Subject: Re: [xsl] most efficient way to get XML source's parent dir path From: Robert Koberg <rob@xxxxxxxxxx> Date: Wed, 11 Feb 2009 10:55:44 -0500 |
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.
best, -Rob
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