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

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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: "Michael Kay" <mike@xxxxxxxxxxxx>
Date: Wed, 11 Feb 2009 15:24:47 -0000

> 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/

Current Thread
RE: [xsl] most efficient way to get XML source's parent dir path, (continued) Michael Kay - 10 Feb 2009 17:13:23 -0000 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

Current Thread

RE: [xsl] most efficient way to get XML source's parent dir path, (continued)
- Michael Kay - 10 Feb 2009 17:13:23 -0000
- 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

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