Description:
------------
I wanted to build a website with an structure as easy as possible. So i decided to use a php script displaying different contents on the same site by using querys.

BTW: Sorry for my english, it may be a bit lame, but I am German so just try to understand my special germenglish.

Reproduce code:
---------------
<?php
     if ($site == "home") {
     echo "<h2>Willkommen auf der Webseite von TM-Servers
     <br>Diese Webseite wurde zum Zweck der Information...
     ...tm-server.de.vu ging heute online.";
     if ($site == "server") {
     echo "Serverdaten folgen in K&uuml;rze!";
     }
     if ($site == "members") {
     echo "Mitgliederliste folgt in K&uuml;rze!";
     }
     if ($site == "links") {
     echo "Hier werden demn&auml;chst sehenswerte Links
     gesammelt!";
     }
     if ($site == "partner") {
     echo "Demn?chst werden hier alle Partner und Sponsoren
     von TM-Serversaufgef&uuml;hrt!";
     }
?>

Expected result:
----------------
I expected my website to open regularly and when clicking a link containing a query
(for example: "index.php?site=members" to display the content defined by the echo function (for example: "Mitgliederliste folgt in K?rze!")

Actual result:
--------------
Parse error: parse error, unexpected $ in /home/export/www/vhosts/funnetwork/hosting/cheetah7/index.php on line 97


an important information is, that line 97 is the very end of index.php and just contains: "</html>" The php tag was already closed many lines before!