If no connection is made, there is no mysql error. You would need some
custom error message output when the mysql_connect fails...
Bastien
From: Aarno Syvänen [EMAIL PROTECTED]
To: David Robley [EMAIL PROTECTED]
CC: php-db@lists.php.net
Subject: Re: [PHP-DB] Problem with creating parallel
actually, you may want to check the source of the page. if the url does not
start with http://; (or https, or ftp, etc.), browsers will assume the link is
on the current server. how are you storing the urls? as full urls, as
domain/path/file.htm, ? if all of your urls should start with
On 02 December 2005 13:06, Bastien Koert wrote:
?php for ($j=ord('A'); $j = ord('Z'); $j++) {
echo | ba
href='alpha.php?artist=.chr($j).'.chr($j)./a/b ;
} ?
You need to use the ORD function to get the numerical ascii
equivalent of the letter and the CHR function to go back the
I am getting link from visitor using form, that means I don't have control
what they would type. e.g they may type
http://www.suggestedlink.com/myfav.wav, www.suggestedlink.com/myfav.ram and
http://suggestedlink.com/myfav.mp3 or so I am looking general version which
can handle all.
and on top of
you should then check the strings provided by your users. you can do this
before or after the link is added to your database.
use a regex to verify that the url provided is a full url (starting with
'http://', etc.). if it is not, then prepend 'http://'. i would recommend
doing this before
$url = str_replace(www.mydomain.com,,$url)
bastien
From: Mohamed Yusuf [EMAIL PROTECTED]
To: php-db@lists.php.net
Subject: [PHP-DB] problem of retrieving urls from mysql
Date: Tue, 6 Dec 2005 09:51:18 -0800
I would like to store and retrieve urls, but I have problem which is I get
my url +
?php for ($j=ord('A'); $j = ord('Z'); $j++) {
echo | ba href='alpha.php?artist=.chr($j).'.chr($j)./a/b ;
} ?
You need to use the ORD function to get the numerical ascii equivalent of
the letter and the CHR function to go back the other way.
Bastien
From: Mohamed Yusuf [EMAIL
Using the code below, when you click on the link for A it will go to
alpha.php?artist=A
In alpha.php, you should be able to query $_GET['artist'] to get the
letter clicked on. You could also check $_REQUEST['artist'] for the same
info.
Is that what you needed or are you having other
Check your permissions on the folder...you may need to make them 777 to
allow read/write by all for the upload
Bastien
From: Mohamed Yusuf [EMAIL PROTECTED]
To: php-db@lists.php.net
Subject: [PHP-DB] Problem of Upload media file
Date: Thu, 17 Nov 2005 08:58:15 -0800
I am trying to upload
the permissions are right, I mean they are all setted to 777.
On 11/17/05, Bastien Koert [EMAIL PROTECTED] wrote:
Check your permissions on the folder...you may need to make them 777 to
allow read/write by all for the upload
Bastien
From: Mohamed Yusuf [EMAIL PROTECTED]
To:
Keep in mind you're referring to a filesystem path, and a relative one to
boot, so what you're telling php is, that your upload directory is:
/home/somally/public_html/music/music/media/
Which doesn't sound like what you intend.
Also don't trust the php line numbers if you have significant
here is the complete code, so you can check it
?
$uploadDir = 'www/media/';
if(isset($_POST['upload']))
{
$fileName = $_FILES['userfile']['name'];
$tmpName = $_FILES['userfile']['tmp_name'];
$fileSize = $_FILES['userfile']['size'];
$fileType =
Well, the error doesn't match the code, but the code says to upload the file
to:
/home/somally/public_html/music/www/media/
Is that correct?
On Thursday 17 November 2005 9:41 am, Mohamed Yusuf wrote:
here is the complete code, so you can check it
so should I change the directory?
where I want to upload file is
www.somalilyrics.com/music/media/http://www.somalilyrics.com/music/media/,
therefore what about if I say
*$uploadDir = 'music/media/'; *is that fine?
if it is not fine would you please show me how I have to do it?
On 11/17/05,
No. It's not okay.. you're already in the music directory.. if you want it in
the media subdir, just put 'media/'
On Thursday 17 November 2005 10:40 am, Mohamed Yusuf wrote:
so should I change the directory?
where I want to upload file is
Many thanks for your help. it works.
On 11/17/05, Micah Stevens [EMAIL PROTECTED] wrote:
No. It's not okay.. you're already in the music directory.. if you want it
in
the media subdir, just put 'media/'
On Thursday 17 November 2005 10:40 am, Mohamed Yusuf wrote:
so should I change the
brError Message: . @mysql_error
the '@' symbol suppresses the error message...remove it
bastien
From: Jason Gerfen [EMAIL PROTECTED]
To: php-db@lists.php.net
Subject: [PHP-DB] problem with mysql_error()
Date: Thu, 27 Oct 2005 08:43:13 -0600
I am not sure why this is not returning any
As far as I know the Unknown error related to the mail function
indicates that PHP could not connect to the SMTP server specified in
php.ini.
You could load phpinfo(); and check to see what php.ini file you are
using. Then check that for the SMTP setting.
FYI: doing a google search on 'PHP
I'm not sure what should be in the php.ini.
Where can I find out what needs to be in the SMTP section?
-Original Message-
From: Chris [mailto:[EMAIL PROTECTED]
Sent: Sonntag, 18. September 2005 05:10
To: php-db@lists.php.net
Subject: Re: [PHP-DB] Problem with mail funktion
As far as I
: [PHP-DB] Problem with mail funktion
As far as I know the Unknown error related to the mail function
indicates that PHP could not connect to the SMTP server specified in
php.ini.
You could load phpinfo(); and check to see what php.ini file you are
using. Then check that for the SMTP setting
2005 05:10
To: php-db@lists.php.net
Subject: Re: [PHP-DB] Problem with mail funktion
As far as I know the Unknown error related to the mail function
indicates that PHP could not connect to the SMTP server specified in
php.ini.
You could load phpinfo(); and check to see what php.ini file you
Am Do, den 23.06.2005 schrieb Bomgardner, Mark A um 22:44:
I upgrade the php-mysql extension from 4.3.2-8 to 4.3.2-23, which is the
most current rpm and still no luck. I now get the message that it can't
find mysqli extension.
mysqli extension is part of PHP5, not of PHP4.
/Georg
--
Shawn Singh wrote:
that was very helpful...Thank you. One question I have is that I want
to ensure that my admin page cannot get accessed unless a variable
that was registered upon a successful login has been passed into the
session...what can I do to ensure this?
There are several ways to do
From the PHP help page on session_register()
If your script uses session_register(), it will not work in
environments where the PHP directive register_globals is disabled.
I'm assuming since you compiled and installed PHP 5.0.4 that your
register_globals is disabled. I wouldn't recommend
The browser has already sent headers on line 13 of your code- line 25 must be
the session_start - it has to come first and be at the very top of your code
Shawn Singh [EMAIL PROTECTED] 05/04/05 03:13PM
Hey All,
I'm fairly new to PHP Programming. I have compiled and installed
postgres version
Why dont'n you use soma classes from www.phpclasses.com about User
Management!!
There are great classes in this site!!
-Original Message-
From: Shawn Singh [mailto:[EMAIL PROTECTED]
Sent: Miércoles, 04 de Mayo de 2005 03:14 p.m.
To: php-db@lists.php.net
Subject: [PHP-DB] Problem Using
that was very helpful...Thank you. One question I have is that I want
to ensure that my admin page cannot get accessed unless a variable
that was registered upon a successful login has been passed into the
session...what can I do to ensure this?
Thank you,
Shawn
On 5/4/05, Patel, Aman [EMAIL
From: NIPP, SCOTT V (SBCSI) [mailto:[EMAIL PROTECTED]
I am attempting to use a simple foreach loop on a query result,
and I am only getting the first element of the array back.
Here is the
code...
$exclude_query = SELECT hostname FROM exclusion;
$exclude_results =
Try this
$exclude_query = SELECT hostname FROM exclusion;
$exclude_results = mysql_query($exclude_query, $Prod);
while ($exclude = mysql_fetch_array($exclude_results)) {
echo $exclude[hostname].\n;
}
-Wendell
-Original Message-
From: NIPP, SCOTT V (SBCSI) [mailto:[EMAIL PROTECTED]
Depending on what you do, you may need to go to things like replication and
queries (selects only) off the replicated server. You may also want to look
caching the results to limit the number of sql queries. Have a look at high
performance mysql (www.highperformancemysql.com) for more info.
Change yourr code to place the sql into a variable and the echo the variable
out to see what the problem might be. You can also cut'n'paste the variable
into the the mysql gui of your choice to see if there are other errors
Bastien
From: [EMAIL PROTECTED]
To: php-db@lists.php.net
Subject:
Bastien, Thanks for your help!.
I think I found the answer to my problem. My system is a windows based
system and it is not windows based. The system where the code doesn't work is
Unix
and is case sensitive. I went online to the mysql reference and found this
link,
Hi,
Not certain what is happening try the following script. On my machine
the form retains the session ID. Note the sending the ID as a GET is
only done for cross checking. You can remove it and there will be no
change in the display.
?php
session_start();
?
html
body
?php
$id =
thanx, i did use the GET method to send a form. with the POST i works!!!
thanks again!
- Original Message -
From: graeme [EMAIL PROTECTED]
To: Henk Jan Wils [EMAIL PROTECTED]
Cc: php-db@lists.php.net
Sent: Saturday, January 08, 2005 11:36 AM
Subject: Re: [PHP-DB] Problem using session
Bastien Koert [EMAIL PROTECTED] wrote:
have you set the permissions in the folders? Right click on the folder and
choose Properties...then set the appropraiate permissions
bastien
[snip]
Thanks, Bastien.
Altering permissions doesn't appear to work, at least not the way it seems
to be
make sure in the general tab that the read only checkbox is NOT checked
bastien
From: Ian [EMAIL PROTECTED]
To: php-db@lists.php.net
Subject: Re: [PHP-DB] Problem with permissions Win32 dba_open
Date: Thu, 30 Dec 2004 12:54:49 -
Bastien Koert [EMAIL PROTECTED] wrote:
have you set
[EMAIL PROTECTED]
To: php-db@lists.php.net
Subject: Re: [PHP-DB] Problem with permissions Win32 dba_open
Date: Thu, 30 Dec 2004 12:54:49 -
Bastien Koert [EMAIL PROTECTED] wrote:
have you set the permissions in the folders? Right click on the folder
and
choose Properties...then set
Bastien Koert [EMAIL PROTECTED] wrote
is your ms firewall turned on? ms sp2 screws a lot of stuff up..
bastien
Thanks bastien,
I'm running Norton IS 2005, with Windows Firewall switched off.
Whether or not I turn off the Norton firewall results in the same error:
?php $db = dba_open(
have you set the permissions in the folders? Right click on the folder and
choose Properties...then set the appropraiate permissions
bastien
From: Ian [EMAIL PROTECTED]
To: php-db@lists.php.net
Subject: Re: [PHP-DB] Problem with permissions Win32 dba_open
Date: Wed, 29 Dec 2004 09:15:01 -
is your ms firewall turned on? ms sp2 screws a lot of stuff up..
bastien
From: Ian [EMAIL PROTECTED]
To: php-db@lists.php.net
Subject: [PHP-DB] Problem with permissions Win32 dba_open
Date: Tue, 28 Dec 2004 23:36:08 -
Hi all, Season's Greetings!
I am testing my PHP scripts locally under XP
On Saturday 13 November 2004 20:25, GH wrote:
Explain Bind Vars please... Thanks
It is a technique for preparing an SQL statement with placeholders which can
then be substituted with values when the statement is executed. It will be
unavailable to you unless you are using MySQL 4.1.3 or above
what mysql version are you running? versions under 4.1 DONT support
subqueries
bastien
From: GH [EMAIL PROTECTED]
Reply-To: GH [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Problem with an insert query
Date: Sat, 13 Nov 2004 01:29:58 -0500
In my mySQL database I have the following
You also appear to have a quote in the value that contains the word
members'
You will need to escape this or use bind vars.
Simon
On Saturday 13 November 2004 06:29, GH wrote:
In my mySQL database I have the following tables:
+-+
| Tables_in_AHRC |
+-+
|
Explain Bind Vars please... Thanks
On Sat, 13 Nov 2004 19:08:59 +, Simon Rees [EMAIL PROTECTED] wrote:
You also appear to have a quote in the value that contains the word
members'
You will need to escape this or use bind vars.
Simon
On Saturday 13 November 2004 06:29, GH wrote:
from the brief error message your provided, I think its blowing up on the
single quote contained in the word can't. you need to handle this by
choosing one a few of these methods
replace all single quotes with two single quotes
$text = str_replace(','', $text);
addslashes function
$text =
On Tue, 2004-11-02 at 10:41 -0600, [EMAIL PROTECTED]
wrote:
Query failed: 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
't read card','1099413551')' at line 1
My initial guess is that you do not have
Just looking at the error message shows why it's failing in this case: 't read
card','1099413551')' at line 1 One of the fields you're trying to insert is likely
'Can't read card', and since in the query you are surrounding it with single quotes,
and it's coughing on the word Can't , which has
Try echoing the actual query being run. It will give you more info to work
with.
Peter
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED]
Sent: 02 November 2004 16:42
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Problem with script
Hi all, greetings from
Dylan Barber wrote:
Okay maybe its late or something but this should work, however I get this
#1064 - 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
'SELECT tblVideos.PerformerID
FROM tblVideos
WHERE 1 =1 AND
AM
To: Dylan Barber
Cc: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] Problem with this query?
Dylan Barber wrote:
Okay maybe its late or something but this should work,
however I get
this
#1064 - You have an error in your SQL syntax. Check the
manual that
corresponds to your MySQL
To: Dylan Barber
Cc: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] Problem with this query?
Dylan Barber wrote:
Okay maybe its late or something but this should work,
however I get
this
#1064 - You have an error in your SQL syntax. Check the
manual that
corresponds to your MySQL
: water_foul [EMAIL PROTECTED]
Date: Wednesday, June 23, 2004 8:58 am
Subject: Re: [PHP-DB] problem
I checked em all they were right
Daniel Clark [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
Sounds like it doesn't like your SQL statement. Perhaps a field
or table
name
23, 2004 8:58 am
Subject: Re: [PHP-DB] problem
I checked em all they were right
Daniel Clark [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
Sounds like it doesn't like your SQL statement. Perhaps a field
or table
name is incorrect?
Warning
Just from looking at it, it seems to be because you are redefining a
variable, which destroys it, then calling for a mysql resource that has
been destroyed. What is the error?
I can't gaurantee that that is what is happening.
Cole
why doesn't this work:
$pic=mysql_query('SELECT Rune,
What error are you getting?
why doesn't this work:
$pic=mysql_query('SELECT Rune, username FROM RuneRunner
RuneRunner_1 WHERE (User_ID = 3)',$connection);
$pic=mysql_fetch_array($pic);
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit:
I tried that it didn't work
Cole S. Ashcraft [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
Just from looking at it, it seems to be because you are redefining a
variable, which destroys it, then calling for a mysql resource that has
been destroyed. What is the error?
I can't
I checked em all they were right
Daniel Clark [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
Sounds like it doesn't like your SQL statement. Perhaps a field or table
name is incorrect?
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL
result
resource in
PROTECTED]
Date: Wednesday, June 23, 2004 8:58 am
Subject: Re: [PHP-DB] problem
I checked em all they were right
Daniel Clark [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
Sounds like it doesn't like your SQL statement. Perhaps a field
or table
name is incorrect
single quotes around the digit 3.
- Original Message -
From: water_foul [EMAIL PROTECTED]
Date: Wednesday, June 23, 2004 8:58 am
Subject: Re: [PHP-DB] problem
I checked em all they were right
Daniel Clark [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
Sounds like
? And
also wrap single quotes around the digit 3.
- Original Message -
From: water_foul [EMAIL PROTECTED]
Date: Wednesday, June 23, 2004 8:58 am
Subject: Re: [PHP-DB] problem
I checked em all they were right
Daniel Clark [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED
If you are able to place the Access database on a Windows machine that can
be accessed by the Linux machine, then you can use odbtp. You can get it
at http://odbtp.sourceforge.net.
-- bob
On Wed, 5 May 2004 [EMAIL PROTECTED] wrote:
Hi,
I try to configure a web server under Debian/GNU/Linux
I try to configure a web server under Debian/GNU/Linux Sarge. All
fonctionning
perfectly except odbc connect.
I have to read datas in an ACCESS2000 mdb file to import in Mysql.
Actually this run correct under a windows 2000 workstation with
apache/PHP/Mysql
I installed on my
If you are able to place the Access database on a Windows machine that can
be accessed by the Linux machine, then you can use odbtp. You can get it
at http://odbtp.sourceforge.net.
-- bob
My problem is i can't modify my PHP code. Actually the web site runs on a
windows2000
$result=mysql_db_query(usr_172_1,UPDATE artikel SET Preis='\$preis\'
WHERE ID='\$id\');
//mysql_error($result);
Here you're escaping the $ and then the ', have you tried
$result=mysql_db_query(usr_172_1,UPDATE artikel SET Preis='{$preis}'
WHERE ID='{$id}');
with those escapes this is the
Jeffrey N Dyke [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
$result=mysql_db_query(usr_172_1,UPDATE artikel SET Preis='\$preis\'
WHERE ID='\$id\');
//mysql_error($result);
Here you're escaping the $ and then the ', have you tried
$result=mysql_db_query(usr_172_1,UPDATE
try writing it like
$sql = UPDATE round1
SET game1=' . $game1 . ',
game2=' . $game2 . ',
game3=' . $game3 . ',
game4=' . $game4 . ',
game5=' . $game2 . ',
game6=' . $game6 . ',
game7=' . $game7 . ',
On Mar 21, 2004, at 3:44 AM, JeRRy wrote:
Hi,
I have not used this site in a while and have a
problem. I can't get anything to save to the
database. :( Not sure what is wrong, need a fix fast
as the site needs to be active in a few days by the
latest so hopefully I can get help in that time.
snip
$sql = UPDATE round1 SET game1='$game1' ,
game2='$game2' , game3='$game3' , game4='$game4'
, game5='$game5' , game6='$game6' , game7='$game7'
,
game8='$game8' , misc='y'
WHERE username= \$sidarray[0]\;
This only updates an existing record, where the
username is equal to
it looks like the use of double and single quotes could be at fault
try writing it like
$sql = UPDATE round1
SET game1=' . $game1 . ',
game2=' . $game2 . ',
game3=' . $game3 . ',
game4=' . $game4 . ',
game5=' . $game2 . ',
Will wrote:
Hello all,
When I try to connect I get this message: #1045 - Access denied for
user: [EMAIL PROTECTED]' (Using password: YES).
Can anyone help me? Do I need to do anything in the php.ini file or
anywhere else.
sounds like you need to set a password in the phpMyAdmin config file
That was already done!!! I knew that.
It is still doing it. MySQL in installed in a window XP machine.
What else is there??
~WILL~
Jochem Maas wrote:
Will wrote:
Hello all,
When I try to connect I get this message: #1045 - Access denied for
user: [EMAIL PROTECTED]' (Using password: YES).
I might add that I can access the database through C: prompt.
~WILL~
Will wrote:
That was already done!!! I knew that.
It is still doing it. MySQL in installed in a window XP machine.
What else is there??
~WILL~
Jochem Maas wrote:
Will wrote:
Hello all,
When I try to connect I get this
When I bring up \mysql\bin\mysqladmin status proc is says that the user
is ODBC, is that right?? Should I have another user in there??
Please help!! :)
~WILL~
Will wrote:
I might add that I can access the database through C: prompt.
~WILL~
Will wrote:
That was already done!!! I knew
I GOT IT YEAH!
Thanks everyone.
~WILL~
Will wrote:
When I bring up \mysql\bin\mysqladmin status proc is says that the user
is ODBC, is that right?? Should I have another user in there??
Please help!! :)
~WILL~
Will wrote:
I might add that I can access the database through C:
Your query resource was detached from the connection. Can you supply the
code prior to calling odbtp_fetch_row() and starting from odbtp_query()?
-- bob
On Tue, 17 Feb 2004, Sascha Kaufmann wrote:
l
Hello everyone
I try to port an application from Windows/ODBC to Linux/ODBTP, I have a
MSSQL
Normally, you get that warning when you've already sent display information
to the browser before you attempt to use the header() function. You need to
make sure you haven't sent ANY HTML to the browser prior to the header()
line in your code. You can check the PHP documentation regarding the
- Original Message -
From: Phil Matt [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Tuesday, January 27, 2004 17:08
Subject: [PHP-DB] Problem with passing variables
Hello to all.
I am a relative newcomer to PHP. I am trying to pass the value of a
variable from a
MySQL database from
If you have something like this:
echo 'tda href=email.php?recip=$myrow[0]img src=graphics/mail.gif
class=imglink alt=Click to send mail to this person //a/td/tr';
It wont work. Use instead:
echo 'tda href=email.php?recip='.$myrow[0].'img
src=graphics/mail.gif class=imglink alt=Click to send
On 27 Jan 2004 at 15:56, Ricardo Lopes wrote:
If you have something like this:
echo 'tda href=email.php?recip=$myrow[0]img src=graphics/mail.gif
class=imglink alt=Click to send mail to this person //a/td/tr';
It wont work. Use instead:
echo 'tda href=email.php?recip='.$myrow[0].'img
; [EMAIL PROTECTED]
Sent: Tuesday, January 27, 2004 4:27 PM
Subject: Re: [PHP-DB] Problem with passing variables
On 27 Jan 2004 at 15:56, Ricardo Lopes wrote:
If you have something like this:
echo 'tda href=email.php?recip=$myrow[0]img src=graphics/mail.gif
class=imglink alt=Click to send
In article [EMAIL PROTECTED], [EMAIL PROTECTED]
says...
*Sorry if this is a duplicate*
Using mySQL db to store my data. Using a select statement on my PHP page.
When it lists out the data it seems to be skipping the first row. Any ideas?
Without seeing your code, I'll have to guess that
hi,
i think $PHP_AUTH_USER will only get unset the next time session_start() is
called. try to use a header(location: ...) to direct your logout page to
your index page and then print $PHP_AUTH_USER right after session_start()..
by right, you should not have any value...
i faced the same problem
A bit difficult to debug this without the file included (config.php);
providing the error message would also be helpful.
At first glance, I'm just wondering what does the dot mean in the table
name used in the FROM clause:
FROM school.physics_chris_rockets
It shouldn't generate a php syntacs
On Thu, 06 Nov 2003 05:46:03 +0100
Boyan Nedkov [EMAIL PROTECTED] wrote:
A bit difficult to debug this without the file included (config.php);
providing the error message would also be helpful.
At first glance, I'm just wondering what does the dot mean in the
table name used in the FROM
My Comment bellow:
On Wednesday 15 October 2003 14:38, Kirk Babb wrote:
Hi,
I'm having trouble with a query inside a function. I thought I'd written
this so that the function would fail if email and zipcode supplied were not
in the same row (trying to use those two things to identify the
From: Luis M Morales C [EMAIL PROTECTED]
My Comment bellow:
On Wednesday 15 October 2003 14:38, Kirk Babb wrote:
Hi,
I'm having trouble with a query inside a function. I thought I'd
written
this so that the function would fail if email and zipcode supplied were
not
in the same row
PROTECTED]
Sent: Thursday, October 16, 2003 10:57 AM
To: [EMAIL PROTECTED]; [EMAIL PROTECTED]
Subject: Re: [PHP-DB] problem - query inside a function
From: Luis M Morales C [EMAIL PROTECTED]
My Comment bellow:
On Wednesday 15 October 2003 14:38, Kirk Babb wrote:
Hi,
I'm having
Kirk,
First, I'd recommend troubleshooting this function by doing the following:
[code snippet]
function getEditData($email,$zipcode) {
$sql = SELECT * FROM contact_info WHERE zipcode='$zipcode' AND
email=\{$email}\;
echo $sql; //ADD THIS LINE
//$query = mysql_query($sql);
On 08 Oct 2003 17:26:55 +0200
Ruprecht Helms [EMAIL PROTECTED] wrote:
Hi,
I've problems with a http-uploadscript. When I run the htmlfile and
the script replies an output of the filename but I can't find the file
in the destinationdirectory.
Why this can happen. Addes you can see the
- Original Message -
From: Sean Burlington [EMAIL PROTECTED]
Newsgroups: php.db
To: [EMAIL PROTECTED]
Sent: Sunday, September 07, 2003 3:26 PM
Subject: Re: [PHP-DB] Problem in executing linux command from PHP
Gnanavel wrote:
$output=exec(cp file1 file2);
echo pre$output/pre;
does
Gnanavel wrote:
$output=exec(cp file1 file2);
echo pre$output/pre;
does not works. can any one help me out of this problem
When I was executing the cp command it doesn't return anything.
But it returned the name of the last file when i executed the ls
`cp file1 file2` doesn't return anything -
- Original Message -
From: Sean Burlington [EMAIL PROTECTED]
Newsgroups: php.db
To: [EMAIL PROTECTED]
Sent: Tuesday, September 02, 2003 1:11 PM
Subject: Re: [PHP-DB] Problem in executing linux command from PHP
Gnanavel wrote:
I have problem in executing linux command
$output=exec
I have problem in executing linux command
$output=exec(ls -a);
echo pre$output/pre;
the above coding works, but
$output=exec(cp file1 file2);
echo pre$output/pre;
does not works. can any one help me out of this problem
When I was executing the cp command it doesn't return anything.
But it
Gnanavel wrote:
I have problem in executing linux command
$output=exec(ls -a);
echo pre$output/pre;
the above coding works, but
$output=exec(cp file1 file2);
echo pre$output/pre;
does not works. can any one help me out of this problem
When I was executing the cp command it doesn't return
Gnanavel wrote:
I have problem in executing linux command
$output=exec(ls -a);
echo pre$output/pre;
the above coding works, but
$output=exec(cp file1 file2);
echo pre$output/pre;
does not works. can any one help me out of this problem
When I was executing the cp command it doesn't return
Quoting [EMAIL PROTECTED]:
From: Gnanavel [EMAIL PROTECTED]
I have problem in executing linux command
$output=exec(ls -a);
echo pre$output/pre;
the above coding works, but
$output=exec(cp file1 file2);
echo pre$output/pre;
does not works. can any one help me out of
From: Gnanavel [EMAIL PROTECTED]
I have problem in executing linux command
$output=exec(ls -a);
echo pre$output/pre;
the above coding works, but
$output=exec(cp file1 file2);
echo pre$output/pre;
does not works. can any one help me out of this problem
Riddle me this... exec()
is it possible that there is no referer. what happens if you link to this
page from another, ensuring there is a referer, will it display it then?
Jeff
-Original Message-
From: Ahmed Abdelaliem [mailto:[EMAIL PROTECTED]
Sent: 15 July 2003 08:54
i have a problem with starting a session in the page that
validates the user
input and sends it tothe database,
when the user clicks register he gets this error
Warning:
PROTECTED], [EMAIL PROTECTED]
Subject: RE: [PHP-DB] problem with starting a session
Date: Tue, 15 Jul 2003 10:29:06 +0100
-Original Message-
From: Ahmed Abdelaliem [mailto:[EMAIL PROTECTED]
Sent: 15 July 2003 08:54
i have a problem with starting a session in the page that
validates the user
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