Question

How to hide an Image Source?

Asked by: RenatoCoto

I need to hide the Image source address, to prevent it to show where it's stored.

There is no need to protect the image at all, but the source of the image is relevant, so I need to find a way how to display the image with out placing the real address on the source.

thanks

Renato Coto

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Asked On: 2002-11-20 at 13:10:41ID20403645
Tags: hide
,; image
,; source
Topic: PHP Scripting Language
Participating Experts: 5
Points: 50
Comments: 20

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Answers

by: VGRPosted on 2002-11-21 at 00:04:41ID: 7477053

almost impossible

It will ALWAYS be deductible by looking properly at the HTML code

You may have a "first level" protection (hiding, rather) by using the status bar (TITLE, label, etc)

I would suggest to put a fake index.html page in the directory of the images, so that indelicate users are redirected to your real index page somewhere. Use Header("Location: ...");

by: PHPaulPosted on 2002-11-21 at 01:45:57ID: 7477353

You could make a page like this: images.php.
And refer to this page in the image tag (src):

<img src="images.php?imagename=mypicture.gif">

If you make the images.php file output as an image and load the image named $imagename, no one could see where your images are placed. You will need to put all the images in one folder (to make it more simple, it's not nessecary).

So if someone looks at the properties of an image, he/she will only see the src as images.php?imagename=mypicture.gif. They cannot find out where the image is actually located :-)
I would also use the index.html solution as VGR said. But I would use a index.php with a "page not found 404" header. This way people will think the folder doesn't excist.

Hope this helps,

--Paul

by: us111Posted on 2002-11-21 at 03:08:33ID: 7477632

something like that
<img src="images.php?f=pictures.gif">

<?
$PATH = "/somewhere";
if (strstr($f, ".gif"))
{ $d = fopen("PATH/$f","r");
if (!$d)
{ // If file cannot be read or not available ...
header("Location: index.html");
}
else
{
header('Content-Type: image/gif');
header("Content-Length: ".filesize("$PATH/$f"));
header('Content-Disposition: inline; filename=$f);

$data="";
while (!feof($d)) $data.=fread($d,12400);
fclose($d);
print $data;
}
}
else
header("Location: index.html");
?>

be careful :
- $PATH must be specified in this file
- check the file type
otherwise you can do something like :
<img src="images.php?f=/etc/password">

The better is to have an array which contains all of your pictures and then check into the array:

$PICTURE["img1.gif"] = true;
$PICTURE["img2.gif"] = true;
$PICTURE["img3.gif"] = true;

if ($PICTURE["$f"] == true)
allow the display of the picture.
else
looks like a hacker

by: PHPaulPosted on 2002-11-21 at 05:51:52ID: 7478187

That is indeed what I meant :-D

--paul

by: KarveRPosted on 2002-11-21 at 07:52:39ID: 7478857

I would suggest validation of REQUEST_URI as well, not a local request, don't serve it.

by: RenatoCotoPosted on 2002-11-22 at 06:12:54ID: 7483452

Hello to all,

and thanks for your comments.

I'm trying your suggestions, but I don't seem to be able to make <img src="images.php?f=pictures.gif"> instruction to display the image.

Here's the complete example code I'm using to test, all in the same http root directory.

test.htm page:
[it shows the same picture to test, the first normal call does shows it, but not the suggested call]

<html>
<body>

<img src="images/is.gif">
 
<img src="images.php?f=is.gif">

</body>
</html>

images.php page:
[just modified the $PATH to the images directory]

<?php
$PATH = "/images";
if (strstr($f, ".gif"))
{ $d = fopen("PATH/$f","r");
if (!$d)
{ // If file cannot be read or not available ...
header("Location: index.htm");
}
else
{
header('Content-Type: image/gif');
header("Content-Length: ".filesize("$PATH/$f"));
header('Content-Disposition: inline; filename=$f);

$data="";
while (!feof($d)) $data.=fread($d,12400);
fclose($d);
print $data;
}
}
else
header("Location: index.htm");
?>

I really thanks all your further help.

Renato Coto

by: PHPaulPosted on 2002-11-22 at 06:48:54ID: 7483608

This works for me:

images.php:
<?php
//images.php
$path = "images";

$fd = fopen ("$path/$imagename", "rb", 1);
$data = fread($fd, filesize("$path/$imagename"));
fclose ($fd);
print $data;
?>

index.html:
<html>
<body>
<img src="images/image.gif">
 
<img src="images.php?imagename=image.gif">
</body>
</html>

Hope this helps

--Paul

by: us111Posted on 2002-11-22 at 08:10:52ID: 7484033

my code was just an idea.
the same code without errors....

<?php
$PATH = ".";
if (strstr($f, ".gif"))
{ $d = fopen("$PATH/$f","r");
if (!$d)
{ // If file cannot be read or not available ...
header("Location: index.htm");
}
else
{
header ("Content-type: application/octet-stream");
header("Content-Disposition: inline; filename=$PATH/$f");

readfile("$PATH/$f");
}
}
else
header("Location: index.htm");
?>

by: RenatoCotoPosted on 2002-11-22 at 13:49:18ID: 7485566

Ok, I tested both ways and I was able to make it to work with the following code.

--- Please, if you can add the KarveR suggest validation using REQUEST_URI to only serve local requests, or requests from an specific page ---

test.htm page:

<html>
<body>
<img src="images.php?imagename=one.gif">
 
<img src="images.php?imagename=two.gif">
 
<img src="images.php?imagename=three.gif">
</body>
</html>

images.php page:

<?php
$path = "images";

$PICTURE["one.gif"] = true;
$PICTURE["two.gif"] = true;
$PICTURE["three.gif"] = true;

if ($PICTURE["$imagename"] == true){

$fd = fopen ("$path/$imagename", "rb", 1);
$data = fread($fd, filesize("$path/$imagename"));
fclose ($fd);
print $data;
}
?>

RenatoCoto

by: PHPaulPosted on 2002-11-22 at 14:32:05ID: 7485716

I see that you have used my way...

Why do you want the request uri check??, it's allready _pretty_ safe as it is!
If you don't want anybody to be able to look in your images folder just insert a page index.php in that folder, with the following code in it:

<?php
header("HTTP/1.0 404 Not Found");
?>

This will return a 'Page Not Found 404 error'. This way people will _NOT_ be able to see your images in any way, or ever find out the folder they are in.

The tips we gave will ensure that nobody will find out where your images are located!

Btw. Just curious, why do you want all this protection, anyone can just right-click on your images (in the lay-out of your pages) to save them to their hard-disk ;-)

I hope this is finally satisfying :-)

--Paul

by: RenatoCotoPosted on 2002-11-22 at 16:33:37ID: 7486019

The thing is Paul,
I run this online network that share some automatically generated images, each time there is a browser request.
And since they are the same for several sites, I don't want to reveal the real and only source of generation.

Otherwise, I'd have to implement the image creator engine on each site.

Now the, I need also the local only URI check to also protect the images being requested through the images.php from another location. You see, you can still link to this page and generate the images.

Thanks a lot for the code you provided, it's very useful and I'd like to finish this question by adding the URI check if you know how.

Cheers.

RenatoCoto

by: RenatoCotoPosted on 2002-11-22 at 16:41:51ID: 7486046

by: RenatoCotoPosted on 2002-11-22 at 17:05:09ID: 7486116

Something I just found.

This code is working only with locally stored images.
In my case, if I need to read an image that is on a different server, accross the network, where the path is an internet path such as: http://mydomain/images then this script will fail.

Is there a way to do it?

thanks

RenatoCoto

PS. Sorry for the previous doble comment, I don't know what happend.

by: PHPaulPosted on 2002-11-23 at 04:22:04ID: 7486935

Nope, at least not like we showed you.
You can't read a file from another server, if this were possible, you could read the contents of a .php file :-) (the actual code)

And we don't want that, do we ;-)

--Paul

by: us111Posted on 2002-11-23 at 04:31:59ID: 7486942

use my last code posted and change the path by a URL

by: hagermanPosted on 2002-11-24 at 22:17:45ID: 7492380

The best way to do this is probably using an Apache .htaccess file. There is an article available at:

http://apache-server.com/tutorials/ATimage-theft.html

by: RenatoCotoPosted on 2002-11-25 at 06:49:52ID: 7494009

Yes us111,
you got the remote image path solved with your answer.

Thanks a lot for the code you provided, and I'd like to finish this question by adding the requesting URI check to verify it's an specific or local page calling the script and I'm sure you know how.

RenatoCoto

by: RenatoCotoPosted on 2002-11-25 at 06:59:59ID: 7494072

by: us111Posted on 2002-11-25 at 07:05:51ID: 7494104

$thisfile = $_SERVER["PHP_SELF"];

if (basename($_SERVER["REQUEST_URI"] == $thisfile)
{
ok
}

you can also test :
- $_SERVER["SERVER_NAME"]
- $_SERVER["HTTP_REFERER"]

<?phpinfo()?> if you want to see all server's variables.

by: RenatoCotoPosted on 2002-11-25 at 09:01:56ID: 7494808

here's the final code we all in this question were able to solve.
This example will help other to better undestand the issue.

Cheers
RenatoCoto

page.htm code:

<html>
<body>
<img src="images.php?imagename=one">
 
<img src="images.php?imagename=two">
</body>
</html>

images.php code:

<?php
if ($_SERVER["HTTP_REFERER"] == "http://www.domain.com/page.htm"){

$PATH = "http://www.domain.com/images";

switch(@$imagename){
case "one":
$imagename = "imagetest1.png";
break;
case "two":
$imagename = "imagetest2.png";
break;
}

if (strstr($imagename, ".png")){
$d = fopen("$PATH/$imagename","r");
if (!$d){ // If file cannot be read or not available ...
header("Location: index.htm");
}
else{
header ("Content-type: application/octet-stream");
header("Content-Disposition: inline; filename=$PATH/$imagename");

readfile("$PATH/$imagename");
}
}
else
header("Location: index.htm");

}
?>

20120131-EE-VQP-002

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How to hide an Image Source?

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PHP Scripting Language

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