Subject: Re: [xsl] catching the last node still satisfying a condition From: "Dimitar Peikov" <mitko@xxxxxxx> Date: Thu, 13 Sep 2001 16:11:27 +0300 |
> 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@xxxxxxxxxxxxxx> > 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
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