Re: [xsl] Problem with generating references.

["Joris Gillis" <roac@xxxxxxxxxx>]

Tempore 14:57:07, die 07/05/2005 AD, hinc in  
xsl-list@xxxxxxxxxxxxxxxxxxxxxx scripsit Lakshmi narayana  
<lchintala@xxxxxxxxxxxx>:

> for each element so that I can identify each element uniquely).
>
> 1. Divide the elements into groups.
> 2. While dividing, create the references to the elements in the current
> group which are going to be placed in another group.

This stylesheet generates the ouput from your sample. It will need some  
adaption though, to work in a real situation.

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform";  
version="1.0">
<xsl:output method="xml" indent="yes"/>
<xsl:key name="element" match="*" use="substring(local-name(),1,1)"/>

<xsl:template match="/">
	<xsl:for-each  
select="//*[generate-id()=generate-id(key('element',substring(local-name(),1,1)))]">
		<xsl:element name="{substring(local-name(),1,1)}-group">
			<xsl:for-each select="key('element',substring(local-name(),1,1))">
				<xsl:copy>
					<xsl:for-each select="*">
						<reference><xsl:attribute name="id"><xsl:number count="*"  
level="any"/></xsl:attribute></reference>
					</xsl:for-each>
				</xsl:copy>
			</xsl:for-each>
		</xsl:element>
	</xsl:for-each>
</xsl:template>

</xsl:stylesheet>


regards,
--
Joris Gillis (http://users.telenet.be/root-jg/me.html)
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