I am trying to dynamically create page links based on URL variables.

So if they come to "http://www.visual-directory.com/finance/carinsurance.php?sid=bob", I want the "sid=bob" to be added to some of my link url on the page.

That's the error that I get:

"Parse error: parse error, unexpected '.' in /home/virtual/site133/fst/var/www/html/finance/carinsurance.php on line 95"

Here is my code:

PHP Code:
  <img src="../images/stars/review-50.gif" width="54" height="12"><br> 
<? echo <a href="../tracking/count.php?url=www.anrdoezrs.net/rh79r09608OQTXYWPUOQPSVQYRS?sid="&"$sid" target="_top" onmouseover="window.status='https://www.insureme.com/';return true;" onmouseout="window.status=' ';return true;">; ?>
<img src="../images/affiliates/finance/insureme-car.gif" alt="Car insurance  discounters" width="120" height="60" border="0" align="left"></a> 
<p> <? echo <a href="../tracking/count.php?url=www.anrdoezrs.net/rh79r09608OQTXYWPUOQPSVQYRS?sid="&"$sid"  target="_top" onmouseover="window.status='https://www.insureme.com/';return true;" onmouseout="window.status=' ';return true;" class="titleadword">;?>Provides Free Auto Insurance Quotes - will receive up to 5 quotes.</a>
Can anyone tell me what I did wrong?

I am a newbi at PHP so I'd appreciate it if you could tell me, using simple terms.