รบกวนช่วยดูโค๊ดให้ทีครับ image cannot displayed because contains errors ป๊อปอัพรูป ผิดพลาด ไม่แสดงรูป จะแก้ไขยังไงครับ

<?php
ob_start();
include("ADMIN/Utils.php");
if(!is_numeric($id)) die("");
$flag = true;
$image_types = Array 
(
		array("image/jpeg","jpg"),
		array("image/pjpeg","jpg"),
		array("image/bmp","bmp"),
		array("image/gif","gif"),
		array("image/x-png","png")
);	

if($id == 0 || $id == 1 || $id == -1|| $id == -2)
{
	Header ("Content-type: image/gif");
	$handle = fopen("images/no_pic.gif", "rb");
	$contents = fread($handle, filesize("images/no_pic.gif"));
	fclose($handle);
	print $contents;
	$flag=false;
}

if($flag)
{
				foreach($image_types as $image_type)
				{
					if(file_exists("uploaded_images/".$id.".".$image_type[1])) 
					{
								Header ("Content-type: ".$image_type[0]);
								$handle = fopen("uploaded_images/".$id.".".$image_type[1], "rb");
								$contents = fread($handle, filesize("uploaded_images/".$id.".".$image_type[1]));
								fclose($handle);
								print $contents;
								$flag=false;
					}
				}
}

if($flag)
{

				mysql_connect("$DBHost","$DBUser","$DBPass");
				mysql_select_db ($DBName) or die ("DB does not exist or access is denied!");
				
				$sql = "SELECT * FROM ".$DBprefix."image WHERE image_id=$id";
				
				$result = mysql_query ($sql);
				
				if (mysql_num_rows ($result)>0) 
				{
					$row = @mysql_fetch_array ($result);
					$image_type = $row["image_type"];
					$image = $row["image"];
					Header ("Content-type: $image_type");
					print $image;
				}
				mysql_free_result($result);
				mysql_close();
}
ob_end_flush();
?>