RE: [PHP] arg....

At 12:31 PM 1/26/2001, Krznaric Michael wrote: sorry - here you all go.. // number crunching time $count++; $time = date("Y-m-d H:i:s"); $result = mysql_query("INSERT INTO links (count) VALUES $count WHERE lid=$id"); echo $result;

        You need to be a little more specific about DB and problem type.

You may have to commit the transaction if commit is not implict (ex Oracle). There could be many resons including your SQL statement. Mike -----Original Message----- From: Kurth Bemis [mailto:kurth@usaexpress.net] Sent: Friday, January 26, 2001 12:23 AM To: php-general@lists.php.net Subject: [PHP] arg.... i'm having a horrible time updating 2 fields in the same db. I don't get an error but the fields aren't updated....can anyone send me a snippet for the to learn from? ~kurth -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: php-general-unsubscribe@lists.php.net For additional commands, e-mail: php-general-help@lists.php.net To contact the list administrators, e-mail: php-list-admin@lists.php.net