You have an error in your PHP code, just as the error says. Including only the query, which you've already tested, isn't very helpful. The error message is telling you that the result "link identifier" used in your fetch statement isn't the same as returned by the query statement OR there was an error returned by the query. It's all in the manual. Doug On Sun, 3 Aug 2003 09:59:42 -0400, Aaron Wolski wrote: >Hi guys, > >I have the following query which returns FINE through the console but >not in PHP > > >select t.id, t.colour, t.colourID, t.price, t.type, g.thread_index, >g.groupNameUrl FROM kcs_threads t LEFT JOIN kcs_threadgroups g ON t.id = >g.thread_index WHERE g.groupNameUrl = '0-500' > > >The error I am getting on the browser is: > >Warning: mysql_fetch_array(): supplied argument is not a valid MySQL >result resource in http://www.martekbiz.com/KCS/utils.inc on line 52 > > > >Anyone have an idea as to what it up? > >Thanks! > >Aaron > -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
