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Re: Grouping, Sorting on Aggegrated sum of the groupings, for Top XX Groupings

2005-06-13 20:38:54
Hello Darin,
  I have a collection of XSLT grouping problems at -
http://gandhimukul.tripod.com/grouping.html .. Hope it
might help you ;)

Regards,
Mukul

--- Darin <darind(_at_)nc(_dot_)rr(_dot_)com> wrote:

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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