Re: [xsl] position of arbitrary parent

Subject: Re: [xsl] position of arbitrary parent
From: David Carlisle <davidc@xxxxxxxxx>
Date: Tue, 17 Jul 2001 19:05:47 +0100
> which leaves me back with needing to know the
> position() of the current <chapter> element.


Probably you need to pass the chapter node as a parameter down to you
html templates.

<xsl:for-each select="chapter">
		<xsl:apply-templates select="$chapterTemplate/html" >
                    <xsl:with-param name="c" select="."/>
                <xsl:apply-templates/>
</xsl:for-each>

and then declare c param in your templates that handle your html doc,

David

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