Re: [xsl] finding position() in xpath 1.0

[Abel Braaksma <abel.online@xxxxxxxxx>]

Frank Marent wrote:
> david, abel
>
> > I think I mis read your requirement. the code I posted works so long as
> > there is at most one node with test="yes. If you have multiple such
> > nodes and need to select the preceding cousin of all of them,
> > Then...
> >
> > //CELL[@test='yes']/../preceding-sibling::ROW[1]/CELL[position()=count(../following-sibling::ROW[1]/CELL[@test='yes']/preceding-sibling::CELL)+1] 
>
> it's always a good feeling among xml-giants... i was posting the point 
> you've detected yourself - and abel. your statement reads for me like:
>
> //CELL[@test='yes']magic()/but()/it()/works()/halleluja()/preceding-sibling::CELL)+1] 

:))

Note that I tried to correct David with removing the '+1' with a 
self-test. But that's not needed. David's last solution is perfect and 
comes down to about the same I invented (it almost looked like David and 
I were in the same room: even the sample input and test template about 
matched!), but mine is a bit overly complicated.

(the reason that you can use +1 safely is that you are already walking 
from the path CELL[@test='yes'], which means that once the path matches, 
you are certain the +1 is correct)

About the 'magic', I usually do a mind-treewalk while dissecting a path. 
I imaging being some place on the XML tree and then walk the steps I 
wrote down in the XPath. Then suddenly the 'magic' goes and it all seems 
so logical.

Cheers,
-- Abel