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

On Jan 8, 2008, at 1:29 PM, Stephen Johnson wrote:

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.

I wasn't able to make it work with mysqli_insert_id() but you did give me the idea of using mysqli_stmt_error() instead. Which doesn't return anything if it was successful... So a simple: $check = mysqli_stmt_error(); if($check = "") redirect; } else { exit(); } works great! Thanks for your help! -- Jason Pruim Raoset Inc. Technology Manager MQC Specialist 3251 132nd ave Holland, MI, 49424 www.raoset.com japruim@raoset.com