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Re: XPath: position of the parent ?
- From: Al Snell <firstname.lastname@example.org>
- To: Gerd Mueller <email@example.com>
- Date: Fri, 17 Aug 2001 16:58:38 +0100 (BST)
On Fri, 17 Aug 2001, Gerd Mueller wrote:
> How do I express in XPath the position of the parent of a node ? I guess it
> must be something like '../position()', but this don't work.
That would get you the position of the parent node in the list of parent
nodes of the current context element, wich would always be "1", right? :-)
What you may want is the position of that parent node in the list of
children of the grandparent node... urgh... not sure off of the top of my
head how to say that in plain XPath.
<xsl:variable name="parent" select="parent::node()" />
<xsl:value-of select="position(../../child::*[. = $parent])" />
...or something like that? There may be gotchas lurking there, I can't
test it right now...
> Best Regards,
Alaric B. Snell
http://www.alaric-snell.com/ http://RFC.net/ http://www.warhead.org.uk/
Any sufficiently advanced technology can be emulated in software