[xsl] How to transform xml and xsl to a php format

Subject: [xsl] How to transform xml and xsl to a php format
From: "Ard Schrijvers" <Ard@xxxxxxxx>
Date: Thu, 23 Jan 2003 10:59:38 +0100
I am serializing my xml input with an xsl , and want to get an output like:

<?		
if(!auth()){
	header("Location: /Home/index.html"); 
	exit; 
}
function auth(){  
	$valid = false;
  	session_start();  
	if (!isset($_SESSION['valid'])) {
    	$valid = false;
	} else {
	    $valid = $_SESSION['valid'];
	}
    return 	$valid;
}  
?>

<html>
	<head>
		etc.............

So, at the beginning , I want to have <? ... some code ?>

I tried it in various way, ending up with essentially the same problem:
When i try it , like for example below:

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:hc="http://www.hippo.nl/xml/hippocontent"; xmlns:xhive="http://hippo.nl/xhive/1.0"; xmlns:xsl="http://www.w3.org/1999/XSL/Transform"; xmlns:fo="http://www.w3.org/1999/XSL/Format"; xmlns:res="http://www.hippo.nl/xml/resource"; xmlns:xhtml="http://www.w3.org/1999/xhtml";>
	<xsl:param name="current"/>
	<xsl:param name="php"/>
	<xsl:template match="/">
		<xsl:if test="$php = 'true'">
<![CDATA[
<?		
if(!auth()){
	header("Location: /Home/index.html"); 
	exit; 
}
function auth(){  
	$valid = false;
  	session_start();  
	if (!isset($_SESSION['valid'])) {
    	$valid = false;
	} else {
	    $valid = $_SESSION['valid'];
	}
    return 	$valid;
}  
?>
]]> 



		</xsl:if>
		<html xmlns="http://www.w3.org/1999/xhtml";>
			<xsl:comment>  ETC................................

The output generated, will look like : &gt;? if(!auth()){
												header("Location: /Home/index.html"); 
					exit; 
					........
					etc.....

					?&lt;
	and then the html code. I am not able to get the output like: <? ......... ?>. 

Is there anybody who knows how to do this??? 

kind regards,

Ard Schrijvers

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Fax:   0031-(0)20-6345179
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