[xsl] passing xsl:param-values to xsl:include

Subject: [xsl] passing xsl:param-values to xsl:include
From: "Markus Arndt" <markus.arndt@xxxxxxxxxxxxx>
Date: Tue, 26 Jun 2001 12:35:22 +0200

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:param name="xml_owner"/>
<xsl:include="master.xsl"/> //this matches the standard templates of all
<xsl:include="{$xml_owner}_styles.xsl"/> // this should match
templates of the owner or 2) overrriding rules for templates of the


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


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