Re: Is it possible to know position of ancestor?

Try

select="count(../preceding-sibing::line + 1)"

Kind regards,

James Carlyle

-----Original Message-----
From: owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx
[mailto:owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx]On Behalf Of
evgeniy.strokin@xxxxxxxxxxxxxxxx
Sent: 17 October 2002 20:11
To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
Subject: Re: Is it possible to know position of ancestor?


Yes, you are so right (how do you know what I meant if I didn't know!?!? :)
But here is also the problem:
Let say we have another XML:
<root>
<some_tag/>
<line>
<a/>
<b/>
</line>
<line>
<a/>
<b/> - we are here
</line>
<root>

In your example select="count(../preceding-sibing::* + 1)" we get 3 because it will count "some_tag" too. But we need to count only "line" elements.
How we can solve this problem?

Jenya


David Carlisle writes:

> >> We are in tag "b",
> note that xslt works on elements (element nodes) not tags, and importat
> distinction. > >> We want to find out what is position of our ancestor >> in their ancestor.
> > I think you mean parent rather than ancestor: > > select="count(../preceding-sibing::* + 1)" > > > David


XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list

XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list