RE: [xsl] A challenge.. Group Periods of Data (1..5, 2..8, 4..9) (10..12; 10..14)

["Andrew Welch" <ajwelch@xxxxxxxxxxxxxxx>]

Here my quick attempt at this one...

(I've been off-list for a while and deleted most of the early posts of
this long-thread, so apologies in advance if I've missed anything)

It uses a nested for-each to find the groups, rather than recursion.
It's not as elegant as David's, or as performant, but possibly easier to
understand :)

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

<xsl:output indent="yes"/>

<xsl:template match="A">
  <result>

    <!-- Get the end of each group -->
    <xsl:for-each select="B[@period_end &lt;
following-sibling::B[1]/@period_begin or not(following-sibling::B)]">
      <xsl:variable name="end" select="@period_end"/>
      <xsl:variable name="endOfGroup" select="generate-id()"/>

    	<!-- Get the start for this end -->
      <xsl:for-each select="(preceding-sibling::B[@period_begin &gt;
preceding-sibling::B[1]/@period_end or
not(preceding-sibling::B)])[last()]">
        <period begins="{@period_begin}" end="{$end}">
          <!-- Copy all elements between start and end -->
          <xsl:copy-of select=".|following-sibling::B[generate-id() =
$endOfGroup or following-sibling::B[generate-id() = $endOfGroup]]"/>
        </period>
      </xsl:for-each>

    </xsl:for-each>
  </result>
</xsl:template>

</xsl:stylesheet>


Cheers
andrew