Question

I want to show image with php script. Here is my current code:

<?php
if (isset ($_GET['id'])) $id = $_GET['id'];
$t=getimagesize ($id) or die('Unknown type of image');

switch ($t[2])
{
    case 1:
    $type='GIF';
    $img=imagecreatefromgif($path);
    break;
    case 2:
    $type='JPEG';
    $img=imagecreatefromjpeg($path);
    break;
    case 3:
    $type='PNG';
    $img=imagecreatefrompng($path);
    break;
}

header("Content-type: image/".$type);
echo $img;
?>

But it doesn’t show the image. What is the right way instead of echo $img?

Answer 1

header("Content-type: image/jpeg");
imagejpeg($img);