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Re: Comparing nodes in axis

2004-12-17 12:46:20
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(_at_)sapo(_dot_)pt>:

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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