Grouping XML by Muenchian Method - Urgent :-(

xrow · September 27th, 2004, 09:16 AM

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.

My input XML looks like:

[Input XML]

Code:

<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>

What I basically want is to filter Article elements where info="sub", create parent node "sub" and group them under
parent node.

Same thing with with Article elements where info="main", create parent node "main" and group them under
parent node.

One parent node for each category. (see in Filtered XML - how I want output to look like).

[Output XML]

Code:

<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>

[base.xslt]

Code:

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

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">

     
        <xsl:include href="filter.xslt"/>

     <xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>
     <xsl:template match="node() | @*">
          <xsl:copy>
               <xsl:apply-templates select="node() | @*"/>
          </xsl:copy>
     </xsl:template>
</xsl:stylesheet>

[filter.xslt]

Code:

<?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:template match="Doc[not(@info='main')][not(@info='sub')]"/>
</xsl:stylesheet>

If this can't be done the way I suggested, could you give me an example on how it can be done?

Thank you in advance!

.Michael

joefawcett · September 27th, 2004, 09:54 AM

I think this is more of a grouping problem:

http://www.jenitennison.com/xslt/gro...muenchian.html

--

Joe (Co-author Beginning XML, 3rd edition)

xrow · September 27th, 2004, 12:32 PM

Hi,

I have tried to implement the grouping by using the Muenchian Method according to http://www.jenitennison.com/xslt/gro...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,

/M

joefawcett · September 28th, 2004, 03:21 AM

So what does your version of Jeni's grouping stylesheet look like? You have to define your own key etc.

--

Joe (Co-author Beginning XML, 3rd edition)

xrow · September 28th, 2004, 03:50 AM

I need to modify xslt stylesheet file "XSLT stylesheet" so it can give me an XML output-file that looks similar to "Wanted XML Output":

Input XML

Code:

<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>

Wanted XML Output

Code:

<records>
    <contact id="0001" surname="Jones">
        <doc forename="Amy" title="Dr"/>    
        <doc forename="Brian" title="Mr"/>
    </contact>
    <contact id="0002" surname="Smith">
        <doc forename="Fiona" title="Ms"/>    
        <doc forename="John" title="Mr"/>
    </contact>
</records>

XSLT stylesheet

Code:

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

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">

     
    <xsl:key name="contacts-by-surname" match="contact" use="surname" />
     
    <xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>
     
    <xsl:template match="records">
        <xsl:for-each select="contact[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>

mhkay · September 28th, 2004, 04:07 AM

Your code appears to use "surname" where it should use "doc/@surname".

Michael Kay
http://www.saxonica.com/