[xsl] Re: Why is my xsl:param empty? (passed with xsl_with-param)

Subject: [xsl] Re: Why is my xsl:param empty? (passed with xsl_with-param)
From: Dimitre Novatchev <dnovatchev@xxxxxxxxx>
Date: Fri, 15 Jun 2001 02:44:14 -0700 (PDT)
Hi Rene,

In your template:
	<xsl:template match="/">
			<xsl:with-param name="trythis" select="8"/>

you are applying the templates to all children (and there's always one -- the top
element) of the root node.

It happens to be a node named "root", for which you haven't provided a matching

Then the XSLT default processing is carried on this node and as part of it a
xsl:apply-templates (without parameters!!!) is initiated for the children of "root".

But notice -- the default processing doesn't know about any parameters that need be
passed -- therefore your template matching "x" is applied without any parameters
being provided.

This explains the output you received.

Dimitre Novatchev.

Rene de Vries wrote:

I'm playing around with xsl:with-param and xsl:param, but I cant get my param
passed. What do I do wrong?
I expect:

$trythis: 8
$trythis: 8

but I get



<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform";;>
	<xsl:template match="/">
			<xsl:with-param name="trythis" select="8"/>

	<xsl:template match="x">
		<xsl:param name="trythis"/>
			$trythis: <xsl:value-of select="$trythis"/>


Greetings Rene

Do You Yahoo!?
Spot the hottest trends in music, movies, and more.

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

Current Thread