Re: [xsl] xsl-for-each-group in xslt 2.0

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Subject: Re: xsl-for-each-group in xslt 2.0
From: Martin Honnen <Martin.Honnen@xxxxxx>
Date: Mon, 28 Sep 2009 17:03:00 +0200

a kusa wrote:

My XML is as follows:
<list1>
<listitem>A</listitem>
<listitem>B</listitem>
<list2>
<listitem>C</listitem>
<listitem>D</listitem>
</list2>
</list1>
Every <listX><listitem> combination corresponds to one step.

So I want this to transform into:
<step1>A B</step1>
<step2>C D</step2>

I am not sure you need xsl:for-each-group, for the above input (wrapped in a root element) the following stylesheet creates above output (wrapped in a root element):

<xsl:stylesheet
  xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
  version="2.0">

<xsl:output indent="yes"/>

  <xsl:template match="root">
    <xsl:copy>
      <xsl:apply-templates/>
    </xsl:copy>
  </xsl:template>

  <xsl:template match="*[matches(local-name(), 'list[0-9]+')]">
    <xsl:element name="step{substring-after(local-name(), 'list')}">
      <xsl:value-of select="listitem"/>
    </xsl:element>
    <xsl:apply-templates/>
  </xsl:template>

<xsl:template match="text()"/>

</xsl:stylesheet>

--

	Martin Honnen
	http://msmvps.com/blogs/martin_honnen/

Current Thread
xsl-for-each-group in xslt 2.0 a kusa - 28 Sep 2009 14:45:36 -0000 Andrew Welch - 28 Sep 2009 15:01:28 -0000 Martin Honnen - 28 Sep 2009 15:03:29 -0000 <=

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Re: xsl-for-each-group in xsl, Andrew Welch	Thread	XSL-FO beginner's question, mark bordelon
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