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[xsl] Transforming XML-Groups to HTML-Lists

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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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