Re: [xsl] question about identity transform

["Dimitre Novatchev" <dnovatchev@xxxxxxxxx>]

I think this has been discussed quite often in this list and should
probably be well-reflected in the FAQ.

A node-set is a set.

A set, by definition, is unordered.

Therefore, it doesn't matter whether an XPath expression is written as:

 A | B

or

B | A


the result is the same set.


--
Cheers,
Dimitre Novatchev
---------------------------------------
Truly great madness cannot be achieved without significant intelligence.
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To invent, you need a good imagination and a pile of junk
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You've achieved success in your field when you don't know whether what
you're doing is work or play




On 10/31/06, Abel Braaksma <abel.online@xxxxxxxxx> wrote:
> Dimitre Novatchev wrote:
> >> If you would refers it, and make it, say: ` node ( ) | @* ' the nodes
> >> would first be copied, resulting in the attributes ending up illegally:
> >> you may not create an attribute node after you create an element node in
> >> the result tree. However, I'm not sure if this is really enforced by the
> >> specs.
> >
> >
> > This is not true.
> >
> > The order in which an <xsl:apply-templates> instruction is applied on
> > the nodes selected by the expression as specified in its select
> > attribute is not defined and may be in any order.
>
> Thanks for pointing that out. Indeed, I always forget about this
> processing order, and tend to think in terms of result tree order...
>
> >
> > What is important, is that the results of these applications are
> > combined/ordered based on the document order of the processed nodes.
>
> I see. Meaning that though it can be processes in any order, the actual
> order of the output is determined by a certain set of rules, which is
> based on the original document order and not the order of the xpath in
> the template or the template itself. And I was always so careful putting
> the right order for the " | " unioned items ;-)
>
> -- Abel