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

2004-09-28 01:15:19
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(_at_)cking(_dot_)be>
Reply-To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
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]


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