[xsl] using exsl:node-set()

Subject: [xsl] using exsl:node-set()
From: "Markus Hanel" <markus.hanel@xxxxxx>
Date: Mon, 26 Jul 2004 11:58:11 +0200 (MEST)
hallo,
I make something wrong! I want to use the exsl:node-set() function but at
www.exsl.org there is no download of the exsl.node-set.xsl stylesheet. But
in the main download there are two stylesheets: exsl.node-set.1.xsl and
exsl.node-set.2.xsl, but this not works.

markus

<?xml version="1.0" encoding="iso-8859-1"?>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
xmlns:exsl="http://exslt.org/common"; extension-element-prefixes="exsl">

<xsl:import href="exsl.node-set.1.xsl" />
<xsl:import href="exsl.node-set.2.xsl" />

<xsl:template match="/">
<html>
<head><title></title></head>

<body>
  <xsl:apply-templates select="./table" />
</body>
</html>
</xsl:template>

<xsl:template match="table">
<xsl:variable name="sorted-rows-rtf">
    <xsl:for-each select="./row">
      <xsl:sort select="sum(./cell/@points)" data-type="number"
order="descending" />
      <xsl:copy-of select="." />
    </xsl:for-each>
  </xsl:variable>

  <xsl:variable name="sorted-rows"
select="exsl:node-set($sorted-rows-rtf)/row" />
  <xsl:value-of select="sum($sorted-rows[position() &lt;= 4]/cell/@points)"
/>
</xsl:template>
</xsl:stylesheet>

Current Thread