$sql = "SELECT * FROM `table` WHERE `status` = 1 ORDER BY `rand()` LIMIT 5";
$query="select * from `commont` where menu='maghale' ORDER BY `rand()` LIMIT 5;";
$query="select * from `commont` where `menu`='maghale' ORDER BY `rand()` LIMIT 5;";
$result=mysql_query('the query') or die(mysql_error());
باز این یه مشکل جدیده!Warning: mysql_query() [function.mysql-query]: A link to the server could not be established
or die(mysql_error());