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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
Re:  Re: [xslt transform & grouping] Using the Muenchia
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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