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RE: [xsl] node is in a nodelist?

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Depends on the processor. Measure it and see.

At one time Jeni did some measurements and found that the count() method and
the generate-id() method for doing set intersection had very variable
performance on different XSLT processors, and that neither was universally
better than the other.

Michael Kay
http://www.saxonica.com/
 

> -----Original Message-----
> From: Kai Hackemesser [mailto:kaha@xxxxxx] 
> Sent: 19 June 2007 12:28
> To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> Subject: Re: [xsl] node is in a nodelist?
> 
> I understand. But with key() as nodelist this is less 
> performant than David's solution, isn't it?
> 
> Ciao!
> kai
> 
> 
> 
> Michael Kay schrieb:
> >> How do I know in a easy way if a node is in a nodelist 
> (for example 
> >> returned by key())?
> >>     
> >
> > In XSLT 2.0: exists($node intersect $nodelist)
> >
> > In 1.0: count($node | $nodelist) = count($nodelist)
> >
> > Michael Kay
> > http://www.saxonica.com/

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