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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