kts Posted December 20, 2007 Share Posted December 20, 2007 Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in column_left2.php on line 27 $parents_cat_id = mysql_query("SELECT * FROM categories WHERE parent_id = ".$catpath)or die(mysql_error()); $parents_cat_id = mysql_num_rows($parents_cat_id) or die(mysql_error()); while ($parent_c = mysql_fetch_array($parents_cat_id)){ $new_parents_cat_id = $parent_c['categories_id']; } Quote Link to comment https://forums.phpfreaks.com/topic/82564-solved-simple-mysqlphp-warningerror/ Share on other sites More sharing options...
revraz Posted December 20, 2007 Share Posted December 20, 2007 You are overwriting your result variable with the num_rows function. Quote Link to comment https://forums.phpfreaks.com/topic/82564-solved-simple-mysqlphp-warningerror/#findComment-419734 Share on other sites More sharing options...
kts Posted December 20, 2007 Author Share Posted December 20, 2007 wow, it was more simple then i could have ever imagined... thanks been staring too long haha Quote Link to comment https://forums.phpfreaks.com/topic/82564-solved-simple-mysqlphp-warningerror/#findComment-419737 Share on other sites More sharing options...
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