Just add a line for each parameter in the xsl stylesheet
<?xml version="1.0" encoding="ISO-8859-1"?> <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"; > <xsl:param name="someParam"/> <xsl:param name="otherParam"/> <xsl:template match="abct"> <xsl:if test="$someParam='cocoon:/patternC' "> <dosomthing /> </xsl:if> </xsl:template> UI -----Ursprüngliche Nachricht----- Von: ypomonh [mailto:[EMAIL PROTECTED] Gesendet: Freitag, 11. Mai 2007 16:54 An: [email protected] Betreff: (newbie) Access parameter passed to an XSL transformation from stylesheet I have a simple pipelineQ <map:pipeline> <map:match pattern="patternA"> <map:generate src="cocoon:/patternB"/> <map:transform src="style.xsl"> <map:parameter name="someParam" value="cocoon:/patternC"/> <map:parameter name="otherParam" value="cocoon:/patternC"/> </map:transform> <map:serialize type="xml"/> </map:match> </map:pipeline> and I want to handle the parameter "someParam" in my style.xsl: <xsl:variable name="myVariable"> <xsl:value-of select="someParam/nodeA"/> </xsl:variable> but it returns empty. What is the correct way to access a parameter passed to an XSL transformation from within the XSL stylesheet..? --------------------------------------------------------------------- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] --------------------------------------------------------------------- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED]
