Question

Given the following XML:

<?xml version="1.0"?>
<Things>
  <Thing>
    <Thing ID="0002"/>
    <Name>Bob</Name>
  </Thing>
  <Thing>
    <Thing ID="0003"/>
    <Name>Alice</Name>
  </Thing>
  <Thing>
    <Thing ID="0001"/>
    <Name>Carol</Name>
  </Thing>
</Things>

I want to output the same XML sorted by the ID attribute. The following stylesheet does what I want. The commented-out xsl:sort (and several variations that I tried) doesn’t work — I just get an unsorted copy of the original document. (And no error messages.)

How can I specifically select the ID attribute on the Thing element (to avoid using ID attributes on other elements that may be present in a larger document)?

<xsl:stylesheet
    xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
    version="1.0">

 <xsl:template match="node()|@*">
   <xsl:copy>
     <xsl:apply-templates select="node()|@*" />
   </xsl:copy>
 </xsl:template>

 <xsl:template match="Things">
   <xsl:copy>
     <xsl:apply-templates>
       <xsl:sort select="*/@ID" data-type="number"/>

       <!-- I don't understand why this doesn't work:

         <xsl:sort select="Thing/Thing[@ID]" data-type="number"/>
       -->
     </xsl:apply-templates>
   </xsl:copy>
 </xsl:template>

</xsl:stylesheet>

Answer 1

You want

    <xsl:sort select="Thing/@ID" data-type="number"/>

At the point where this occurs, the context is already at each first-level Thing, so the sort key is the @ID attribute of the second-level Thing.