Err... I think that you mean:
$a[(_at_)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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