Display PHP image by URL

Question

I have this php code (image.php):

if($_SERVER['SERVER_NAME'] == 'site.tld') {

    header('Content-type: image/jpeg');
    echo file_get_contents("image1.jpg");

} else {

    header('Content-type: image/jpeg');
    echo file_get_contents("image2.jpg");

}

*This page (image.php) is hosted on site.tld;

Now... This code will be called by <img src='http://site.tld/image.php' />.

I'd like to dispay image1.jpg if image is called on site.tld and image2.jpg if not.

That code displays image1.jpg whatever domain name where code is called.

Any solutions ? Thanks.

Edit: The output of $_SERVER['SERVER_NAME'] is site.tld (my server) on any websites...

That's the problem because the script not works properly.

There is no errors. Problem is that code displays always image1.jpg whatever domain name where code is called. — vlady
– vlady, Commented Oct 24, 2012 at 20:48
Please, Show us the echo $_SERVER['SERVER_NAME'] result and check the if condition. — Mate
– Mate, Commented Oct 24, 2012 at 20:48

Mate · Accepted Answer · 2012-10-24 22:27:48Z

4

Try this:

    $imgPath = "image1.jpg";

    if(stripos($_SERVER["HTTP_REFERER"], 'http://site.tld/')=== false){

       $imgPath = "image2.jpg";

    }

    //echo $_SERVER["HTTP_REFERER"]."<br/>";
    //echo $imgPath."<br/>";

    header('Content-type: image/jpeg');
    echo file_get_contents($imgPath);

Other option (based on PWhite comments) :

All calls you make from http://site.tld/ add a querystring key, and then evaluates this data too.

Inside site.tld :

<img src='http://site.tld/image.php?myfookey=secret' />

Outside site.tld :

<img src='http://site.tld/image.php' />

image.php

$imgPath = "image1.jpg";

if (!(isset($_REQUEST["myfookey"]) &&  $_REQUEST["myfookey"] == 'secret')){
    $imgPath = "image2.jpg";
} 

header('Content-type: image/jpeg');
echo file_get_contents($imgPath);

edited Oct 24, 2012 at 22:27

answered Oct 24, 2012 at 21:02

Mate

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6 Comments

Patrick White Over a year ago

From the PHP documentation: "The address of the page (if any) which referred the user agent to the current page. This is set by the user agent. Not all user agents will set this, and some provide the ability to modify HTTP_REFERER as a feature. In short, it cannot really be trusted."

vlady Over a year ago

Yes, something like that :) But this example has some errors. There is an alternative?

vlady Over a year ago

Not "physical" errors. I mean some "bugs". Eg: From site2.tld : right click > view image, and image is now displayed on site1.tld but it remains the same.

Mate Over a year ago

sorry, I do not quite understand. If you navigate from Server site2.tld or site1.tld the image you show is image2.jpg? To be sure to uncomment the echos and check if some data is wrong

Mate Over a year ago

I added another alternative, please let me know if you have solved @PWhite

|

antho · Accepted Answer · 2012-10-24 20:50:53Z

1

That is because

$_SERVER['SERVER_NAME']

shows the name of your server (i.e. the server that executes the PHP).

What you want is the URL of the page that contains <img src='http://site.tld/image.php' /> and I'm not sure the client's browser sends such information to your server.

/edit: If you're trying to prevent third party websites from showing your picture, you want to do that at the server configuration level, e.g. http://www.davidairey.com/stop-image-theft-hotlinking-htaccess/

answered Oct 24, 2012 at 20:50

antho

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2 Comments

Patrick White Over a year ago

Most of the time it does, as $_SERVER['HTTP_REFERER'] but this can be spoofed.

vlady Over a year ago

The output of $_SERVER['SERVER_NAME'] is site.tld (my server) in all websites... Looking for a way to take the current url using the picture.

jaco · Accepted Answer · 2012-10-24 20:50:13Z

0

First off, check the $_SERVER['SERVER_NAME'] output for the if condition. Then use readfile() instead of file_get_contents().

answered Oct 24, 2012 at 20:50

jaco

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1 Comment

vlady Over a year ago

The output of $_SERVER['SERVER_NAME'] is site.tld (my server) in another websites... That's the problem because the script not works properly.

addiedx44 · Accepted Answer · 2012-10-24 20:56:56Z

0

If you're using Apache, try setting UseCanonicalName Off in your virtual host config (probably need to reload/restart Apache).

answered Oct 24, 2012 at 20:56

addiedx44

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