Hi Mike

See if you cant get some sense out of this: http://www.jenitennison.com/xslt/grouping/muenchian.html

Another method would be to call a second template to do the year logic based on some passed parameters.

Let us know how you do :)

For whomever stumbles upon this at a future date, I've seemingly had success with the "Grouping with xsl variable" outlined here: http://docstore.mik.ua/orelly/xml/xslt/ch06_02.htm

I had actually encountered the same problem with the "Our First Attempt" mentioned in the article, which it did a good job of explaining. If you're working with several nodes, I'd recommend you follow the suggestion of the article and try for the Muench method (similar to what Ernst Utvik posted above - thanks Ernst).

Anyways, here's my XSLT:

<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE xsl:stylesheet [ <!ENTITY nbsp "&#x00A0;"> ]>
<xsl:stylesheet
  version="1.0"
  xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
  xmlns:msxml="urn:schemas-microsoft-com:xslt"
  xmlns:umbraco.library="urn:umbraco.library" xmlns:Exslt.ExsltCommon="urn:Exslt.ExsltCommon" xmlns:Exslt.ExsltDatesAndTimes="urn:Exslt.ExsltDatesAndTimes" xmlns:Exslt.ExsltMath="urn:Exslt.ExsltMath" xmlns:Exslt.ExsltRegularExpressions="urn:Exslt.ExsltRegularExpressions" xmlns:Exslt.ExsltStrings="urn:Exslt.ExsltStrings" xmlns:Exslt.ExsltSets="urn:Exslt.ExsltSets"
  exclude-result-prefixes="msxml umbraco.library Exslt.ExsltCommon Exslt.ExsltDatesAndTimes Exslt.ExsltMath Exslt.ExsltRegularExpressions Exslt.ExsltStrings Exslt.ExsltSets ">


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

<xsl:param name="currentPage"/>
<xsl:variable name="parentNode" select="1410"/>

    <xsl:template match="/">
      <h2>News Archive</h2>
        <xsl:for-each select="umbraco.library:GetXmlNodeById($parentNode)/newsArticle [@isDoc]">
          <xsl:sort select="date" order="descending"/>
            <xsl:variable name="previousYear" select="umbraco.library:FormatDateTime(date, 'yyyy')"/>
            <xsl:if test="not(preceding-sibling::newsArticle[umbraco.library:FormatDateTime(date, 'yyyy')=$previousYear])">
              <xsl:value-of select="umbraco.library:FormatDateTime(date, 'yyyy')"/>
              <xsl:for-each select="umbraco.library:GetXmlNodeById($parentNode)/newsArticle[umbraco.library:FormatDateTime(date, 'yyyy')=$previousYear]">
                <xsl:sort select="date" order="descending"/>
                <div class="newsArticle">
                  <span class="title"><a href="{umbraco.library:NiceUrl(@id)}"><xsl:value-of select="title"/></a></span>
                  <span class="date"><xsl:value-of select="umbraco.library:FormatDateTime(date, 'MMMM d, yyyy')"/></span>
                </div>
              </xsl:for-each>
            </xsl:if>
       </xsl:for-each>
    </xsl:template>
</xsl:stylesheet>

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