อยากทราบวิธีการส่งฟังก์ชันกลับไปยัง window.parent เพื่อไป alert ค่าให้ user ทราบ

<?
session_start();
require("common.php");

$hd_code = $_POST["hd_code"];
//$picture_detail = $_POST["picture_detail"];


$file = $_FILES['fileupload']['name'];
$typefile = $_FILES['fileupload']['type']; 
$sizefile = $_FILES['fileupload']['size']; 

if($_FILES['fileupload']["size"] < 200000){
	$foldername = "PICTURE/";
	CheckFolderExit($foldername);

	$year = date('Y',time())."/";

	$foldername = $foldername.$year;
	CheckFolderExit($foldername);

	//--------- COPY File ---------//
	$t = time();
	$fileName = $foldername.$t."_".$file;


	if($_FILES['fileupload']['name'] != ""){
		copy($_FILES['fileupload']['tmp_name'],$fileName); 
	}else{
		$fileName = "";
	}



	$con = mysql_connect( $mysql_host,$mysql_user,$mysql_pass );
	mysql_select_db($mysql_db);


	$sql = "INSERT INTO table_picture (
			id ,
			code ,
			picture_url , 
			upload_date ,  
			upload_time 
			)
			VALUES (
			NULL , '$hd_code', '$fileName', CURDATE(), CURTIME() 
			)";
	if($_FILES['fileupload']['name'] != ""){

		mysql_query($sql);
	}

	mysql_close($con);
?>
<script language="JavaScript">
<!--
	window.parent.uploadok('<? echo $hd_code; ?>');
//-->
</script>
<?
}else{
	echo "<script language=\"JavaScript\">";
	echo "window.parent.uploaderror('$hd_code');";	
	echo "</script>";
}



?>