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RE: [xsl] most efficient way to get XML source's parent dir path

2009-02-11 10:25:16

There are XPath 2.0 functions specifically provided to do the job.

select="doc(resolve-uri(@ref,base-uri(.)))"


And in fact the document() function does this by default if the first
argument is a node:

    select="document(@ref)"

should work fine.

Michael Kay
http://www.saxonica.com/


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