Mr_Burton

Woa, that solved it! Silly me, I altered the table locally, your solution pointed me straight to the missing field, thanks a mllion for your quick response! Probably the most useful mysql-syntax, and I didn't even know it Great forum!

Hey guys, I'm quite new at MySQL and finally managed to get the following query working: $results = mysql_query("SELECT prod.group, prod.name, sub.name, list.name FROM prod LEFT JOIN sub ON prod.group = sub.id LEFT JOIN list ON sub.list = list.id"); and to loop through the results: while($result = mysql_fetch_array($results)) { //do stuff } However the query only seems to work locally, when on a webhost it gives the following error: Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in (..) on line 21. Where line 21 is the "mysql_fetch_array"-part. Any help would be greatly appreciated!