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

["Markus Arndt" <markus.arndt@xxxxxxxxxxxxx>]

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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