How to count different data on the basis of date for distinct users in SQL Server - sql

Question

I have the following schema :
Table T
|ITEM_ID| |USER_ID| | DUE_DATE |
1 1 2018-01-2
1 1 2018-01-2
1 1 2018-01-3
1 2 2018-01-2
2 1 2018-01-2
2 2 2018-01-2
What I want is to get the count of actions performed by a unique user on that particular day, one week.
This is the select query which I'm using however it does not takes care of the fact about unique user.
select ITEM_ID,
sum(case when cast(due_date as date) = cast(getdate() as date) then 1 else 0 end) as today,
sum(case when cast(due_date as date) >= cast(getdate() - 7 as date) then 1 else 0 end) as week,
sum(case when cast(due_date as date) >= dateadd(month, -1, cast(getdate() as date)) then 1 else 0 end) as month
from t
group by ITEM_ID;
How can I the select count of actions performed by a unique user on that particular day, one week.
For example : let's suppose today is 2018-01-2
So the expected result should be :
ITEM_ID, COUNT_TODAY, COUNT_TOMORROW
1 2 1
2 1 1
What I need is the unique user performing the transaction of that particular item type for that day.
That is in the above example, the user having id 1 performed the transaction of type 1 2 times on 2018-01-2. So it should be counted as 1 instead of 2.
How can I achieve this?

Answer 1

Is this what you want.
select USER_ID, ITEM_ID,
sum(case when cast(due_date as date) = cast(getdate() as date) then 1 else 0 end) as today,
sum(case when cast(due_date as date) >= cast(getdate() - 7 as date) then 1 else 0 end) as week,
sum(case when cast(due_date as date) >= dateadd(month, -1, cast(getdate() as date)) then 1 else 0 end) as month
from t
group by USER_ID, ITEM_ID;