Re: Why is my xsl:param empty? (passed with xsl_with-p

Hi Rene,

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

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
template.

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.

Cheers,
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

$trythis: 
$trythis: 

xml:
<root>
	<x>bla</x>
	<x>blabla</x>
</root>

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

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

</xsl:stylesheet>

Greetings Rene



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