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

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
On Feb 11, 2009, at 10:24 AM, Michael Kay wrote:


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.

Well, I tried to simplify the example. In reality, I am using the unparsed-text function which has no base-uri argument () and the resolver is provided with a full java.net.URI so I needed a way to create the correct path in the XPath. I should have been more clear.

Am I still missing something?

best,
-Rob

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