shunut Posted October 6, 2006 Share Posted October 6, 2006 I wasn't sure if this was the right place for this or if this should be in php help but I'll start here. I have an online paintball store. The site was a presetup site from osCommerce I just had to change paths and add my stuff. I know html but know nothing about php coding or mysql. I came across this error the other day and posted it on a osCommerce site but I have received no help. If anybody could help me or point me in the right direction I would appreciate it. I'm up for learning what I need to do to fix this problem but don't know where to look.Here is the error I get:Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/content/d/u/r/durrell12/html/catalog/admin/includes/functions/database.php on line 100Warning: reset(): Passed variable is not an array or object in /home/content/d/u/r/durrell12/html/catalog/admin/includes/classes/object_info.php on line 17Warning: Variable passed to each() is not an array or object in /home/content/d/u/r/durrell12/html/catalog/admin/includes/classes/object_info.php on line 18I went into each file and here are the lines it is calling:Line 100 - database.php (first error)[code]return mysql_fetch_array($db.query, MYSQL_ASSOC);[/code]Line 17 - object_info.php (second error)[code]reset ($object_array);[/code]Line 18 - object_info.php (third error)[code]while (list($key, $value) = each($object_array)) {[/code]If anybody could help me at all I would greatly appreciate it. If you need more info, please ask. Quote Link to comment Share on other sites More sharing options...
fenway Posted October 6, 2006 Share Posted October 6, 2006 Well, I'm guessing it all stems from the first issue, which is that you're not passing a valid result set. Quote Link to comment Share on other sites More sharing options...
shunut Posted October 6, 2006 Author Share Posted October 6, 2006 Thats what I figured. Now more importantly what do I need to do to fix it? Quote Link to comment Share on other sites More sharing options...
fenway Posted October 9, 2006 Share Posted October 9, 2006 What do you think is supposed to be in $db.query? Quote Link to comment Share on other sites More sharing options...
shunut Posted October 9, 2006 Author Share Posted October 9, 2006 I have no idea. I thought that was just telling it to look at my database. I don't do php, html is all I've learned so far. Quote Link to comment Share on other sites More sharing options...
fenway Posted October 10, 2006 Share Posted October 10, 2006 Well, apparently, the script doesn't either... somewhere there's a MySQL query, and it supposed to be in this variable. Quote Link to comment Share on other sites More sharing options...
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