[PHP]<?php
$video_id=addslashes($_GET["id"]);
if($video_id==Null||!is_numeric($video_id))
{
die("قابل مشاهده نيست.");
}
$num_comments=$videos->get_num_comments($video_id);
$comments_result=$videos->get_front_comments($num_comments,$video_id);
while($comments_set=mysql_fetch_array($comments_re sult))
{
?>[/PHP]
خطا :
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in
