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

[Pep Coll <pcoll at globalia-sistemas dot com>]

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>
>
>
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 XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list