Unique Count of YTD per month - sql

Question

I'm trying to get a YTD count for each of unique employees who have had any revenue in the current or preceding months
Table1
Month Employee Revenue
01-04-18 A 867
01-04-18 B
01-04-18 C
01-04-18 D
01-05-18 A 881
01-05-18 B
01-05-18 C 712
01-05-18 D
01-06-18 A 529
01-06-18 B 456
01-06-18 C
01-06-18 D 878
Expected Output
Month Count
01-04-18 1
01-05-18 2
01-06-18 4
In the 1st month only A had any revenue so the count is 1, in the 2nd month A & C had revenue till date so the count is 2 and finally in the 3rd month A, B, C & D have had revenue in the current or preceding months (C had revenue in month 2 but not month 3) so the count is 4.
Is there any way to get this result?
Thank you for your help

Answer 1

This is tricky, because you have an aggregation and a window function. I would go for the approach of marking the first month where a use has revenue and then using that information:
select month,
sum(sum(case when seqnum = 1 and revenue is not null then 1 else 0 end)) over (order by month)
from (select t.*,
row_number() over (partition by employee order by (case when revenue is not null then month end) nulls last) as seqnum
from t
) t
group by month;
The row_number() is enumerating the months for each employee putting the ones with revenue first. So, if there is a month with revenue, it goes first.
The outer aggregation then does a cumulative sum check both for the sequence and whether the revenue is not null.

Answer 2

I'd take a slightly different approach, still using an aggregate of an analytic function inside an inline view, but sticking to count() as I think the intent is slightly cleaeer:
select month,
count(has_revenue) as result
from (
select month, employee,
case when count(revenue)
over (partition by employee order by month) > 0
then employee end as has_revenue
from table1
)
group by month
For the inline view, the analytic count for each month/employee uses the default window of unbounded preceding to current row, so it ignore any rows in future months; and only gives a not-null response if that count is non-zero. The outer count ignore nulls in that generated column expression.
Demo with your sample data in a CTE:
with table1 (month, employee, revenue) as (
select date '2018-04-01', 'A', 867 from dual
union all select date '2018-04-01', 'B', null from dual
union all select date '2018-04-01', 'C', null from dual
union all select date '2018-04-01', 'D', null from dual
union all select date '2018-05-01', 'A', 881 from dual
union all select date '2018-05-01', 'B', null from dual
union all select date '2018-05-01', 'C', 712 from dual
union all select date '2018-05-01', 'D', null from dual
union all select date '2018-06-01', 'A', 529 from dual
union all select date '2018-06-01', 'B', 456 from dual
union all select date '2018-06-01', 'C', null from dual
union all select date '2018-06-01', 'D', 878 from dual
)
select month,
count(has_revenue) as result
from (
select month, employee,
case when count(revenue)
over (partition by employee order by month) > 0
then employee end as has_revenue
from table1
)
group by month
order by month;
MONTH RESULT
---------- ----------
2018-04-01 1
2018-05-01 2
2018-06-01 4
This is cumulative over all rows in your data set, but you only showed data from one year. If your data has multiple years, and you aren't filtering to a single year already, then add the year into the partitioning:
select month, employee,
case when count(revenue)
over (partition by employee, trunc(month, 'YYYY') order by month) > 0
then employee end as has_revenue
from table1

Answer 3

In this case I'd use a compound table expression to pull the distinct months from your table, then use COUNT(DISTINCT to count the distinct employees, using the appropriate join criteria. Or, in other words:
WITH cteMonths AS (SELECT DISTINCT MONTH
FROM TABLE1)
SELECT m.MONTH, COUNT(DISTINCT t1.EMPLOYEE)
FROM cteMonths m
INNER JOIN TABLE1 t1
ON t1.MONTH <= m.MONTH AND
t1.REVENUE IS NOT NULL
GROUP BY m.MONTH
ORDER BY m.MONTH;
SQLFiddle here
Best of luck.