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

[evgeniy.strokin@xxxxxxxxxxxxxxxx]

But if you don't know name of element "line"? It should be the same name 
only this you know. 

Thanks,
Jenya 

James Carlyle writes: 

> 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 


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