XSLT: How to select multiple sub nodes into a single list

Question

How can I extract sub nodes to a single list for processing in a template?

Consider the following XML.

<document>
  <menu></menu>
  <feature>
    <header>Header 1</header>
    <article>article 1</article>
    <article>article 2</article>
  </feature>
  <feature>
    <header>Heading 2</header>
    <article>article 1a</article>
    <article>article 2a</article>
  </feature>  
</document>

I'd like to extract all the article nodes to a single list for processing in a template.

I need article nodes to be available at once, because I need to do calculations based on the number of articles there are.

Joel M. Lamsen · Accepted Answer · 2014-03-27 08:06:06Z

1

you can try the following stylesheet:

<xsl:stylesheet version="1.0"
    xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
    xmlns:ext="http://exslt.org/common"
    exclude-result-prefixes="ext">

    <xsl:output indent="yes" omit-xml-declaration="yes"/>

    <xsl:template match="/">
        <xsl:variable name="list">
            <articles>
                <xsl:copy-of select="descendant::article"/>
            </articles>
        </xsl:variable>

        <xsl:variable name="vPass1" select="ext:node-set($list)"/>

        <xsl:apply-templates select="$vPass1/*"/>

    </xsl:template>

    <xsl:template match="articles">
        <xsl:copy>
            <xsl:text>Number of articles: </xsl:text><xsl:value-of select="count(article)"/>
        </xsl:copy>
    </xsl:template>
</xsl:stylesheet>

when applied your input, it produces:

<articles>Number of articles: 4</articles>

answered Mar 27, 2014 at 8:06

Joel M. Lamsen

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1 Comment

Reuben Over a year ago

Accepting this answer, but what I ended up doing was a select on the document element, and a for-each on feature/article. No reliance on extensions, low tech, but I understand that for-each can be frowned upon.

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