Subject: Re: [xsl] Equivalence between XSL and XPath expression From: xptm <xptm@xxxxxxx> Date: Sat, 11 Dec 2004 11:12:04 +0000 |
<xsl:variable name="position"> <xsl:number level='any' count="*"/> </xsl:variable>
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@xxxxxxx> 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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