Subject: Re: [xsl] passing xsl:param-values to xsl:include From: Johannes Döbler <jd@xxxxxxxxxxxxxx> Date: Tue, 26 Jun 2001 13:27:58 +0200 |
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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