RE: [xsl] Re: Is it possible to know position of ancestor?

["James Carlyle" <james.carlyle@xxxxxxxxxxxx>]

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: [xsl] 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