Re: a new(?) grouping problem

["Nikolai Grigoriev" <grig at renderx dot com>]

Mike,

> I want to group consecutive days with the same hours together, and just
> print the first and last day in each group.

The same solution as posted before, but slightly shorter ;-)

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

<xsl:output method="html" version="4.0"/>

<!-- Root template: just create a table. -->
<xsl:template match="Hours">
  <table><xsl:apply-templates/></table>
</xsl:template>

<!-- Exclude holidays from processing -->
<xsl:template match="Holidays" priority="2"/>

<!-- Single days, except for holidays. -->
<!-- A modeless template creates the row -->
<xsl:template match="Hours/*">
  <xsl:variable name="hours" select="text()"/>
  <xsl:if
      test="not(preceding-sibling::*[not(self::Holidays)][1][text()=$hours])">
    <tr>
      <td>
        <xsl:value-of select="name()"/>
        <xsl:apply-templates mode="end"

select="following-sibling::*[not(self::Holidays)][1][text()=$hours]"/>
      </td>
      <td><xsl:value-of select="."/></td>
    </tr>
  </xsl:if>
</xsl:template>

<!-- A day closes the period if there's no better candidate -->
<xsl:template match="Hours/*" mode="end">
  <xsl:variable name="hours" select="text()"/>
  <xsl:choose>
    <xsl:when
        test="following-sibling::*[not(self::Holidays)][1][text()=$hours]">
      <xsl:apply-templates mode="end"
          select="following-sibling::*[not(self::Holidays)][1]"/>
    </xsl:when>
    <xsl:otherwise>
      <xsl:text> - </xsl:text><xsl:value-of select="name()"/>
    </xsl:otherwise>
  </xsl:choose>
</xsl:template>

</xsl:stylesheet>




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