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Re: Comparing node for identity using union

Subject: Re: Comparing node for identity using union
From: António Mota <amsmota@xxxxxxxxx>
Date: Wed, 19 Jan 2005 19:08:29 +0000
processor.transformtodocument
Neverthless, just for the record, please note

       Sarissa.setXslParameter(xslDoc, "pos", pos+"");
       xml.transformNodeToObject(xslDoc, xmlResult);   ----> Works OK

       Sarissa.setXslParameter(xslDoc, "pos", pos);
       xml.transformNodeToObject(xslDoc, xmlResult);   ----> DOES NOT

       processor.setParameter(null, "pos", pos);            
       var xmlResult = processor.transformToDocument(xmlDoc);     
----> Works OK

       processor.setParameter(null, "pos", pos+"");
       var xmlResult = processor.transformToDocument(xmlDoc);     
----> DOES NOT


Works OK = $pos is recognized as a number so (//Menu)[$pos] gives the
Menu as position $pos

DOES NOT =  $pos is NOT recognized as a number so (//Menu)[$pos] gives
ALL the Menu  nodes, as just //Menu does.




On Wed, 19 Jan 2005 18:10:20 -0000, Michael Kay <mike@xxxxxxxxxxxx> wrote:
> >
> > > > The $pos is defined like
> > > >
> > > >   <xsl:param name="pos" select="0"/>
> > >
> > > That's certainly a number
> >
> > I think it's only a number as long as it doesn't get overridden - as
> > soon as you pass in the value a parameter it becomes a string.
> 
> Of course: I missed that.
> 
> Shows how useful it is to declare your types in XSLT 2.0.
> 
> Michael Kay
> http://www.saxonica.com/

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