RE: [xsl] sort / unsort

["Michael Kay" <michael dot h dot kay at ntlworld dot com>]

> Interesting,
> Could you show me?  I couldn't see a way of getting
> a sorted nodeset into a variable. Wouldn't this approach require that?

It needs the node-set() extension function.

<xsl:variable name="n">
  <xsl:for-each>
    <xsl:sort>
    <xsl:copy-of select="."/>
  </
</

<xsl:for-each select="xx:node-set($n)">
 ...


Michael Kay
Software AG
home: Michael.H.Kay@ntlworld.com
work: Michael.Kay@softwareag.com 

> 
> -- Mike Berrow
> 
> ----- Original Message ----- 
> From: "Mike Brown" <mike@skew.org>
> To: <xsl-list@lists.mulberrytech.com>
> Sent: Thursday, August 08, 2002 10:09 PM
> Subject: Re: [xsl] sort / unsort
> 
> 
> Mike Berrow wrote:
> > Does anyone see a better way?
> 
> I was going to try the Kay/Becker/Holman technique from the FAQ.
>   http://www.dpawson.co.uk/xsl/sect2/muench.html#d197e74
> 
> That is, find the difference between all the nodes and the 
> sorted nodes to get the remaining nodes. No need for 
> generate-id(); count() and != ought to work nicely.
> 
>    - Mike 
> ______________________________________________________________
> ______________
>   mike j. brown                   |  xml/xslt: http://skew.org/xml/
>   denver/boulder, colorado, usa   |  resume: 
> http://skew.org/~mike/resume/
> 
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