Making a php driven granted button

cemlouis

I asked this first in devshed forums but no result anyway,

Hello,

I am trying to send a granted or failed image to a remote domain which is using the html code to trigger the php code on host domain. Let me explain:

There are 2 domains:

www.hostdomain.com (hosting the php code)
www.remotedomain.com (running/calling the php code which is in hostdomain.com)

Below is the code which is on hostdomain.com -> grantedorfailed.php

<?php

// File name: grantedorfailed.php
// ***Remote domain checker***
// Displays a granted image if the script is running on www.remotedomain.com else displays a failed image

if ($_SERVER['HTTP_HOST'] == "www.remotedomain.com")
{
	$html = '
		<div align="center"><a href="http://www.hostdomain.com"><img src="http://www.hostdomain.com/granted.gif" border="0"></a></div>
		';
}
else{
	$html = '
		<div align="center"><a href="http://www.hostdomain.com"><img src="http://www.hostdomain.com/failed.gif" border="0"></a></div>
		';
}

echo $html;
?>

And below is the html code which is on remotedomain.com , Let's say the file is index.php(doesn't matter)

<html>
<head>
</head>
<body>
Status: <a href="http://www.hostdomain.com"><img src="http://www.hostdomain.com/grantedorfailed.php" border="0"></a>
</body>
</html>

So what I am trying to do is: when someone displays the index.php on remotedomain.com the file will trigger the grantedorfailed.php on hostdomain.com and the code should display granted image file which is also in hostdomain.com. If the domain name is not www.remotedomain.com then should display the failed image

The above returns a white screen no syntax errors or etc, probably the $_SERVER['HTTP_HOST'] isn't the right thing to get the domain name or? any ideas?

Thank you,

ClarkF1

Well from what I can see, running that you might get

<html> 
<head> 
</head> 
<body> 
Status: <a href="http://www.hostdomain.com"><img src="<div align="center"><a href="http://www.hostdomain.com"><img src="http://www.hostdomain.com/granted.gif" border="0"></a></div>" border="0"></a> 
</body> 
</html>

which would make no sense as you have more than the link to the filename as the src of the file.

It might be that it's actually expecting an image to be produced by grantedorfailed.php

What is the source code you actually get when you run the script and view the source of the resulting page....ie. what's in the img src tag

cemlouis

When I run mine, code isn't generating the image. A broken picture with the correct link to hostdomain.com appears. The source:

<a href="http://www.hostdomain.com"><img src="http://www.hostdomain.com/grantedorfailed.php" border="0"></a>

I think the php can't parsing the image?

By the way I noticed that this is something like the stat buttons or pagerank buttons. I searched a sample code on google but no result...

Anyway, any ideas?

Weedpacket

Well, <img> tags expect an image to be on the other end of their src URL, not HTML. If you just passed the GIF file itself (and why not a PNG? Nevermind...) instead of putting a URL to it in another <img> tag inside a <div>, you'd probably have more success. Don't forget the appropriate Content-Type [man]header[/man].