Subject: Re: [xsl] xsd:import and namespaces From: Peter Moore <pmoore@xxxxxxx> Date: Thu, 24 Apr 2003 09:27:50 -0500 |
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.
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