RE: easy distinctive sorting?

Subject: RE: easy distinctive sorting?
From: "Albert, Kevin" <kjalbert@xxxxxxxxxxxxxxxxxxxxx>
Date: Mon, 31 Jul 2000 12:04:11 -0400
Here is a stylesheet that was posted to the list a while back by Mark Volkman, which I think will help you .... the key is in testing whether the context node is equivalent to preceding-sibling.

Regards,
Kevin Albert

-----------------------------
<?xml version="1.0"?>

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

  <xsl:template match="/">
    <h2>Original List</h2>
    <xsl:apply-templates select="list"/>

    <h2>Sorted List w/ No Duplicates</h2>
    <xsl:apply-templates select="list" mode="sort-nodup"/>
  </xsl:template>

  <xsl:template match="list">
    <xsl:apply-templates select="item"/>
  </xsl:template>

  <xsl:template match="list" mode="sort-nodup">
    <xsl:apply-templates select="item[not(.=preceding-sibling::item)]">
      <xsl:sort/>
    </xsl:apply-templates>
  </xsl:template>

  <xsl:template match="item">
    <div>
      <xsl:value-of select="."/>
    </div>
  </xsl:template>

</xsl:stylesheet>
-----------------------------








-----Original Message-----
From: Hugo de Groot [mailto:H.de.Groot@xxxxxxxxx]
Sent: Monday, July 31, 2000 9:32 AM
To: xsl-list@xxxxxxxxxxxxxxxx
Subject: easy distinctive sorting?


Hi,

One of my XSL stylesheets should output sorted elements, with an extra difficulty that there can be several identical elements, which should be included only once. If I would use SQL I'd probably use the statement 'SELECT DISTINCT'. I haven't found an easy way of doing this in XSL. Does anyone have a suggestion (apart from writing a custom template)?

Hugo De Groot
X punt X b.v. IT Services



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