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

Cart

XML Editor - Download a Free Trial >

See What's New >

Buy Now >

[Home] [By Thread] [By Date] [Recent Entries]

Subject: Re: most efficient way to get XML source's parent dir path
From: Owen Rees <owen.rees@xxxxxx>
Date: Wed, 11 Feb 2009 15:17:33 +0000

--On Wednesday, February 11, 2009 09:40:48 -0500 Robert Koberg wrote:

<xsl:apply-templates select="doc(concat($parent-dir, '/', @ref))/*"/>

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

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

If you want relative references resolved as described in the W3C specifications, use these functions.

Relative reference resolution is not as simple as just replacing whatever follows the last '/'.

--
Owen Rees; speaking personally, and not on behalf of HP.
========================================================
Hewlett-Packard Limited.   Registered No: 690597 England
Registered Office:  Cain Road, Bracknell, Berks RG12 1HN

Current Thread

Re: 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

<- Previous	Index	Next ->
Re: most efficient way to get, Robert Koberg	Thread	RE: most efficient way to get, Michael Kay
Re: Problem with "except" ope, Andrew Welch	Date	Re: Problem with "except" ope, G. Ken Holman
	Month

XML Editor - Download a 15 Day Free Trial Now >

See What's New in Stylus Studio >

Buy Stylus Studio - XML Editor - Now >