Re: catching the last node still satisfying a conditio

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@xxxxxxxxxxxxxx>
GPG key http://lis.snv.jussieu.fr/~rousse/gpgkey.html

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