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RE: Why is my xsl:param empty? (passed with xsl_with-param)
- To: <xsl-list at lists dot mulberrytech dot com>
- Subject: RE: [xsl] Why is my xsl:param empty? (passed with xsl_with-param)
- From: "Michael Kay" <mhkay at iclway dot co dot uk>
- Date: Fri, 15 Jun 2001 10:39:25 +0100
- Reply-To: xsl-list at lists dot mulberrytech dot com
> I'm playing around with xsl:with-param and xsl:param, but I
> cant get my param passed. What do I do wrong?
Your first template rule doesn't call the second one directly, it does so
via the built-in template for the <root> element. Built-in templates don't
pass their parameters on to the templates they call.
Mike Kay
> 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!"
>
>
>
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