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(_dot_)the(_dot_)cynical(_dot_)robot(_at_)xxxxxxxxx>
Date: Tue, 12 Oct 2004 15:43:27 -0500
Hardy,
Have you tried $section_legislators/legistor ?
Regards,
Kenneth
On Tue, 12 Oct 2004 14:42:17 -0400, Hardy Merrill
<hmerrill(_at_)xxxxxxxxxxxxxxxx> wrote:
Here's a snippet from an XSL style sheet:
<xsl:template name="display_section">
<xsl:param name="section_legislators"/>
<tr><td colspan="4">Count section_legislators:<xsl:value-of
select="count($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>