Re: KindTest for namespaces

Thank you for this explanation. Coming from an XQuery standpoint, I was 
not aware of the namespace axis in XPath, nor that it was deprecated. I 
originally got stuck on this point because fn:node-kind states it can 
return "namespace" and I could find no way to get at a namespace node 
(since even computed namespaces must be within an element).

--Sarah

On Oct 6, 2003, at 3:17 PM, Kay, Michael wrote:

> > However, writing it inline gives the impression that it could be
> > matched as an attribute, such as $alt/@xmlns:metric. From what I
> > understand, this is not the case. So count($alt/attribute()) would
> > return 0, but count($alt/node()) would return 1. If this is the case,
> > then it seems to be an omission not to have a namespace kind test
> > (especially when there is one for every other kind of node).
>
> You can retrieve namespace nodes using the namespace axis, 
> count(namespace::*) will return the number of namespace nodes. But we 
> have deprecated the namespace axis in XPath 2.0 because we want 
> implementations to have maximum freedom in how they represent the 
> information that modelled formally in the Data Model using namespace 
> nodes. In XQuery, and in XPath 2.0 if the namespace axis is not 
> provided, there is no expression that returns a namespace node.
>
> There is no need for a KindTest for namespace nodes, because they are 
> found only on the namespace axis, so looking for all nodes on the 
> namespace axis works fine.
>
> >
> > I can see no other way to do such matching/resolving other than using
> > get-in-scope-namespaces then resolving each prefix.
>
> get-in-scope-namespaces() was indeed provided as a replacement for the 
> namespace axis. It provides the information that's needed (the set of 
> namespace prefix/uri pairs for each element) without imposing 
> semantics that aren't needed (namespace nodes have distinct identity, 
> they have parents, they have an ordering in document order).
>
> Michael Kay