RE: constructing the Node Sets

David Carlisle wrote:

> > Err...  I think that you mean:
> >     $a[@obid = $b/@obid]
>
> No I meant what I wrote. $a and $b in the original posting were sets of
> attribute nodes so @obit in your expression would always be an empty
> node set.

My bad.  I misread the original poster's variable setting.


> > Well, except possibly for order, as I noted earlier.
> Not at all. Node sets are sets and as such have no ordering property.
> If two node sets have the same members they are indistinguishable in
> Xpath.

I will address this in a separate message.


> > Right, in the first expression all the "Class" elements will
> have a "name"
> > attribute with the value "Asm".  In the second expression all
> the "Class"
> > elements will have a "name" attribute with the value "Cmp".
>
> as both $a and $b just contain obid attributes the subsets will
> similarly contain just these attribute nodes, but they will
> be different attribute nodes (but with the same value)

Right.  Again, my mistake in reading the original poster's variable setting.

-- Roger Glover
   glover_roger@xxxxxxxxx



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