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RE: Grouping, Sorting on Aggegrated sum of the groupings, for Top XX Group

2005-06-14 07:55:12
Darin,

Here's an approach using the extension function "nodeset". MSXML has it as "node-set". This can be done without nodeset, but AFAICT those approaches are longer. Thus, on your input (below), this XSL:

<?xml version="1.0" encoding="iso8859-1"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
   xmlns:xalan="http://xml.apache.org/xalan";
   exclude-result-prefixes="xalan">

   <xsl:output method="xml" indent="yes"/>

   <xsl:template match="x">
       <xsl:variable name="sorted-y">
           <xsl:for-each select="y">
<xsl:sort select="@Value" order="descending" data-type="number"/>
               <xsl:copy-of select="."/>
           </xsl:for-each>
       </xsl:variable>

       <xsl:copy>
<xsl:apply-templates select="xalan:nodeset($sorted-y)/y[position() &lt;= 5]"/>
           <xsl:call-template name="sum-y">
<xsl:with-param name="n" select="xalan:nodeset($sorted-y)/y[position() &gt; 5]"/>
           </xsl:call-template>
       </xsl:copy>
   </xsl:template>

   <xsl:template match="y">
       <xsl:copy-of select="."/>
   </xsl:template>

   <xsl:template name="sum-y">
       <xsl:param name="n" select="/.."/>
       <xsl:param name="total" select="0"/>

       <xsl:choose>
           <xsl:when test="count($n) &gt; 0">
               <xsl:call-template name="sum-y">
                   <xsl:with-param name="n" select="$n[position() != 1]"/>
<xsl:with-param name="total" select="$total + $n[1]/@Value"/>
               </xsl:call-template>
           </xsl:when>
           <xsl:otherwise>
               <remainder Value="{$total}"/>
           </xsl:otherwise>
       </xsl:choose>
   </xsl:template>

</xsl:stylesheet>

produces:

<?xml version="1.0" encoding="UTF-8"?>
<x>
 <y Id="2" Name="Label1" Value="2760"/>
 <y Id="1" Name="Label" Value="2151"/>
 <y Id="14" Name="Label6" Value="193"/>
 <y Id="13" Name="Label5" Value="118"/>
 <y Id="25" Name="Label9" Value="100"/>
 <remainder Value="551"/>
</x>

Regards,

--A



From: "Darin" <darind(_at_)nc(_dot_)rr(_dot_)com>
Reply-To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
To: <xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com>
Subject: [xsl] Grouping, Sorting on Aggegrated sum of the groupings, for Top XX Groupings
Date: Mon, 13 Jun 2005 16:14:39 -0400

I am using MSXML 4.0 running on an IIS 5.0 server running classic ASP.

I'm trying to transform the xml below grouped by the Id attribute, and sort
(descending) on the aggregated sum(Value).  I'm also trying to limit the
amount of groupings to 6, where the first 5 groupings are the highest
aggregated sum values, and the 6th grouping is an aggregation of the
remaining groups.

Here is an example of the XML:

<x>
    <y Id="1" Name="Label" Value="2151"/>
    <y Id="2" Name="Label1" Value="2760"/>
    <y Id="2" Name="Label1" Value="50"/>
    <y Id="3" Name="Label2" Value="99"/>
    <y Id="4" Name="Label3" Value="9"/>
    <y Id="4" Name="Label3" Value="55"/>
    <y Id="5" Name="Label4" Value="56"/>
    <y Id="13" Name="Label5" Value="118"/>
    <y Id="14" Name="Label6" Value="193"/>
    <y Id="17" Name="Label7" Value="54"/>
    <y Id="23" Name="Label8" Value="76"/>
    <y Id="23" Name="Label8" Value="76"/>
    <y Id="23" Name="Label8" Value="76"/>
    <y Id="25" Name="Label9" Value="100"/>
</x>

I would like the xml to look like:

<x>
<!-- Summation of the Top 5 entries aggregated sum (Desc) -->
    <y Id="2" Name="Label1" Value="2810"/>
    <y Id="1" Name="Label" Value="2151"/>
    <y Id="23" Name="Label8" Value="228"/>
    <y Id="14" Name="Label6" Value="193"/>
    <y Id="13" Name="Label5" Value="118"/>

<!-- Summation of the remaining entries -->
    <y Id="" Name="Other" Value="373"/>

<!-- <y Id="3" Name="Label2" Value="99"/>
    <y Id="4" Name="Label3" Value="64"/>
    <y Id="5" Name="Label4" Value="56"/>
    <y Id="17" Name="Label7" Value="54"/>
    <y Id="25" Name="Label9" Value="100"/> -->
</x>


I have tried to use Keys and the XPath position function, but with no
success.

Thanks in an advance for any help!

Darin

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