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Re: Equivalence between XSL and XPath expression

2004-12-10 18:12:36
On Sat, 11 Dec 2004 00:54:19 +0000, xptm <xptm(_at_)sapo(_dot_)pt> wrote:
So basically you're saying that the root element doesn't have the self::
axis, besides the obvious ancestor, parent and preceding. Is that so?

Any node, including "/"    "has a self axis".

However, the expression you suggested:
    
       count(./ancestor-or-self::*)+count(./preceding::*)

evaluates to 0 in the case when the context node is the document node.

The reason?

   The principal node kind for the self axis is the element-node kind.

Therefore, 

       self::*

selects the context node only if the context node is an element. This
is not the case with the root (document) node.

Correct the above to:

    self::node()

and it now selects the context node always, regardless of its node-kind.


Cheers,
Dimitre.

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