RE: setStylesheetParameter() question

Woops - you're right.  I kinda misunderstood the structure of things - an
xsl:param is not a template parameter, it's a stylesheet paramater.  :)

Thanks,

Brian

> -----Original Message-----
> From: Steve Tinney [mailto:stinney@xxxxxxxxxxxxx]
> Sent: Wednesday, February 23, 2000 10:46 AM
> To: xsl-list@xxxxxxxxxxxxxxxx
> Subject: Re: setStylesheetParameter() question
> 
> 
> >   xsltProcessor.setStylesheetParam("testing", "'testing 
> val'");  //note the
> > single quotes
> > 
> > <xsl:template match="/">
> >   <xsl:apply-templates>
> >     <xsl:with-param name="testing"/>
> >   </xsl:apply-templates>
> > </xsl:template>
> > 
> > <xsl:template match="someelement">
> >   <xsl:param name="testing"/>
> >   <xsl:value-of select="$testing"/>
> > </xsl:template>
> 
> Looks to me like it might help to:
> 
>  (a) declare a top-level xsl:param name="testing"
>  (b) remove the declaration of the param in someelement
>  (c) remove the xsl:with-param from the apply-templates
> 
>  Steve
> 
> 
>  XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list
> 


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