Re: [NEWBIE] Confused about usage of $_SERVER['DOCUMENT_ROOT']

Andre, Mirco, BDT, Thank you for your responses. Here is some test code <?php echo "<img src=\"inline/topimage.jpg\" >"; echo "<img src=\"/inline/topimage.jpg\" >"; echo "<img src=\"" . $_SERVER['DOCUMENT_ROOT'] . "/inline/topimage.jpg\" >"; ?>: Theoretically, they should all output the same image. However: <img src="inline/topimage.jpg" > ->IMAGE DISPLAYS <img src="/inline/topimage.jpg" > ->IMAGE DISPLAYS <img src="/home/sites/sitexxx/web/inline/topimage.jpg" > IMAGE DOES NOT DISPLAY Of course, the first instance only works if the script is in the correct place relative to the /inline directory. However, the second case fails if I use it in the following function (assuming I specify $dir as "/inline"). It returns a false result, as if it couldn't find the directory. function random_img($dir,$type='random') { global $errors,$seed; if (is_dir($dir)) { $fd = opendir($dir); $images = array();

     while (($part = @readdir($fd)) == true) { 
         if ( eregi("(gif|jpg|png|jpeg)$",$part) ) {
             $images[] = $part;
         }
     }

// adding this in case you want to return the image array if ($type == 'all') return $images; if ($seed !== true) {

     mt_srand ((double) microtime() * 1000000);
     $seed = true;

}

         $key = mt_rand (0,sizeof($images)-1);

return $dir . $images[$key]; } else {

     $errors[] = $dir.' is not a directory';
     return false;

} } -- Dave Gutteridge dave@tokyocomedy.com/