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XSL distinct group by date

Subject: XSL distinct group by date
From: "Mindy McCutchan" <karma@xxxxxxxxxxxxxx>
Date: Sat, 23 Apr 2005 21:49:25 -0500
distinct group by
Hi Everyone,
I'm very new to XML/XSL, so I'm struggling with something that seems it
should be fairly straightforward. With the following XML:

<documents>
	<casestudies>
		<study id="">
			<title />
			<date />
			<description><![CDATA[ ]]></description>
			<pdf />
			<logo />
		</study>
	</casestudies>
	<knowledgeadmin>
		<pressreleases>
			<pressrelease id="">
				<title />
				<date />
				<description><![CDATA[ ]]></description>
				<pdf />
			</pressrelease>
		</pressreleases>
		<whitepapers>
			<whitepaper id="">
				<id />
				<title />
				<date />
				<description><![CDATA[ ]]></description>
				<pdf />
			</whitepaper>
		</whitepapers>
	</knowledgeadmin>
	<news>
		<newsitem id="">
			<title />
			<date />
			<description><![CDATA[ ]]></description>
			<fulltext><![CDATA[ ]]></fulltext>
		</newsitem>
	</news>
</documents>

I want to be able to get the different groups (newsitem, whitepaper,
pressrelease) each grouped by year. Using news as example, I want to
generate a list of distinct years in news/newsitem/date and make that a URL
that can be clicked to get the list of newsitems that were in that year. I'm
open to suggestions on how the date should be formatted. I'm at a stage
where I can restructure the XML as well, if that appears necessary.

I'm on Win 2K3 Web Edition server using VBScript. Here is my transformation
code, which should show the processors/versions used:

<%
' load list of news dats from xml
Dim oXSL
Dim myTemplate
Dim myProc
Dim oXML
				
set oXML = server.createobject("Microsoft.XMLDOM")
oXML.async = false
oXML.load( dbPath )
				
Set oXSL = Server.CreateObject("Msxml2.FreeThreadedDOMDocument.4.0")
oXSL.async = false
oXSL.load Server.MapPath("xsl/date_list.xsl")

' compliled XSL template
Set myTemplate = Server.CreateObject("Msxml2.XSLTemplate.4.0")
myTemplate.stylesheet = oXSL

Set myProc = myTemplate.createProcessor() myProc.input = oXML

myProc.output = Response
myProc.transform()
%>

I'm on a very tight deadline and having been trying/researching for several
hours for a solution to this, so any help would be much appreciated.

Thank you in advance!

Mindy

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