On Tue, 12 Oct 2004 10:38:51 -0400, Hardy Merrill
<hmerrill(_at_)dhcr(_dot_)state(_dot_)ny(_dot_)us> wrote:
I like the idea, but I don't know how to do that. Can you take pity on
an XSL newbie and kindly show me how to do that?
Hi,
Using XSL 2.0, in a single program :
<?xml version="1.0"?>
<xsl:stylesheet version="2.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
<xsl:template match="/">
<xsl:apply-templates />
</xsl:template>
<xsl:template match="legislators">
<xsl:variable name="sorted">
<xsl:for-each select="legislator">
<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>
<xsl:text>
</xsl:text>
<xsl:for-each select="$sorted/legislator">
<xsl:variable name="districtNo" select="district_no" />
<xsl:choose>
<xsl:when
test="count(preceding-sibling::*[district_no = $districtNo]) = 0">
<xsl:value-of select="$districtNo" />
</xsl:when>
<xsl:otherwise>
<xsl:text> </xsl:text>
</xsl:otherwise>
</xsl:choose>
<xsl:text> </xsl:text>
<xsl:value-of select="district_type" />
<xsl:text> </xsl:text>
<xsl:value-of select="full_name" />
<xsl:text>
</xsl:text>
</xsl:for-each>
</xsl:template>
</xsl:stylesheet>
And this produces :
<?xml version="1.0" encoding="UTF-8"?>
001 A Doe, John J.
A Smith, Jane
002 A Jones, Daniel
If you are using an XSLT 1.0 processor, the following line in the
above code :
<xsl:for-each select="$sorted/legislator">
...would need to use the appropriate nodeset extension function
for the processor for this to work. For example, for Xalan, the line
would say :
<xsl:for-each select="xalan:nodeset($sorted)/legislator">
....assuming of course, that the "xalan" namespace prefix has been
defined correctly.
Regards,
Kenneth