Hi Vasu,
Yes, he could do it like that, but why? Why resort to extensions and build a
RTF, when he could just sort by item and filter by position - why make things
the hard way?
Cheers,
Jarno
-----Original Message-----
From: ext Vasu Chakkera [mailto:vasucv(_at_)hotmail(_dot_)com]
Sent: 18. October 2002 12:04
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: Re: [xsl] Top n values
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()<=
$n-value]">
<xsl:value-of select="."/><br/>
</xsl:for-each>
</xsl:template>
</xsl:stylesheet>
This works with XALAN..
From: Dan Diebolt <dandiebolt(_at_)yahoo(_dot_)com>
Reply-To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
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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