Re: [xsl] XPath attribute expression

[Florent Georges <darkman_spam@xxxxxxxx>]

"Hintz, David L" wrote:

  Hi

> I should have noted that this is not an XSLT application.
> So, I need a pure XPath expression -- probably XPath 1.0.

  I think it is not possible in plain XPath 1.0 unless you can bind a
variable or a function to get the @name of the context node.  In XPath
2.0, you can use something like (if I understand correctly, you are not
sure you are limited to 1.0):

    for $n in @name return
      /topLevel/ents/ent[@name eq $n]/value

  Regards,

--drkm























	

	
		
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