Re: [xsl] Top n values

Subject: Re: [xsl] Top n values
From: "Vasu Chakkera" <vasucv@xxxxxxxxxxx>
Date: Fri, 18 Oct 2002 09:03:41 +0000
Following is an example of how you can work with node-set()..

<?xml version="1.0"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"; xmlns:xalan = "http://xml.apache.org/xalan"; exclude-result-prefixes="xalan">
<xsl:template match="/">
<xsl:variable name="sorted-items">
<xsl:for-each select="/root/item">
<xsl:sort select="." order="descending" data-type="number"/>
<xsl:copy-of select="."/>
</xsl:for-each>
</xsl:variable>
<xsl:variable name="n-value" select="'3'"/>
<xsl:for-each select="xalan:nodeset($sorted-items)/item[position()&lt;= $n-value]">
<xsl:value-of select="."/><br/>
</xsl:for-each>
</xsl:template>
</xsl:stylesheet>


This works with XALAN..






From: Dan Diebolt <dandiebolt@xxxxxxxxx>
Reply-To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
Subject: [xsl] Top n values
Date: Thu, 17 Oct 2002 12:30:04 -0700 (PDT)

I need to process the top n values in a list where n
is dynamic. So if n=3 and I had the following XML:

<root>
 <item>1</item>
 <item>5</item>
 <item>3</item>
 <item>2</item>
 <item>6</item>
 <item>4</item>
</root>

I need to process <item>'s with values 6,5,4. If I had
captured n in a variable/parameter, what is the most
efficient way to do this? What I want to avoid is walking
the <item> nodset multiple times as there are a lot of
<item>s (thousands). Thanks.

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