Need to get Running Aggregate
My report should be like below. Type Invoice # Quantity Progressive Travel 1 50.00 2 75.00 Total 125.00 Sub 1 25.00 2 10.0 3 30.00 Total 190.00 MISC 1 3.0 Total 193.00 I want to add the current aggregate “Quantity Progressive “. When I add total for the group, it gives me group total not the running aggregate. Then I used Running value as below but got the same value. Runningvalue(Fields!Quantity Progressive, Sum, “Group1”) Is there any way to get sum as in the picture. For an example At the MiSC the “Quantity Progressive” total gives Previous total ( 190.00) + Misc total (3.00) Appreciate your help on this
May 17th, 2010 8:31pm

Assuming that your table looks like this: Type Invoice Price Travel 1 50 Travel 2 75 Sub 1 25 Sub 2 10 Sub 3 30 MISC 1 3 Then this is how I would do it. Create your dataset with this SQL statement: select Type, Invoice, Price from mytable order by Type, Invoice Create a table on the design palette with two columns, Invoice and Price Click on the frame of the table. This will display the Row Groups section below the Design palette. Create a group on the Type column by dragging Type down to the Row Group and dropping it above the "Details" row. This will create a Row group probably named "Type". Now, in the Row Groups section, click on "Details" again, and "Add Total--After". This will create a "Sum(value))" function in each detail cell of the row. Remove all the Sum()s and replace it with your RunningValue function. =RunningValue(Fields!Price.Value,Sum,Nothing) Done! I hope this helps. --DanDJAnsc
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May 17th, 2010 11:53pm

Thank you Very much Dan !!. Yes It's working.
May 24th, 2010 5:27pm

Hi, Mr. DJAnsc Nice very good Answer with proper logic. i been to this link to get RunningValue solution which got it very perfect as i have used to get the Quantity Vs hours graph to show hourly output and found it very helpfull. and used as sum of Quantity in Field Value expression =RunningValue(Sum(Fields!Quantity.Value),Sum,Nothing) and X axis are running hours. and finally solve problem. Thanks very much for the nice and proper answer. Eric Ericcisco
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January 17th, 2011 7:30am

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