Re: [xsl] arguments for xsl:call-template

["Charly" <cohana at investacorp dot com>]

Works great .
Thank you so much

----- Original Message -----
From: "Thomas B. Passin" <tpassin@mitretek.org>
To: <xsl-list@lists.mulberrytech.com>
Sent: Friday, December 07, 2001 7:13 AM
Subject: Re: [xsl] arguments for xsl:call-template


> [Charly]
>
> > 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>
> >
> Just change to one set of quotes in the select attribute and the select
will
> return a set of nodes, just what you asked for.  In other words, your
> attribute value as you wrote it is a string (because of the extra set of
> quotes), and you want it to be a node-set.
>
> Tom P
>
> > 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
>
>


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