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Re: Equivalence between XSL and XPath expression

2004-12-11 03:58:47
So, in conclusion,

The equivalent on

<xsl:variable name="position">
        <xsl:number level='any' count="menu"/>
</xsl:variable>

is

count(./ancestor-or-self::*)+count(./preceding::*)



but the generalization to every node on a tree of my menu use

count(./ancestor-or-self::menu)+count(./preceding::menu)

is
count(./ancestor-or-self::node())+count(./preceding::node())


Ok, thanks peolple...





Dimtre Novatchev wrote:

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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