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[(_at_)filter='food']"></xsl:param>
<xsl:template match="Documents">
<xsl:apply-templates select="*[(_at_)filter=$filter]"/>
<Documents>
<xsl:for-each
select="Document[(_at_)filter=$filter]/Article[count(.|key('by-info',@info)[1])=1]">
<Document name="{(_at_)info}">
<xsl:copy-of
select="key('by-info',@info)[(_at_)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(_at_)nag(_dot_)co(_dot_)uk>
Reply-To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
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[(_at_)filter='food']"
David
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