PapaFace.com

<img src="http://www.papaface.com/blog/wp-content/themes/papaface/images/beer.png" />
 

<img src="http://www.papaface.com/blog/somephpfile.php" />
 

<img src="http://www.papaface.com/blog/resources/dynamicimage/example.php?id=22451" />
 

<?php  
       
        $linkid = $_GET['id']; //this is the link id that the script needs to look for in the database
       
        //you would open a mysql connection here
       
        $get = @mysql_query("SELECT `link` FROM `tbllinks` WHERE `linkid`='".mysql_real_escape_string($linkid)."'"); //@ suppresses the error message (this is usually a bad coding practice, but in this case we do not want anything other than an image displayed to the user.

        list($url) = mysql_fetch_array($get);
        /*The code below checks the link. If it is a valid and working link it will return a 200 code.*/
        $ch = curl_init($url);
        curl_setopt($ch, CURLOPT_NOBODY, true);
        curl_setopt($ch, CURLOPT_HEADER, true);
        curl_setopt($ch, CURLOPT_FOLLOWLOCATION, true );
        curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
        curl_exec($ch);
        $info = curl_getinfo($ch, CURLINFO_HTTP_CODE);
        curl_close($ch);
        if      ($info == "200")
                $colour = "green";   //link is valid
        else
                $colour = "red"; //link is NOT valid
        }

    //code above decided if it should be green or red and produced $colour
   
        header("Content-type: image/jpg");
        $im     = imagecreatefromjpeg($colour.".jpg"); //tells the script what image to source locally (green.jpg/red.jpg are actual images in the directory)
        $orange = imagecolorallocate($im, 220, 210, 60);
        $px     = (imagesx($im) - 7.5 * strlen($string)) / 2;
        imagestring($im, 3, $px, 9, $string, $orange);
        imagepng($im);
        imagedestroy($im);

?>