Re: Re: RE: Re: RE: Re: XPath riddle

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Subject: Re: Re: RE: Re: RE: Re: XPath riddle
From: Wendell Piez <wapiez@xxxxxxxxxxxxxxxx>
Date: Thu, 05 Jul 2001 11:54:11 +0100

At 01:28 PM 7/5/01, Dimitre wrote:

Now, if I understand you well, you want every D/C, for which the first ancestor from a given list of ancestors (the list of elements defining D and its child C as descendents) to be "A" and nothing else.

Then this is returned by:

//D/C[name(ancestor::*[name()='A' or name()='F'][1])='A']

Or

//D/C[(ancestor::A|ancestor::F)[last()][self::A]]

Grouping the two location paths ancestor::A and ancestor::F creates a node set to be evaluated in document order; the last() predicate selects the latest (deepest) of these ancestors; the self::A makes sure it's an A element (by throwing it out if it's not).

You can add other ancestors to the mix by adding to the (ancestor::A|ancestor::F) group, or alter the condition in other ways (for example, just leaving out any C with an F grandparent (not just an F ancestor at any level), by grouping (ancestor::A|../parent::F).

Also, remember if this expression is going in a match you don't need the initial "//" (a detail that not everyone seems to have assimilated).

Cheers,
Wendell


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Wendell Piez                            mailto:wapiez@xxxxxxxxxxxxxxxx
Mulberry Technologies, Inc.                http://www.mulberrytech.com
17 West Jefferson Street                    Direct Phone: 301/315-9635
Suite 207                                          Phone: 301/315-9631
Rockville, MD  20850                                 Fax: 301/315-8285
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Re: RE: Re: RE: Re: XPath riddle Dimitre Novatchev - Thu, 5 Jul 2001 08:33:27 -0400 (EDT) Wendell Piez - Thu, 5 Jul 2001 11:58:49 -0400 (EDT) <=

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