RE: [PHP] Supplied argument is not a valid MySQL result resource

Try collection_id ='$id' AND group_image = '$gi' and check your Table and Column names.... ________________________________________ Sjoerd van Oosten Digitaal vormgever svoosten@esites.nl Datamex E-sites B.V. http://www.esites.nl Minervum 7368 Telefoon: (076) 5 730 730 4817 ZH BREDA Telefax: (076) 5 877 757 _______________________________________ -----Oorspronkelijk bericht----- Van: Dan McCullough [mailto:McWC@rocketmail.com] Verzonden: donderdag 8 november 2001 16:43 Aan: PHP General List Onderwerp: [PHP] Supplied argument is not a valid MySQL result resource Why isn't this a valid argument, does it have to do with the AND statement? $gi = "Y"; $sqlf = "SELECT * FROM products WHERE collection_id = $id AND group_image = $gi ORDER BY sort_order ASC"; $resultf = mysql_query($sqlf); while ($group_images = mysql_fetch_array($resultf)) { $list_images .= "<a href=test3.php?id=".$group_images["prod_id"]."><img src=\"images/product/".$group_images["product_image"]."\" border=\"0\" alt=\"".$group_images["product_name"]."\"></a><br>"; } dan __________________________________________________ Do You Yahoo!? Find a job, post your resume. http://careers.yahoo.com -- 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