Re: [xsl] Re: [xslt transform & grouping] Using the Muenchian Method?

Subject: Re: [xsl] Re: [xslt transform & grouping] Using the Muenchian Method?
From: "Michael PG" <xrow@xxxxxxx>
Date: Tue, 05 Oct 2004 12:44:31 +0000
Nope. It's not working.


<?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:param name="filter" select="Article[@filter='food']"></xsl:param>

<xsl:template match="Documents">

<xsl:apply-templates select="*[@filter=$filter]"/>

<Documents>
<xsl:for-each select="Document[@filter=$filter]/Article[count(.|key('by-info',@info)[1])=1]">
<Document name="{@info}">
<xsl:copy-of select="key('by-info',@info)[@filter=$filter]"/>
</Document>
</xsl:for-each>
</Documents>
</xsl:template>


</xsl:stylesheet>


XML:


<?xml version="1.0" encoding="utf-8"?>

<Documents>
  <Document chapter="1" title="title 1" href="file1.xml" filter="food">
        <Article title="1.1" info="sub" filter="food"/>
        <Article title="1.2" info="main" filter="drink"/>
   </Document>
  <Document chapter="2" title="title 2" href="file2.xml" filter="food">
        <Article title="2.1" info="sub" filter="drink"/>
        <Article title="2.2" info="main" filter="food"/>
   </Document>
</Documents>


From: David Carlisle <davidc@xxxxxxxxx>
Reply-To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
Subject: Re: [xsl] Re: [xslt transform & grouping] Using the Muenchian Method?
Date: Tue, 5 Oct 2004 13:24:06 +0100


    Anyhow, after I corrected to select="food" the result is the same.
    No output. Notice original XML input file,


as I explained, that would select <food> elements and you have none, so it selects nothing, you probably want

select="Article[@filter='food']"

David

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