[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] RE: 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
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