Re: MySQL error... but it works!!??

Hi, So far the best way to troubleshoot mysql+php for me has been to add an "or die" after the sql exec. $crdate = date("Y-m-d"); $result = mysql_query("SELECT * FROM sites WHERE creation_date = '$crdate' AND status = 'T'") or die ('CAN NOT EXEC SQL'); Or you can do $sql = "SELECT * FROM sites WHERE creation_date = '$crdate' AND status = 'T'"; echo $sql; $result = mysql_query($sql) or die .....; and try to run your sql against mysql directly to make sure your sql is fine. one thing to always check is your connection to the database..... did you connect? Simon Kimber <simon@funny.co.uk> wrote: Does anyone have any idea why this is giving me a "Warning: Supplied argument is not a valid MySQL result resource in..." error? The funny thing is that apart from that error message it works perfectly!! --------------------------- $crdate = date("Y-m-d"); $result = mysql_query("SELECT * FROM sites WHERE creation_date = '$crdate' AND status = 'T'"); while ($sitedata = mysql_fetch_array($result)) { echo $sitedata['name'] . " "; } --------------------------- Cheers Simon -- 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 Mehmet Erisen http://www.erisen.com --------------------------------- Do You Yahoo!? Send FREE video emails in Yahoo! Mail.