RE: [xsl] how select all siblings?

Subject: RE: [xsl] how select all siblings?
From: "Hardy Merrill" <HMerrill@xxxxxxxxxxxxxxxx>
Date: Wed, 13 Oct 2004 09:37:19 -0400
Kenneth, yes I have tried that - I still get 

    Count section_legislators:151
    Count sorted_legislators: 1

Remember that <xsl:param name="section_legislators"/>
is a node set containing 151 "legislator" nodes. 

Any idea why the "sorted_legislators" variable only contains 1 node,
and
not 151 ???

Thanks.

Hardy Merrill

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Subject: Re: [xsl] how select all siblings?
From: Kenneth Stephen <marvin.the.cynical.robot@xxxxxxxxx>
Date: Tue, 12 Oct 2004 15:43:27 -0500
 

Hardy,

    Have you tried $section_legislators/legistor ? 

Regards,
Kenneth

On Tue, 12 Oct 2004 14:42:17 -0400, Hardy Merrill
<hmerrill@xxxxxxxxxxxxxxxx> wrote:
> Here's a snippet from an XSL style sheet:
> 
> <xsl:template name="display_section">
>         <xsl:param name="section_legislators"/>
> 
>         <tr><td colspan="4">Count section_legislators:<xsl:value-of
> select="count($section_legislators)"/></td></tr>
>         <xsl:variable name="sorted_legislators">
>                 <xsl:for-each select="$section_legislators">
>                         <xsl:sort select="district_type" />
>                         <xsl:sort select="district_no"
> data-type="number" />
>                         <xsl:sort select="legislator_active_date"
> data-type="number" order="descending" />
>                         <xsl:copy-of select="."/>
>                 </xsl:for-each>
>         </xsl:variable>
>         <tr><td colspan="4">Count sorted_legislators: <xsl:value-of
> select="count(msxsl:node-set($sorted_legislators))"/></td></tr>

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