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RE: [xsl] xsd:import and namespaces
Subject: RE: [xsl] xsd:import and namespaces From: Sundar Shanmugasundaram <SSHANMUGASUNDARAM@xxxxxxxxxxxxx> Date: Thu, 24 Apr 2003 10:29:05 +0530 |
To access the value of schemaLocation, use the following expression: <xsl:variable name="schema" select="../xsd:import/@schemaLocation"/> regards, sundar -----Original Message----- From: Peter Moore [mailto:pmoore@xxxxxxx] Sent: Thursday, April 24, 2003 3:39 AM To: XSL-List@xxxxxxxxxxxxxxxxxxxxxx Subject: [xsl] xsd:import and namespaces How can I get the value of schemaLocation from the xsd:element/@ref attribute in the following XML Schema sample: ... <xsd:schema ... xmlns:xxx="http://www.yo.com/example.namespace"> <xsd:import namespace="http://www.yo.com/example.namespace" schemaLocation="http://www.yo.com/example.xsd"/> ... <xsd:element ref="xxx:SomeName"/> ... My xsl looks something like this: ... <xsl:template match="xsd:element[@ref]"> <xsl:element name="{@ref}"> <xsl:variable name="schema" select="//xsd:import/@schemaLocation[../@namespace=???]"/> <xsl:apply-templates select="document($schema)//element[@name=current()/@ref]"/> </xsl:element> </xsl:template> ... Am I on the right track? If so, what should replace the ??? in the location path? If I remove the stuff between the xsl:element tags I get the following output: ... <xxx:SomeName xmlns:xxx="http://www.yo.com/example.namespace"/> ... So XSLT knows the namespace I want...I just don't know how to ask it. Thanks. BTW...I'm using LibXSLT and perl for the transform. XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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