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: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 __________________________________________________ Do You Yahoo!? Spot the hottest trends in music, movies, and more. http://buzz.yahoo.com/ XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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