Re: [xsl] Wildcards in namespaces ?

[George Cristian Bina <george@xxxxxxx>]

You can use the local-name [1] function to ignore the namespace:
  UML:targetElement changes to
  *[local-name()='targetElement']

Alternativelly if you have only a couple of possible namespaces you can 
declare them all and use or constructs to cover all the cases:
  UML:targetElement changes to
  *[self::UML1:targetElement or self::UML2:targetElement]
where UML1 and UML2 are mapped to the corresponding namespaces.

[1] http://www.w3.org/TR/xpath#function-local-name

Hope that helps,
George
-----------------------------------------------
George Cristian Bina
<oXygen/> XML Editor & XSLT Editor/Debugger
http://www.oxygenxml.com

Mick Baggen wrote:
> Hello,
>
> I would appreciate some help with the following problem:
>
> I'm writing a XSLT stylesheet for processing of UML XMI files. Alas,
> depending on the UML tool, the elements in the XMI file are placed in a
> different namespace. E.g. Poseidon (www.gentleware.com) uses the
> namespace-URI xmlns:UML = "org.omg.xmi.namespace.UML", and Rational Rose
> uses xmlns:UML = "href://org.omg/UML/1.3". However, all of them generate a
> prefix UML: for their XMI elements.
>
> Now my problem is: how can I define a namespace in my XSLT stylesheet so
> that I can process all XMI files with the SAME stylesheet ? I came across a
> remark on the XSLT FAQ suggesting that it is possible to use wildcards in
> the namespace, eg <xsl:stylesheet xmlns:UML="*UML*">. And indeed, it seems
> that Xalan 2.6.0 recognizes this notation. However, libxslt 1.1.6
> (xmlsoft.org) does not.
>
> I tried to look it up in the XSLT and XML Namespace specifications, but I
> couldn't find whether this is valid. Could someone help me out on this one ?
> If it is NOT valid XSLT, is there another way to achieve the same result ?
>
> Regards,
>
> Mick Baggen
> Principal Consultant
> Technolution B.V.
> E-mail: 	 mick.baggen@xxxxxxxxxxxxxxx
>
>
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