Muenchian Grouping without xsl:for-each

Question

Just for grins I wanted to see if I could rewrite a Muenchian Grouping template in XSLT 1.0 without using <xsl:for-each>. The XML:

<jobs>
  <job>
    <year>2012</year>
    <position>Mayonnaise Maker, Malden Mayonnaise Manufactory, Malden, MA</position>
  </job>
  <job>
    <year>2012</year>
    <position>Twine Twirler, Timmy's Twine, Tyngsboro, MA</position>
  </job>
  <job>
    <year>2013</year>
    <position>Bagel Boiler, Bob's Bagels, Boxboro, MA</position>
  </job>
</jobs>

Standard Muenchian solution, grouping positions by year:

<?xml version="1.0"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:key name="this-year" match="job" use="year"/>
  <xsl:template match="jobs">
    <resume>
      <xsl:for-each select="job[count(. | key('this-year', year)[1]) = 1]">
        <year y="{year}">
          <xsl:for-each select="key('this-year', year)">
            <position><xsl:value-of select="position"/></position>
          </xsl:for-each>
        </year>
      </xsl:for-each>
    </resume>
  </xsl:template>
</xsl:stylesheet>

Result:

<?xml version="1.0"?>
<resume>
  <year y="2012">
    <position>Mayonnaise Maker, Malden Mayonnaise Manufactory, Malden, MA</position>
    <position>Twine Twirler, Timmy's Twine, Tyngsboro, MA</position>
  </year>
  <year y="2013">
    <position>Bagel Boiler, Bob's Bagels, Boxboro, MA</position>
  </year>
</resume>

It's easy to get rid of the outer <xsl:for-each>. This works:

<?xml version="1.0"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:key name="this-year" match="job" use="year"/>

  <xsl:template match="jobs">
    <resume>
      <xsl:apply-templates select="job[count(. | key('this-year', year)[1]) = 1]"/>
    </resume>
  </xsl:template>

  <xsl:template match="job">
    <year y="{year}">
      <xsl:for-each select="key('this-year', year)">
        <position><xsl:value-of select="position"/></position>
      </xsl:for-each>
    </year>
  </xsl:template>
</xsl:stylesheet>

This, however...

<?xml version="1.0"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:key name="this-year" match="job" use="year"/>

  <xsl:template match="jobs">
    <resume>
      <xsl:apply-templates select="job[count(. | key('this-year', year)[1]) = 1]"/>
    </resume>
  </xsl:template>

  <xsl:template match="job">
    <year y="{year}">
      <xsl:apply-templates select="key('this-year', year)"/>
    </year>
  </xsl:template>

  <xsl:template match="key('this-year', year)">
    <position><xsl:value-of select="position"/></position>
  </xsl:template>

</xsl:stylesheet>

...does not:

error
xsltCompileIdKeyPattern : Literal expected
compilation error: file no-dice.xsl line 17 element template
xsltCompilePattern : failed to compile 'key('this-year', year)'

<xsl:template match="key('this-year', year)"> is valid but it throws that error. Is there some way to push the processing on that key out to another template? Or some other trick to make this work?

Martin Honnen Martin Honnen · Accepted Answer · 2015-07-29T11:32:33

Use a mode:

  <xsl:template match="job">
    <year y="{year}">
      <xsl:apply-templates select="key('this-year', year)" mode="group"/>
    </year>
  </xsl:template>

  <xsl:template match="job" mode="group">
    <position><xsl:value-of select="position"/></position>
  </xsl:template>

Of course, as long as you simply want to copy the position elements you can do that with

  <xsl:template match="job">
    <year y="{year}">
      <xsl:copy-of select="key('this-year', year)/position"/>
    </year>
  </xsl:template>

Muenchian Grouping without xsl:for-each

1 Answers