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Re: Selecting node at a certain position from a nodeset

2003-01-22 07:08:58
I've just tried without exsl:node-set, but only parenthesis:
<xsl:apply-templates
select="($sections/section[(_at_)group=1]/item[(_at_)id=1])[$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(_dot_)guazzone(_at_)kerbero(_dot_)com
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(_dot_)guazzone(_at_)kerbero(_dot_)com
www: http://www.kerbero.com
Tel. +39 02 99514.247
Fax. +39 02 99514.399
--------------------------------

On Wed, 22 Jan 2003, David Carlisle wrote:



select="$sections/section[(_at_)group=1]/item[(_at_)id=1][$pos]"/>

the $pos item node that is a child of section[(_at_)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[(_at_)group=1]/item[(_at_)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 [(_at_)id=1] so position() is
always equal to 1.

David


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