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[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] Re: Re: [xslt transform & grouping] Using the Muenchia
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> _________________________________________________________________ Dont just search. Find. Check out the new MSN Search! http://search.msn.click-url.com/go/onm00200636ave/direct/01/
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