[xsl] Template match via xsl:param

Subject: [xsl] Template match via xsl:param
From: Mario Lang <mlang@xxxxxxxxxxx>
Date: 14 Jan 2002 22:03:40 +0100

I am trying to write some generic stylesheets which I can stack in filterproxy.

>From the archive/FAQ I figure that this isnt as easy as one may think, but
I couldnt find a simple example which illustrates how to do it right.

So, here is what I am trying to do:

<?xml version="1.0"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"; version="1.0">
  <xsl:output method="html" indent="yes"
  <xsl:param name="path"></xsl:param>
  <xsl:include href="identity.xsl"/>

  <xsl:template match="$path"/>


Can anyone give me a simple example on how to achieve this without
genereting the xsl file on the fly?

   Mario <mlang@xxxxxxxxxxx>

 XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list

Current Thread