Re: [xsl] Re: RE: Re: RE: Re: XPath riddle

[Wendell Piez <wapiez at mulberrytech dot com>]

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@mulberrytech.com
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