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

[Dimitre Novatchev <dnovatchev@xxxxxxxxx>]

Hi Wendel,

> 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).
> 

Yes, this only will not work in the more general case when the ancestor element
names to check for are generated "on the fly" from a schema.

Then only the second expression I gave will work:

//D/C[name(ancestor::*[contains($listOfDefiningEls, concat(' ', name()))][1])='A']

Also note, that an optimizing XPath processor will stop at the first occurence of
the above expression, while if last() is used, it will have to produce all the
nodes.

Cheers,
Dimitre Novatchev.

> 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).







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