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Re: passing xsl:param-values to xsl:include
- To: xsl-list at lists dot mulberrytech dot com
- Subject: Re: [xsl] passing xsl:param-values to xsl:include
- From: Johannes Döbler <jd at aztecrider dot com>
- Date: Tue, 26 Jun 2001 13:27:58 +0200
- Reply-To: xsl-list at lists dot mulberrytech dot com
The issue is discussed in the FAQ:
http://www.dpawson.co.uk/xsl/sect2/N4760.html#d143e57
Cheers,
Johannes
>Hello,
>
>i found out that constructions linke <xsl:include href="{$param}"/>
>doesn't work.
>I have to pass a parameter form an application to the xsl to include
>various stylesheets in
>dependance of the xml_owner (and his different formatting wishes).
>
>e.g. the xml_owner is no. 101 (from about 200)
>
><xsl:stylesheet>
><xsl:param name="xml_owner"/>
><xsl:include="master.xsl"/> //this matches the standard templates of all
>owners
><xsl:include="{$xml_owner}_styles.xsl"/> // this should match
>1)additional
>templates of the owner or 2) overrriding rules for templates of the
>master.xsl
>
><xsl:stylesheet>
>
>Is there any method to work this out, e.g. to write the line with the
>parameter dynamically? maybe with the help of a Perlscript?
>
>thanks a lot
>
>Markus
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