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

["Michael PG" <xrow@xxxxxxx>]

Hi,

I have tried to implement the grouping by using the Muenchian Method 
according to http://www.jenitennison.com/xslt/grouping/muenchian.html, but I 
still receive errors when I transform.

I am trying to start with Jeni's example, but keep my XML structure. (below)

"Let's take our address book above. We want to group the contacts according 
to their surname, so we create a key that assigns each contact a key value 
that is the surname given in the record. The nodes that we want to group 
should be matched by the pattern in the 'match' attribute. The key value 
that we want to use is the one that's given by the 'use' attribute:"

xml-file structure

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

<Records>
	<Contact id="0001">
		<Doc title="Mr"/>
		<Doc forename="John"/>
		<Doc surname="Smith"/>
	</Contact>
	<Contact id="0002">
		<Doc title="Dr"/>
		<Doc forename="Amy"/>
		<Doc surname="Jones"/>
	</Contact>
</Records>



xslt-file

<?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="contacts-by-surname" match="Doc" use="surname" />

	<xsl:template match="Records">
		<xsl:for-each select="Doc[count(. | key('contacts-by-surname', 
surname)[1]) = 1]">
			<xsl:sort select="surname" />
				<xsl:value-of select="surname" />,<br />
					<xsl:for-each select="key('contacts-by-surname', surname)">
						<xsl:sort select="forename" />
						<xsl:value-of select="forename" /> (<xsl:value-of select="title" 
/>)<br />
					</xsl:for-each>
		</xsl:for-each>
	</xsl:template>


</xsl:stylesheet>


What is the problem? Can anybody see it ?

Thanx,

/Michael


> From: Anton Triest <anton@xxxxxxxx>
> Reply-To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> Subject: Re: [xsl] XSLT, XML and Identity transform
> Date: Mon, 27 Sep 2004 18:49:54 +0200
>
> Hi Michael,
>
> You want to group your Article elements by their info attribute:
> see http://www.jenitennison.com/xslt/grouping/muenchian.html
>
> HTH,
> Anton
>
>
> Michael PG wrote:
>
> > Hello,
> >
> > I am trying to solve following problem:
> >
> > Today I use identity transform. to filter elements from my original XML 
> > file and create XML output file.
> > I have created two XSLTs (base.xslt and filter.xslt), where base.xslt does 
> > the identity transformation and filter.xslt defines filtering rules and 
> > templates.
> >
> > What I basically want is to filter elements of type "sub" and sort them 
> > under one parent node and make selections of type "main" and sort them 
> > under another parent node. One parent node for each category. (see in 
> > Filtered XML how output shoul look like).
> >
> > My input XML looks like:
> >
> > [Original XML]
> >
> > <Documents>
> >    <Document chapter="1" title="title 1" href="file1.xml">
> >          <Article title="1.1" info="sub"/>
> >          <Article title="1.2" info="main"/>
> >     </Document>
> >    <Document chapter="2" title="title 2" href="file2.xml">
> >          <Article title="2.1" info="sub"/>
> >          <Article title="2.2" info="main"/>
> >     </Document>
> > </Documents>
> >
> > [Filtered XML SHOULD LOOK LIKE]
> >
> > <Documents>
> >    <Document name="main">
> >          <Article title="1.2" info="main"/>
> >          <Article title="2.2" info="main"/>
> >     </Document>
> >    <Document name="sub">
> >          <Article title="1.1" info="sub"/>
> >          <Article title="2.1" info="sub"/>
> >     </Document>
> > </Documents>
> >
> > [snip]

_________________________________________________________________
Express yourself instantly with MSN Messenger! Download today - it's FREE! 
http://messenger.msn.click-url.com/go/onm00200471ave/direct/01/