It Possible to do Sum inside Count AGGR Function

[jood_ahmad]

HI All ,

it is possible to do Sum Inside Count Aggr function

COUNT(AGGR(Sum({<  ITEM_GROUP={[XXXXX]} >},QUANTITY)), %Order_Key)

what is wrong in my formula .

Message was edited by: Ahmad Kastero

[Kushal_Chawda]

what is correct Output?

[Kushal_Chawda]

This seems to be correct

1) Count({<%Order_Key={"=Count({<ITEM_FAMILY_GROUP={'Large Pizza'}>}ITEM_FAMILY_GROUP)=1"}>}DISTINCT %Order_Key)

2) Count({<%Order_Key={"=Count({<ITEM_FAMILY_GROUP={'Large Pizza'}>}ITEM_FAMILY_GROUP)=2"}>}DISTINCT %Order_Key)

3) Count({<%Order_Key={"=Count({<ITEM_FAMILY_GROUP={'Large Pizza'}>}ITEM_FAMILY_GROUP)>2"}>}DISTINCT %Order_Key)

[sunny_talwar]

May be this:

1) =Sum(Aggr(If(Count({<ITEM_FAMILY_GROUP = {'Large Pizza'}>}ITEM_FAMILY_GROUP) = 1, 1, 0), ORDER_NUMBER))

27452

2) =Sum(Aggr(If(Count({<ITEM_FAMILY_GROUP = {'Large Pizza'}>}ITEM_FAMILY_GROUP) = 2, 1, 0), ORDER_NUMBER))

24302

3) =Sum(Aggr(If(Count({<ITEM_FAMILY_GROUP = {'Large Pizza'}>}ITEM_FAMILY_GROUP) >= 2, 1, 0), ORDER_NUMBER))

41626

The sum of the three buckets equals 93380 which equals this -> =Count(DISTINCT ORDER_NUMBER)

[sunny_talwar]

Or this without Aggr() function

1) =Count(DISTINCT {<ORDER_NUMBER = {"=Count({<ITEM_FAMILY_GROUP = {'Large Pizza'}>}ITEM_FAMILY_GROUP) = 1"}>} ORDER_NUMBER)

2) =Count(DISTINCT {<ORDER_NUMBER = {"=Count({<ITEM_FAMILY_GROUP = {'Large Pizza'}>}ITEM_FAMILY_GROUP) = 2"}>} ORDER_NUMBER)

3) =Count(DISTINCT {<ORDER_NUMBER = {"=Count({<ITEM_FAMILY_GROUP = {'Large Pizza'}>}ITEM_FAMILY_GROUP) > 2"}>} ORDER_NUMBER)

[Kushal_Chawda]

We don't really know whether he wants to count % order_key or ORDER_NUMBER

[sunny_talwar]

That's true

[jood_ahmad]

i am just check it on data base also what i need it is the Count of %Order_Key because this is combination key  in my model so no need to number of order;

i will get back with the result.

thanks all of you buddy