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

["Pep Coll" <pcoll@xxxxxxxxxxxxxxxxxxxxx>]

YES, you're right!
I wanted to say <xsl:with-param name="path"
select="/report/histo/bar/@value" />

----- Original Message -----
From: "Thomas B. Passin" <tpassin@xxxxxxxxxxxx>
To: <xsl-list@xxxxxxxxxxxxxxxxxxxxxx>
Sent: Friday, December 07, 2001 4:20 PM
Subject: Re: [xsl] arguments for xsl:call-template


> [Pep Coll]
>
> > 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'" />
>
> NO, @bar would only return attributes named "bar" and the original example
> had "bar" elements, not attributes.
>
> Tom P
>
> > 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.
> >
>
>
>
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