[XSL-LIST Mailing List Archive Home]
[By Thread]
[By Date]
[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 |
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 XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
Current Thread |
---|
|
<- Previous | Index | Next -> |
---|---|---|
RE: [xsl] How to propagate links in, Andreas Waechter | Thread | Re: [xsl] passing xsl:param-values , Johannes Döbler |
[xsl] How to propagate links in hie, Nick Vincent | Date | RE: [xsl] How to propagate links in, Andreas Waechter |
Month |
Keywords