I've just tried without exsl:node-set, but only parenthesis:
<xsl:apply-templates
select="($sections/section[@group=1]/item[@id=1])[$pos]"/>
and it works!!
So I think this is the better solution!
Any comment?

--------------------------------
Marco Guazzone
Software Engineer
Kerbero S.r.L. - Gruppo TC
Viale Forlanini, 36
Garbagnate M.se (MI)
20024 - Italy
mail: marco.guazzone@xxxxxxxxxxx
www: http://www.kerbero.com
Tel. +39 02 99514.247
Fax. +39 02 99514.399
--------------------------------

On Wed, 22 Jan 2003, Marco Guazzone wrote:

> Hi David,
>   so how do you solve this problem?
>   Is the exsl:node-set solution the better one (respect to computation
> time and memory consumption) ?
> Regards,
> 
> --------------------------------
> Marco Guazzone
> Software Engineer
> Kerbero S.r.L. - Gruppo TC
> Viale Forlanini, 36
> Garbagnate M.se (MI)
> 20024 - Italy
> mail: marco.guazzone@xxxxxxxxxxx
> www: http://www.kerbero.com
> Tel. +39 02 99514.247
> Fax. +39 02 99514.399
> --------------------------------
> 
> On Wed, 22 Jan 2003, David Carlisle wrote:
> 
> > 
> > 
> > select="$sections/section[@group=1]/item[@id=1][$pos]"/>
> > 
> > the $pos item node that is a child of section[@group=1]
> > and has @id=1.
> > 
> > Since you have uniue ids within each section, if $pos is 1
> > you should get the same as 
> > 
> > 
> > select="$sections/section[@group=1]/item[@id=1]"/>
> > 
> > but id pos is any number other than 1 you should get nothing as 
> > there is only one item in each step satisfying [@id=1] so position() is
> > always equal to 1.
> > 
> > David
> > 
> > 
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