Re: [xsl] Comparing nodes in axis

[António Mota <xptm@xxxxxxx>]

However, in my second example,

ancestor::menu = (//menu)[$pos]

i can not do

(count(ancestor::menu|(//menu)[$pos])=1)

but if my understanding of the union operator is correct, i can make

(count(ancestor::menu|(//menu)[$pos])=(count(ancestor::menu))

Right?


Citando xptm <xptm@xxxxxxx>:

> Michael Kay wrote:
>
> >You are comparing the nodes for equality, whereas your description
suggests
> >you want to compare them for identity. To compare whether $A and $B are
the
> >same node, use ($A is $B) in XPath 2.0, or (count($A|$B)=1) in XPath 1.0.
> >
> >
> >
> That's exactly what i want. In fact, testing for equality could lead to
> potencial errors, cause nothing in my structure forbids the existence of
> two nodes with the same content.
>
> A very large thank you...
>
>





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