php.net |  support |  documentation |  report a bug |  advanced search |  search howto |  statistics |  random bug |  login

go to bug id or search bugs for

Bug #3295 MYSQL_INSERT_ID not working
Submitted: 2000-01-24 10:35 UTC Modified: 2000-01-27 15:18 UTC
From: arni at linux dot is Assigned:
Status: Closed Package: MySQL related
PHP Version: 4.0 Beta 3 OS: Linux
Private report: No CVE-ID: None
View Developer Edit
Welcome back! If you're the original bug submitter, here's where you can edit the bug or add additional notes.
If you forgot your password, you can retrieve your password here.
Password:
Status:
Package:
Bug Type:
Summary:
From: arni at linux dot is
New email:
PHP Version: OS:

 

 [2000-01-24 10:35 UTC] arni at linux dot is 
Hello,

I wrote a PHP program that uses MYSQL_INSERT_ID() function. I begun writing the program under PHP3, but once the program was finished I upgraded my system to PHP4-b3.

Once the upgrade was complete I tried to use the program. But it wouldn't work because of a error MYSQL_INSERT_ID() caused.

The error is the following:
Warning: Supplied argument is not a valid MySQL-Link resource in phpfile.php3 on line 35

I don't know what causes this, my code works fine in PHP3, but this error only occures when I try to use it under PHP4-b3.

Could this be a BUG? :-)




Regards.

Arni Arent
Web Developer
Firmanet ehf.

arni@firmanet.is
http://www.firmanet.is

appel @ EFnet

Patches

Pull Requests

History

AllCommentsChangesGit/SVN commitsRelated reports
 [2000-01-24 14:06 UTC] joey at cvs dot php dot net 
Show example of code.
 [2000-01-24 18:23 UTC] arni at linux dot is 
Here are the 2 lines that matter:

$querymid = @mysql_query("insert into tempmail (userid,toaddress,cc,bcc,subject,priority,body,signature) values ('$uid','$to','$cc','$bcc','$subject','$priority','$body','$signature')");
$mailid = mysql_insert_id($querymid);
 [2000-01-27 15:18 UTC] torben at cvs dot php dot net 
This isn't a bug; you've given mysql_insert_id() the result id from
the query instead of the link id from the connection--i.e., the value
returned by mysql_connect() or mysql_pconnect(). The code shown should 
work if you either supply that value or supply no argument; if none is 
given, the most recently opened link is assumed.
 
PHP Copyright © 2001-2026 The PHP Group
All rights reserved. 		Last updated: Thu Oct 08 13:00:02 2026 UTC