Subject: RE: setStylesheetParameter() question From: Brian Dupras <briand@xxxxxxxxxxx> Date: Wed, 23 Feb 2000 12:20:41 -0700 |
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 > XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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