Subject: RE: [xsl] node is in a nodelist? From: "Michael Kay" <mike@xxxxxxxxxxxx> Date: Tue, 19 Jun 2007 14:11:18 +0100 |
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/
Current Thread |
---|
|
<- Previous | Index | Next -> |
---|---|---|
Re: [xsl] node is in a nodelist?, Kai Hackemesser | Thread | Re: [xsl] node is in a nodelist?, Dimitre Novatchev |
Re: [xsl] Dynamic pipelining in XSL, Florent Georges | Date | Re: [xsl] node is in a nodelist?, Dimitre Novatchev |
Month |