Re: [xsl] Why no namespace node KindTest?

Subject: Re: [xsl] Why no namespace node KindTest?
From: Florent Georges <darkman_spam@xxxxxxxx>
Date: Sat, 26 Aug 2006 22:43:13 +0200 (CEST)
Colin Paul Adams wrote:

> >>>>> "Florent" == Florent Georges writes:

>     Florent>   Hi I wonder why there is no KindTest for
>     Florent> namepace nodes.  In particular, we can't write
>     Florent> something like:

>     Florent>     if ( $arg instance of namespace() ) ...

> Because namespace nodes are deprecated, and a compliant
> processor need not support them.

  Thanks Colin.

  What I understood is that namespace nodes are not
deprecated, but the namespace axis is.  So in plain XPath,
if the namespace axis is not supported, it is no possible to
actually get a namespace node.

  But XPath is designed as an embeded language.  XSLT for
example has xsl:namespace to create namespace nodes "by
hand".  As there is no kind test for namespace nodes, we
can't write something like:

    <xsl:variable name="..." as="namespace()">
      <xsl:namespace .../>

  So again, here, if no namespace axis, it is not possible
to get a namespace node.  But what about an other language?
I feel that setting the namespace axis as deprecated but not
the namespace nodes is not very logical.  But maybe I'm
completely wrong?



 uncompressed/chunked Sat Aug 26 20:13:39 GMT 2006 
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