This is the mail archive of the
xsl-list@mulberrytech.com
mailing list .
Re: catching the last node still satisfying a condition
- To: xsl-list at lists dot mulberrytech dot com
- Subject: Re: [xsl] catching the last node still satisfying a condition
- From: "Dimitar Peikov" <mitko at rila dot bg>
- Date: Thu, 13 Sep 2001 16:11:27 +0300
- Reply-To: xsl-list at lists dot mulberrytech dot com
> Ainsi parlait Dimitar Peikov :
> > You can't without for-each cycle. because [] operands return 1 recordset
> > not an array. You must set exact match that is enshured that only one
> > element could contain this.
> A cycle for treating one only element ? And recordset are ordered: foo[3]
> returns the third foo child.
>
> > You possibly search for :
> >
> > foos/foo[position() = last()]/bar[position() < limit]
> IMHO, this will return all the <bar> child of the last <foo>, whose position
> regarding their <foo> parent is inferior or egal to limit
>
> Considering the following situation
> <foo>
> <foo id="foo1">
> <bar/>
> </foo>
> <foo id="foo2">
> <bar/>
> </foo>
> <foo id="foo3>
> </foo>
> </foo>
> i would like to catch foo2 with limit = 2 or 3, and foo1 with limit = 1
>
foos/foo[position() < limit]/bar
> --
> Guillaume Rousse <rousse@ccr.jussieu.fr>
> GPG key http://lis.snv.jussieu.fr/~rousse/gpgkey.html
>
> XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
>
--
Dimitar Peikov
Programmer Analyst
Globalization Group
"We Build e-Business"
RILA Solutions
27 Building, Acad.G.Bonchev Str.
1113 Sofia, Bulgaria
phone: (+359 2) 9797320
phone: (+359 2) 9797300
fax: (+359 2) 9733355
http://www.rila.com
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list