Re: New years resolution: To get serious with my programming! Anyone wanna help? :)

> From: Jason Pruim <japruim@raoset.com> > Date: Tue, 8 Jan 2008 13:18:40 -0500 > To: Jack Mays <jmays@hrsmart.com> > Cc: php-general General List <php-general@lists.php.net> > Subject: Re: [PHP] New years resolution: To get serious with my programming! > Anyone wanna help? :) > > > > Actually, the problem isn't the query... the query works fine. What I > want is for if it doesn't work, I want it to tell the user that it > didn't work. > > Right now I'm just blindly accepting that the insert succeeded, and I > want to get away from that! :) > > Does that explain it a little better? If you are simply wanting to do a check on whether the insert was successful or not, then grab the id form the insert and check to see if it is there... $success = mysql_insert_id(); If(!$success){ echo "no joy.. Insert failed."; } Because mysql_insert_id() acts on the last performed query, be sure to call mysql_insert_id() immediately after the query that generates the value. -- Stephen Johnson c | eh The Lone Coder http://www.thelonecoder.com continuing the struggle against bad code http://www.thumbnailresume.com --