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: Guillaume Rousse <rousse at ccr dot jussieu dot fr>
- Date: Thu, 13 Sep 2001 14:47:38 +0200
- Organization: =?iso8859-1?q?Universit=E9=20Pierre=20&=20Marie?= Curie
- References: <200109131225.f8DCPN079124@earth.rila.bg>
- 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
--
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