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passing xsl:param-values to xsl:include
- To: XSL-List at lists dot mulberrytech dot com
- Subject: [xsl] passing xsl:param-values to xsl:include
- From: "Markus Arndt" <markus dot arndt at de dot adcore dot com>
- Date: Tue, 26 Jun 2001 12:35:22 +0200
- Organization: Adcore Deutschland AB
- Reply-To: xsl-list at lists dot mulberrytech dot com
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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