Subject: Re: [xsl] arguments for xsl:call-template From: "Charly" <cohana@xxxxxxxxxxxxxxx> Date: Fri, 7 Dec 2001 10:34:21 -0800 |
I'm trying to write a template that generate an SVG graph just by passing the root element. It's working now. All I had to do is to pass only one set of quotes <xsl:with-param name="path" select="/report/histo/bar" /> instead of <xsl:with-param name="path" select="'/report/histo/bar'" /> Thanks for your help ----- Original Message ----- From: "Pep Coll" <pcoll@xxxxxxxxxxxxxxxxxxxxx> To: <xsl-list@xxxxxxxxxxxxxxxxxxxxxx> Sent: Friday, December 07, 2001 7:01 AM Subject: Re: [xsl] arguments for xsl:call-template > Hi Charly! > I don't understand quite well what you want, because you can do this: > <xsl:with-param name="path" select="'/report/histo/bar'" /> > but you can do this , > <xsl:with-param name="path" select="'/report/histo/@bar'" /> > because you are not assigning anything to var. path and also you are doing > histogram just once so the 'for-each' has no reason to be. Explain what's > the purpose of this template. > > At 09:38 07/12/01 -0800, you wrote: > >Hi everyone. > >I'm a beginner with XSL . > >Does anyone of you has any idea how to call a template and passing with a > >node as argument. > >I'm trying the following . > > > ><xsl:template name="histogram"> > > <xsl:param name="path" /> > > <xsl:for-each select="$path"> > > <xsl:value-of select="@value"/> > > </xsl:for-each> > ></xsl:template> > > > ><xsl:call-template name="histogram"> > > <xsl:with-param name="path" select="'/report/histo/bar'" /> > ></xsl:call-template> > > > >my XML looks like . > ><report> > > <histo> > > <bar value="20" /> > > <bar value="30" /> > > <bar value="40" /> > > <bar value="50" /> > > </histo> > > > ></report> > > > > > > XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list > > > 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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