Subject: RE: selecting the nearest preceding sibling
From: Mukul <mukulw3@xxxxxxxxx>
Date: Tue, 13 May 2003 09:22:23 -0700 (PDT)
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Hi Jarno ,
Thanks a lot. I can get the desired o/p with your
syntax..
Regards,
Mukul
--- Jarno.Elovirta@xxxxxxxxx wrote:
> Hi,
> > OK - the node set of all the preceding siblings
> which
> > fit the criterium is given by
> >
> > preceding-sibling::tag[a]
> >
> > The nearest preceding sibling fitting the
> criterium is
> > the last one in that nodeset (the nodes are always
> in
> > document order)... that is,
> >
> > preceding-sibling::tag[a and last()]
>
> Wrong. Simple "last()" will always evaluate to a
> positive number, thus the above will compile to
>
> preceding-sibling::tag[a]
>
> Furthermore, nodes are always *processed* in
> document order, but preceding-sibling is a preceding
> axis and
>
> preceding-sibling::tag[a and position() = last()]
>
> will select the first node in document order. You
> can use
>
> (preceding-sibling::tag[a])[position() = last()]
>
> and then you'd get the first preceding sibling. Evan
> Lenz wrote an article about this stuff
>
<http://www.biglist.com/lists/xsl-list/archives/200201/msg00108.html>.
>
> Cheers,
>
> Jarno - neuroticfish: wakemeup! (club-edit)
>
> XSL-List info and archive:
> http://www.mulberrytech.com/xsl/xsl-list
>
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