Sum distinct CustomerID

[#1]

Hi All,

i am sitting with a requirement to write:

the sum of Revenue where Customer ID is equal to Distinct

but have no idea how to tackle this one, any help would be accepted.

Thanks.

[sunny_talwar]

Would you be able to share some data to see what you are trying to accomplish?

[prma7799]

What you actually want. Can you elaborate more?

[avinashelite]

may be like this:

if (count(Customer ID)=1,sum(Revenue ))

if this doesn't give the desired result , then you  need check the count in the SQL i.e in the script level and set a flag for the same. Finally based on the flag value you can calculate the sum

[tresesco]

May be like:

Sum({<[Customer ID={"=Count([Customer ID])=1"}]>}Revenue)

[Anonymous]

hi

maybe:

sum(aggr(sum(DISTINCT Revenue), Customer ID))

[#7]

Hi All,

im trying to implement most Solutions you guys are giving me,

one more thing that i want is: Distinct CustID and Number of Revenues made each distinct CustID Made.

Thanks.

[sunny_talwar]

Alternative to this could be:

=Sum(Aggr(Only(Revenue), CustID)) or

=Sum(Aggr(Avg(Revenue), CustID))

[avinashelite]

yes, count(Customer ID)=1 will give you you the Distinct custID....once this condition is satisfied..for the distinct Custid you will get the sum 

[tresesco]

Try:

Straight table:

Dim: [Customer ID]

Exp: Sum({<[Customer ID={"=Count([Customer ID])=1"}]>}Revenue)