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Re: Selecting node at a certain position from a nodese

Subject: Re: Selecting node at a certain position from a nodeset
From: Marco Guazzone <sguazt@xxxxxxxxxxx>
Date: Wed, 22 Jan 2003 15:42:03 +0100 (CET)
xsl selecting node position
You're right!
I'm going to replace [$pos] with [position()=$pos].
Thanks for all!

--------------------------------
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:

>   <xsl:apply-templates
>   select="($sections/section[@group=1]/item[@id=1])[$pos]"/>
>   and it works!!
>   So I think this is the better solution!
> 
> Yes, if that's what you want to do, although you would be clearer
> to use [position()=$pos] rather than [$pos].
> [x] is short for [position()=x] if x is a number, this is clear in
> [1] which is [position()=1] but you can not tell in Xpath whether
> a variable is a number or not and if $pos were for example a string
> rather than a number  [$pos] would be short for  [$pos != ''] 
> which would not be what you want. If you use position()=$pos
> then you will force XPath always to do a numeric test.
> 
> David
> 
> 
> 
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