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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 15:31:25 +0200
- Organization: =?iso8859-1?q?Universit=E9=20Pierre=20&=20Marie?= Curie
- References: <200109131311.f8DDBR079587@earth.rila.bg>
- Reply-To: xsl-list at lists dot mulberrytech dot com
Ainsi parlait Dimitar Peikov :
[..]
> > 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
Inside a for-each to catch only the last, that's OK:
<xsl:for-each select="foos/foo[bar and position() < limit]">
<xsl:if test="position()=last()">
<xsl:apply-template select="."/>
</xsl:if>
</xsl:for-each>
Thanks
--
Guillaume Rousse <rousse@ccr.jussieu.fr>
GPG key http://lis.snv.jussieu.fr/~rousse/gpgkey.html
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