RE: constructing the Node Sets

> 
> 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.

> 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. 


> 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)

David

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