RE: [xsl] selecting the nearest preceding sibling

[Mukul <mukulw3@xxxxxxxxx>]

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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