问PHP是为URL存储在数据库中的图像提供服务的更好方法

EN

<img src="http://www.example.com/image/view/93029420" />

$query = "SELECT img_url FROM images WHERE image_id = '93029420'";

$result = DB.findOne($query);

$contents = file_get_contents($result['img_url'] );
header('Content-Type: image/png');
echo $contents;
exit(); 

$query = "SELECT img_url FROM images WHERE image_id = '93029420'";
 $result = DB.findOne($query);
 header('Location: ' $result['img_url'] );
exit();