Are you familiar with
http://www.jenitennison.com/xslt/grouping
which gives the standard XSLT 1.0 approaches to grouping problems?
Michael Kay
http://www.saxonica.com/
-----Original Message-----
From: Mindy McCutchan [mailto:karma(_at_)mindymarie(_dot_)com]
Sent: 24 April 2005 03:49
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: [xsl] XSL distinct group by date
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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