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Re: Re: [xslt transform & grouping] Using the Muenchia
Subject: Re: Re: [xslt transform & grouping] Using the Muenchian Method?
From: Anton Triest <anton@xxxxxxxx>
Date: Thu, 07 Oct 2004 15:13:55 +0200
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Michael PG wrote:
Thanx!
In this example, where I only user Muenchian method to group, elements
where info-attribute doesn't contains anything are also displayed, but
not needed. How do I get rid of those elements here?
<xsl:for-each select="Document/Article[string(@info) and
count(.|key('by-info', @info)[1])=1]">
Anton
<?xml version='1.0' encoding='UTF-8'?>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" version="1.0" encoding="UTF-8"
indent="yes"/>
<xsl:key name="by-info" match="Article" use="@info"/>
<xsl:template match="/Documents">
<Documents>
<xsl:for-each
select="Document/Article[count(.|key('by-info', @info)[1])=1]">
<Document name="{@info}">
<xsl:copy-of select="key('by-info', @info)"/>
</Document>
</xsl:for-each>
</Documents>
</xsl:template>
</xsl:stylesheet>
Thank you!
_m
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