Partition Function COUNT() OVER possible using DISTINCT

Question

I\'m trying to write the following in order to get a running total of distinct NumUsers, like so:

NumUsers = COUNT(DISTINCT [UserAccountKey]) OVER (PARTITION         


        

Answer 1

I use a solution that is similar to that of David above, but with an additional twist if some rows should be excluded from the count. This assumes that [UserAccountKey] is never null.

-- subtract an extra 1 if null was ranked within the partition,
-- which only happens if there were rows where [Include] <> 'Y'
dense_rank() over (
  partition by [Mth] 
  order by case when [Include] = 'Y' then [UserAccountKey] else null end asc
) 
+ dense_rank() over (
  partition by [Mth] 
  order by case when [Include] = 'Y' then [UserAccountKey] else null end desc
)
- max(case when [Include] = 'Y' then 0 else 1 end) over (partition by [Mth])
- 1

An SQL Fiddle with an extended example can be found here.