Re: Group processing / Muench method with MSXML or XSL

so I am looking for a possibility to make the
same transformation only with XSLT 1.0 functions.
http://jenitennison.com/xslt/grouping/index.xml

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Cheers,
Dimitre Novatchev
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Truly great madness cannot be achieved without significant intelligence.
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To invent, you need a good imagination and a pile of junk
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You've achieved success in your field when you don't know whether what
you're doing is work or play

Hi Folks,

I try to transform XML-Tags which can combined as groups to HTML-Lists
(ordered and unordered).

First an example of the XML-Code:

<EinzugListe>
         <Einzug aufz="1." level="0">Some Text 1 - Level 0</Einzug>
                 <Einzug aufz="-" level="1">Some Text 1 - Level 1</Einzug>
                 <Einzug aufz="-" level="1">Some Text 2 - Level 1</Einzug>
                 <Einzug aufz="" level="1">Some more Text</Einzug>
                         <Einzug aufz="-" level="2">Some Text 1 -
Level 2</Einzug>
                         <Einzug aufz="-" level="2">Some Text 2 -
Level 2</Einzug>
         <Einzug aufz="2." level="0">Some Text 2 - Level 0</Einzug>
</EinzugListe>

This should be transformed to the following HTML-Code:

<ol>
         <li>Some Text 1 - Level 0
         <ul>
                 <li>Some Text 1 - Level 1</li>
                 <li>Some Text 2 - Level 1<br />
                 Some more Text
                         <ul>
                                 <li>Some Text 1 - Level 2</li>
                                 <li>Some Text 2 - Level 2</li>
                         </ul>
                 </li>
         </ul>
         </li>
         <li>Some Text 2 - Level 0</li>
</ol>

For the transformation I will use the following XSLT 2.0 code (without
the decision if ul or ol):

<xsl:template match="EinzugListe">
         <ul>
                 <xsl:call-template name="ebene">
                         <xsl:with-param name="liste" select="Einzug"/>
                         <xsl:with-param name="level" select="0"/>
                 </xsl:call-template>
         </ul>
</xsl:template>
<xsl:template name="ebene">
         <xsl:param name="liste"/>
         <xsl:param name="level"/>
         <xsl:for-each-group select="$liste"
group-starting-with="*[xs:integer(@level) eq
$level]">
                 <li>
                         <xsl:value-of select="normalize-space(.)"/>
                         <ul>
                                 <xsl:call-template name="ebene">
                                         <xsl:with-param name="liste"
select="current-group() except ."/>
                                         <xsl:with-param name="level"
select="$level + 1"/>
                                 </xsl:call-template>
                         </ul>
                 </li>
         </xsl:for-each-group>
</xsl:template>

I found the code in the List, modified it and it works fine until I
will use Saxon as transform-processor which supports XSLT 2.0.

But the transformation has to be made on a Windows computer and there
I only have the MSXML as transform-processor (because the
transformation is part of an
VBA-MSAccess2000-project). I read that MSXML does not (and will not)
support XSLT 2.0 and so I am looking for a possibility to make the
same transformation only with XSLT 1.0 functions. Or is there a
possibility to use an other tranform-processor for the VBA-project (I
read something about System.Xml, but I don't know how to include it in
MSAccess2000 so I can use it in my VBA-project).

Thanks for your tipps, hints and code-samples (maybe including the
decision if ul, ol or br (see example above))

Best regards,

Michael

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