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Re: Passing a list of arguments

Subject: Re: Passing a list of arguments
From: "Mukul Gandhi" <gandhi.mukul@xxxxxxxxx>
Date: Fri, 9 Nov 2007 18:46:22 +0530
Re:  Passing a list of arguments
On 11/9/07, Mathieu Malaterre <mathieu.malaterre@xxxxxxxxx> wrote:
> test.xml:
> <?xml version="1.0" encoding="UTF-8"?>
> <article>
> </article>
>
> test.xsl:
> <?xml version="1.0" encoding="UTF-8"?>
> <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="2.0">
>  <xsl:output method="xml" indent="yes" encoding="UTF-8"/>
>  <xsl:variable name="sections-list">
>    <section>
>      <section>C.8.7.1.1.2</section>
>      <section>C.8.14.1.1</section>
>      <!-- ... -->
>    </section>
>  </xsl:variable>
>  <xsl:template match="article">
>    <xsl:param name="extract-section"/>
>    <xsl:message>
>      <xsl:value-of select="$extract-section"/>
>    </xsl:message>
>  </xsl:template>
>  <xsl:template match="/">
>    <xsl:for-each select="$sections-list">
>      <xsl:apply-templates select="article">
>        <xsl:with-param name="extract-section" select="$section"/>
>      </xsl:apply-templates>
>    </xsl:for-each>
>  </xsl:template>
> </xsl:stylesheet>

You seem to be missing something.

I think, the root template should be:

 <xsl:template match="/">
   <xsl:for-each select="$sections-list/section/section">
     <xsl:apply-templates select="article">
       <xsl:with-param name="extract-section" select="."/>
     </xsl:apply-templates>
   </xsl:for-each>
 </xsl:template>

It's not clear to me what exactly is your requirement. But you seem to
be writing a wrong XPath expression.

-- 
Regards,
Mukul Gandhi

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