Aravnah Ornan

If you maintain a Web site, have you ever had a situation in which people, without authorization, use your graphics and don't even bother copying them to their own servers, but instead simply identify the image source on your server? If so, how do you handle this? Do you rename the affected image file and then put up an embarrassing image (e.g., a sad clown) under the old file name? Do you write to that person's Web hosting company and then just wait to be ignored or told off? Do you simply let it slide and accept the implied compliment to your creativity?

Nostalgic Pushhead

If I ever had that problem, that's what I'd do

Pyrrho

I think what you should do depends on who is doing it, and what, exactly, they are doing. You can take legal action against them, which would only be practical if they have money and are in a country that would take the action seriously. I doubt that you are dealing with cases like this.

For a personal web site, you might first try sending them an email. If they don't respond promptly (i.e., within a week), then I would probably rename the file, and put something else as the file to which they have a link, though I would probably use something more embarrassing than a "sad clown". You could have a graphic that insults the person who maintains the web site, though you would want to make sure you insult them in a legal way. It could probably say something like "The person who maintains this site is an idiot", as the referent would really be to yourself, though you would not have it appear on any of your own web pages. It would, however, appear to refer to the thief who is stealing your property.

DarkBronzePlant

Depending on the tools you have at your disposal, you could whip up a system by which your images are renamed (maybe according to the current date or something), and the SRC value of your IMG tags have the value dynamically inserted.

Say every day, you change the name of your images directory; maybe a cron job or something renames it to

/images20030515/

and stores that name somewhere. Then you use PHP or something like that to write out your img tags:

<img src="<? print(img_dir_name) ?>/myimage.gif">

A bit of work, but it would do the trick.

Vorkosigan

This looks like it has development potential...

Behold! A thread is promoted out of Elsewhere

Shadowy Man

I've always wondered how to do this? Could you describe this procedure in more detail?

Sakpo

Figure out how to do what www.somethingawful.com does when someone does that to them.

AquaVita

Lol... SA is brutal to pic leechers.

DarkBronzePlant

Sure. Bear in mind that I haven't done this myself; I'm just outlining what I would do if I found myself running into the problem you describe.

I'm going to assume a setup involving UNIX as the platform and PHP as the server-side development environment/language used. Of course, it can be easily be adapted to other environments. I will also assume that you do not have permission on your machine to set up things like cron jobs, since you very well may not.

First, be sure that all of your images--at least the ones that are being linked to by other sites--are located within a set of subdirectories like:
/img/images-1234/
Of course, you can have subdirectories like:

/img/images-1234/about/
/img/images-1234/contact/

but they should all ultimately have the same parent directories.

Now, create a .php file that you can include in all of the pages on your site (in doing this, all of your pages that contain images must be .php files, or .jsp if you are using Java, or .asp if you are using micro$oft...) We'll call this new file imageget.php .

Within imageget.php, create a function that looks approximately like this (again, I am just quickly whipping this up; since I am not testing it, my code will probably be riddled with mistakes):

<?php
function img_dir() {
_exec( "ls -la /img",$lines,$rc);
_$count = count($lines) - 1;
_for ($i = 1; $i <= $count; $i++) {
_____$type = substr($lines[$i],0,1);
_____$name = strrchr($lines[$i], " ");
_____$name = substr($name,1);
$pos = strpos($name, "images-");
if ($pos == 0) {
return $name;
}
}
return "images";
}
?>

What that does is to list out all of the files/directories within the /img directory. It then loops through them. Given that you (should have) only put your images-1234 subdirectory in there, it won't need to do much looping. Anyhow, as it loops through, it checks for a file or directory whose name begins with "images-" and, once it finds it, returns that name.

So your job will be to occassionally change the name of images-1234 to something else, always ensuring that it begins with "images-". Your PHP method will thus always be able to find it.

Okay, so you make sure to include imageget.php in all of your PHP pages by putting this at the top of every page:

<?php include "imageget.php"; ?>

Now all you need to do is make all of your HTML image tags look like this:

<img src="<?php print(img_dir()); ?>/about/mypicture.jpg">
The correct URL will automatically be printed out by PHP. But any web site who links to your images, well, as soon as you change the image directory's name, they will encounter a broken link.

How you change the image directory's name is up to you, but it should be easy enough for you to do manually once in awhile.

The problem, of course, is that there is a lot of overhead involved in this. There are definitely ways to reduce this overhead, such as maybe listing the name of the images directory in a variable in a PHP file, and then writing a separate script that actually rewrites that PHP file to change the name of the variable periodically. That would eliminate the overhead of listing out the directory contents and looping through them.

Anyhow, hope that wasn't too confusing, and that it helps somewhat.

Shadowy Man

Thanks. I'll have to review it. I haven't worked with PHP stuff before.