K. Gold [EMAIL PROTECTED] 07/08/02 03:40PM
Hi Everybody,
Please forgive the basic nature of this question--I have looked at the
manual, but I found that much of it went over my head.
I'm trying to create a second level of a website that displays course
listings. The first level lists a bunch of courses. I'd like users to
be
able to click on the title of a course to go to a page that contains
details
about that course.
From what I understand, I should do this by making the title of the
course
on the first page (courses.php) a link to the second page
(courseinfo.php)
with a query string attached--so that the link would look like
a href=courseinfo.php?CourseID=12Accounting 101/a
What I'm having trouble doing is coding the second page. Here's what
I've
done. When I run this, I get a blank page in return...if anyone can
help, I
would greatly appreciate it. Thanks in advance.
Matt
?php
$db = @mysql_connect(host,user,pword);
mysql_select_db(dln, $db);
if ( !$CourseID ) {
print (no course id included)
exit;
};
$query = Select * from course, disc, instit, prof where
course.CourseID =
$CourseID and course.DiscID = disc.DiscID and course.InstitID =
instit.InstitID and course.ProfID = prof.ProfID;
$result = mysql_query ($query);
echo MySQL error number . mysql_errno() . : . mysql_error();
print (table border=\0\ cellpadding=\5\ class=\default\\n);
print (tr
bgcolor=\#ff9900\th$CourseID/th/trtrtd$CourseTitle/td/tr
/t
able);
?
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RE: [PHP-DB] passing variables from one page to another
Matthew, have you tried running your query from the command line in MySQL?
If you can do that successfully, that'll prove that the query is functioning
properly and the problem can be tracked elsewhere.
-Original Message-
From: Matthew K. Gold [mailto:[EMAIL PROTECTED]]
Sent: Monday, July 08, 2002 3:40 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] passing variables from one page to another
Hi Everybody,
Please forgive the basic nature of this question--I have looked at the
manual, but I found that much of it went over my head.
I'm trying to create a second level of a website that displays course
listings. The first level lists a bunch of courses. I'd like users to be
able to click on the title of a course to go to a page that contains details
about that course.
From what I understand, I should do this by making the title of the course
on the first page (courses.php) a link to the second page (courseinfo.php)
with a query string attached--so that the link would look like
a href=courseinfo.php?CourseID=12Accounting 101/a
What I'm having trouble doing is coding the second page. Here's what I've
done. When I run this, I get a blank page in return...if anyone can help, I
would greatly appreciate it. Thanks in advance.
Matt
?php
$db = @mysql_connect(host,user,pword);
mysql_select_db(dln, $db);
if ( !$CourseID ) {
print (no course id included)
exit;
};
$query = Select * from course, disc, instit, prof where course.CourseID =
$CourseID and course.DiscID = disc.DiscID and course.InstitID =
instit.InstitID and course.ProfID = prof.ProfID;
$result = mysql_query ($query);
echo MySQL error number . mysql_errno() . : . mysql_error();
print (table border=\0\ cellpadding=\5\ class=\default\\n);
print (tr
bgcolor=\#ff9900\th$CourseID/th/trtrtd$CourseTitle/td/tr/t
able);
?
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Re: [PHP-DB] passing variables from one page to another
It works from the command line if I leave out course.CourseID = $CourseID
in the where line of the select clause. I assume that I'd have to leave
that out when working from the command line because it's defined in the php
file...
But I think that that is where my problem is. As I've mentioned, I'm very
new at this, so the answer could be very basic.
thanks,
Matt
- Original Message -
From: Hutchins, Richard [EMAIL PROTECTED]
To: 'Matthew K. Gold' [EMAIL PROTECTED]; [EMAIL PROTECTED]
Sent: Monday, July 08, 2002 4:35 PM
Subject: RE: [PHP-DB] passing variables from one page to another
Matthew, have you tried running your query from the command line in MySQL?
If you can do that successfully, that'll prove that the query is
functioning
properly and the problem can be tracked elsewhere.
-Original Message-
From: Matthew K. Gold [mailto:[EMAIL PROTECTED]]
Sent: Monday, July 08, 2002 3:40 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] passing variables from one page to another
Hi Everybody,
Please forgive the basic nature of this question--I have looked at the
manual, but I found that much of it went over my head.
I'm trying to create a second level of a website that displays course
listings. The first level lists a bunch of courses. I'd like users to be
able to click on the title of a course to go to a page that contains
details
about that course.
From what I understand, I should do this by making the title of the course
on the first page (courses.php) a link to the second page (courseinfo.php)
with a query string attached--so that the link would look like
a href=courseinfo.php?CourseID=12Accounting 101/a
What I'm having trouble doing is coding the second page. Here's what I've
done. When I run this, I get a blank page in return...if anyone can help,
I
would greatly appreciate it. Thanks in advance.
Matt
?php
$db = @mysql_connect(host,user,pword);
mysql_select_db(dln, $db);
if ( !$CourseID ) {
print (no course id included)
exit;
};
$query = Select * from course, disc, instit, prof where course.CourseID =
$CourseID and course.DiscID = disc.DiscID and course.InstitID =
instit.InstitID and course.ProfID = prof.ProfID;
$result = mysql_query ($query);
echo MySQL error number . mysql_errno() . : . mysql_error();
print (table border=\0\ cellpadding=\5\ class=\default\\n);
print (tr
bgcolor=\#ff9900\th$CourseID/th/trtrtd$CourseTitle/td/tr/t
able);
?
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RE: [PHP-DB] passing variables from one page to another
Just wanted to eliminate one possible problem area. If the query works, then
the problem must be in the PHP or the HTML. What do you see if you right
click and view source on the blank page you get? If you see something there,
copy it into a reply. Might be nothing important, but it'll show us what's
being parsed by the browser.
-Original Message-
From: Matthew K. Gold [mailto:[EMAIL PROTECTED]]
Sent: Monday, July 08, 2002 5:07 PM
To: Hutchins, Richard; [EMAIL PROTECTED]
Subject: Re: [PHP-DB] passing variables from one page to another
It works from the command line if I leave out course.CourseID = $CourseID
in the where line of the select clause. I assume that I'd have to leave
that out when working from the command line because it's defined in the php
file...
But I think that that is where my problem is. As I've mentioned, I'm very
new at this, so the answer could be very basic.
thanks,
Matt
- Original Message -
From: Hutchins, Richard [EMAIL PROTECTED]
To: 'Matthew K. Gold' [EMAIL PROTECTED]; [EMAIL PROTECTED]
Sent: Monday, July 08, 2002 4:35 PM
Subject: RE: [PHP-DB] passing variables from one page to another
Matthew, have you tried running your query from the command line in MySQL?
If you can do that successfully, that'll prove that the query is
functioning
properly and the problem can be tracked elsewhere.
-Original Message-
From: Matthew K. Gold [mailto:[EMAIL PROTECTED]]
Sent: Monday, July 08, 2002 3:40 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] passing variables from one page to another
Hi Everybody,
Please forgive the basic nature of this question--I have looked at the
manual, but I found that much of it went over my head.
I'm trying to create a second level of a website that displays course
listings. The first level lists a bunch of courses. I'd like users to be
able to click on the title of a course to go to a page that contains
details
about that course.
From what I understand, I should do this by making the title of the course
on the first page (courses.php) a link to the second page (courseinfo.php)
with a query string attached--so that the link would look like
a href=courseinfo.php?CourseID=12Accounting 101/a
What I'm having trouble doing is coding the second page. Here's what I've
done. When I run this, I get a blank page in return...if anyone can help,
I
would greatly appreciate it. Thanks in advance.
Matt
?php
$db = @mysql_connect(host,user,pword);
mysql_select_db(dln, $db);
if ( !$CourseID ) {
print (no course id included)
exit;
};
$query = Select * from course, disc, instit, prof where course.CourseID =
$CourseID and course.DiscID = disc.DiscID and course.InstitID =
instit.InstitID and course.ProfID = prof.ProfID;
$result = mysql_query ($query);
echo MySQL error number . mysql_errno() . : . mysql_error();
print (table border=\0\ cellpadding=\5\ class=\default\\n);
print (tr
bgcolor=\#ff9900\th$CourseID/th/trtrtd$CourseTitle/td/tr/t
able);
?
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Re: [PHP-DB] passing variables from one page to another
Thanks for the response. Here's what I see in the source code:
!DOCTYPE HTML PUBLIC -//W3C//DTD HTML 4.0 Transitional//EN
HTMLHEAD
META content=text/html; charset=windows-1252
http-equiv=Content-Type/HEAD
BODY/BODY/HTML
I'm not actually testing the link--I'm just typing the following url into
the browser: courseinfo.php?CourseID=12
Could that be the source of the problem?
thanks,
Matt
- Original Message -
From: Rich Hutchins [EMAIL PROTECTED]
To: Matthew K. Gold [EMAIL PROTECTED]; [EMAIL PROTECTED]
Sent: Monday, July 08, 2002 5:39 PM
Subject: RE: [PHP-DB] passing variables from one page to another
Just wanted to eliminate one possible problem area. If the query works,
then
the problem must be in the PHP or the HTML. What do you see if you right
click and view source on the blank page you get? If you see something
there,
copy it into a reply. Might be nothing important, but it'll show us what's
being parsed by the browser.
-Original Message-
From: Matthew K. Gold [mailto:[EMAIL PROTECTED]]
Sent: Monday, July 08, 2002 5:07 PM
To: Hutchins, Richard; [EMAIL PROTECTED]
Subject: Re: [PHP-DB] passing variables from one page to another
It works from the command line if I leave out course.CourseID = $CourseID
in the where line of the select clause. I assume that I'd have to leave
that out when working from the command line because it's defined in the
php
file...
But I think that that is where my problem is. As I've mentioned, I'm very
new at this, so the answer could be very basic.
thanks,
Matt
- Original Message -
From: Hutchins, Richard [EMAIL PROTECTED]
To: 'Matthew K. Gold' [EMAIL PROTECTED]; [EMAIL PROTECTED]
Sent: Monday, July 08, 2002 4:35 PM
Subject: RE: [PHP-DB] passing variables from one page to another
Matthew, have you tried running your query from the command line in MySQ
L?
If you can do that successfully, that'll prove that the query is
functioning
properly and the problem can be tracked elsewhere.
-Original Message-
From: Matthew K. Gold [mailto:[EMAIL PROTECTED]]
Sent: Monday, July 08, 2002 3:40 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] passing variables from one page to another
Hi Everybody,
Please forgive the basic nature of this question--I have looked at the
manual, but I found that much of it went over my head.
I'm trying to create a second level of a website that displays course
listings. The first level lists a bunch of courses. I'd like users to
be
able to click on the title of a course to go to a page that contains
details
about that course.
From what I understand, I should do this by making the title of the
course
on the first page (courses.php) a link to the second page
(courseinfo.php)
with a query string attached--so that the link would look like
a href=courseinfo.php?CourseID=12Accounting 101/a
What I'm having trouble doing is coding the second page. Here's what
I've
done. When I run this, I get a blank page in return...if anyone can
help,
I
would greatly appreciate it. Thanks in advance.
Matt
?php
$db = @mysql_connect(host,user,pword);
mysql_select_db(dln, $db);
if ( !$CourseID ) {
print (no course id included)
exit;
};
$query = Select * from course, disc, instit, prof where course.CourseID
=
$CourseID and course.DiscID = disc.DiscID and course.InstitID =
instit.InstitID and course.ProfID = prof.ProfID;
$result = mysql_query ($query);
echo MySQL error number . mysql_errno() . : . mysql_error();
print (table border=\0\ cellpadding=\5\ class=\default\\n);
print (tr
bgcolor=\#ff9900\th$CourseID/th/trtrtd$CourseTitle/td/tr/t
able);
?
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Re: [PHP-DB] Passing variables to/from Flash
Yeah, it really worked, only the char I had to put it at the beginning of the string too and not only before the second and every consecutive name-value pair. So the right string I had to output was: var1=value1var2=value2var3=value3 Thank you for your help! Atanas Vassilev [EMAIL PROTECTED] [EMAIL PROTECTED] wrote in message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... afaik it's something like: var1=value1var2=value2var3=value3... -Original Message- From: Atanas Vassilev [mailto:[EMAIL PROTECTED]] Sent: Dienstag, 15. Mai 2001 12:31 To: [EMAIL PROTECTED] Subject: [PHP-DB] Passing variables to/from Flash I'm trying to send variables to a flash movie... In fact the designer of the flash part of the site told me that ActionScript had a built in function getData that had one of its arguments the url from where the data should be pulled out - if he points his function to a *.php page in what form should I output the data within this php script - is it a urlencoded string or can it be an array, or just outputting name-value pairs on separate lines would be enough? Any help will be appreciated. Thanks in advance! -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED] -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED] -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
RE: [PHP-DB] Passing variables to/from Flash
afaik it's something like: var1=value1var2=value2var3=value3... -Original Message- From: Atanas Vassilev [mailto:[EMAIL PROTECTED]] Sent: Dienstag, 15. Mai 2001 12:31 To: [EMAIL PROTECTED] Subject: [PHP-DB] Passing variables to/from Flash I'm trying to send variables to a flash movie... In fact the designer of the flash part of the site told me that ActionScript had a built in function getData that had one of its arguments the url from where the data should be pulled out - if he points his function to a *.php page in what form should I output the data within this php script - is it a urlencoded string or can it be an array, or just outputting name-value pairs on separate lines would be enough? Any help will be appreciated. Thanks in advance! -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED] -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
