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


just change your first template match to 	<xsl:template match="/root">

your param was being passed to the default template match for root and not
to x as you wanted.

hope this helps,

Conor


At 11:04 AM 6/15/01 +0200, you 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
>   { @   @ }
>        ^
>      \__/
>
>"You don't need eyes to see, you need vision!"
>
>
>
> XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list
>
>
---------
Conor Ryan,
XML Workshop Ltd,
10 Greenmount Industrial Estate, Harolds Cross, Dublin 12, IRELAND.
Email: cryan@xmlw.ie
Phone: +353 1 4547811; Fax: +353 1 4731626
Web: http://www.xmlw.ie/


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