ติด warning mysql_fetch_array(): supplied argument is not a valid MySQL result resource

	$sql = "SELECT * FROM `type_id` WHERE genres IN ('Comedy','Romance')";
	   while($arr = fetch($sql)) {
		echo $arr['id'];	   
		echo "<br/>";
	}

function fetch($result) {

   if (isset($result)) {

     $row = mysql_fetch_array($result);// line30

   } else {

     echo "Invalid Query Fetch";

   }

    return $row;
}