Member Avatar for Mrki77

Hi everybody,

I have a specific problem, and cant get over it.

For my latest project I need a simple PHP script that display an image according to its ID sent through URL. Here's the code:

header("Content-type: image/jpeg"); $img = $_GET["img"]; echo file_get_contents("http://www.somesite.com/images/$img");

The problem is that the image doesn't show although the browser recognizes it (i can see it in the page title), instead I get the image URL printed out.

It doesn't work neither on a server with remote access allowed nor with one without. Also, nothing is printed or echoed before the header.

I wonder if it is a content type error, or something else. Thanks in advance.

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  • Latest Post Latest Post by karteek.vemula
Member Avatar for Mrki77

I managed to get the image, and display it in another file within the tag. Eventually the remote access was turned off.

However when I try to send the ID of the image as a variable ex. image.php?img=1 it doesn't display within the tag in the file. Instead I get this when I click on the image source:

<!DOCTYPE HTML PUBLIC "-//IETF//DTD HTML 2.0//EN">
<html><head>
<title>302 Found</title>
</head><body>
<h1>Found</h1>
<p>The document has moved <a href="http://www.somesite.com">here</a>.</p>
</body></html>

So basically I need a method to send the ID of the image from ex. showimage.php and I need image.php to retrieve it form "somesite dot com" and show it.

Member Avatar for HazardTW

In your first example lets assume that $_GET was properly validated and tested to be a legitimate image filename, and another test to verify that the file exists, you could use the GD library.

Instead of the "file_get_contents()" line, use GD's methods: (search GD at php.net)

header("Content-type: image/jpeg");
// test that GET is set and add the path or set image path/name to a default image named "no-image.jpg"
$img = isset($_GET['img']) ? 'images/'.$_GET['img'] : 'images/no-image.jpg';

// good idea to test file existence
is_file($img) or die();

// make gd image object
$requested_image = imagecreatefromjpeg($img);
// stream out the image 
imagejpeg($requested_image);
// do your chores and eat your vegetables
imagedestroy($requested_image);

Might even test file name extensions which I didn't do in the example, NEVER TRUST DATA BEING SENT TO YOUR SCRIPTS, even if you *think* your other scripts are the only ones that should be making requests to these scripts, ALWAYS VALIDATE EVERYTHING!

I don't know if there are better libraries for image handling in PHP, but the GD library is the only one I am familiar with and it has worked very well for me.

I am only taking into consideration your first example where you are sending the image file name to the php script that will stream the image to the browser.

Edited by HazardTW because: n/a
Member Avatar for karteek.vemula

// Get an image based on id;
$id = $_GET;
$image = "Select image_field_name from $table_name where id = '$id' ";
$result = mysql_query($image);

//create an image
header('Content-type: image/jpeg');
$samp_image = imagecreatefromjpeg($result);

//display an image;
imagejpeg($samp_image);

// Free up memory
imagedestroy($im);

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