Re: arguments for xsl:call-template

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:  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