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

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Subject: Re: 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

Current Thread

Re: most efficient way to get XML source's parent dir path, (continued)
- Owen Rees - 11 Feb 2009 09:51:53 -0000
  - Robert Koberg - 11 Feb 2009 14:41:13 -0000
    - Owen Rees - 11 Feb 2009 15:18:10 -0000
    - Michael Kay - 11 Feb 2009 15:25:12 -0000
    - Robert Koberg - 11 Feb 2009 15:56:08 -0000 <=
    - Michael Kay - 11 Feb 2009 16:02:39 -0000
    - Houghton,Andrew - 11 Feb 2009 16:21:55 -0000
    - David Carlisle - 11 Feb 2009 16:37:34 -0000
    - Florent Georges - 11 Feb 2009 16:43:32 -0000

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