
1. Re: calculated field for data at granular level
Andrew Ball Jun 3, 2013 12:48 PM (in response to siddhartha.siddhartha)Hi Siddhartha,
Are you just doing a reverse running sum? Or is E always always equal to E+R ?
The other thing is how do you want this to appear?  is it a whole long list of E, I and R and you want the full sum against each, or just three figures at a top level?
Apologies for lots of questions, but there are many ways of solving your problem, and the best way depends on exactly what you want out at the end.
Andrew

2. Re: calculated field for data at granular level
siddhartha.siddhartha Jun 3, 2013 1:15 PM (in response to Andrew Ball)Hi Andrew
I am not doing the reverse running sum.
Actually its a whole long list of I, E and R. To explain it more its like
I 5
E 6
R 2
I 1
E 1
R 1
So its like for "I" i should get the entire sum i.e. 5+6+2+1+1+1. For E its E+R and for R its only R.
i Hope this helps.
Thanks
Sid

3. Re: calculated field for data at granular level
Andrew Ball Jun 3, 2013 1:25 PM (in response to siddhartha.siddhartha)Hi Sid,
You could do this using 3 new measures (well, you only need 2, but I'd create 3 for completeness):
Measure "I"
[Amount]
Use SUM(I) in the view
Measure "E"
IF [Event Name] = "I" THEN null ELSE [Amount] END
Use SUM(E) in the view
Measure "R"
IF [Event Name] = "R" THEN [Amount] END
Use SUM(R) in the view
This method is assuming you don't want to see the sum of all I, E and R against every I. Let me know if this is not the case.
Andrew

4. Re: calculated field for data at granular level
siddhartha.siddhartha Jun 3, 2013 2:04 PM (in response to Andrew Ball)
Thanks Andrew. It was of great help.