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 : >? if(!auth()){ header("Location: /Home/index.html"); exit; ........ etc..... ?< 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 ------------------------------------------------------ Professional Services Department Hippo Webworks Grasweg 35 1031 HW Amsterdam The Netherlands Tel: 0031-(0)20-6345173 Fax: 0031-(0)20-6345179 http://www.hippo.nl ------------------------------------------------------ XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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