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

There is a very simple solution using dense_rank()

dense_rank() over (partition by [Mth] order by [UserAccountKey]) 
+ dense_rank() over (partition by [Mth] order by [UserAccountKey] desc) 
- 1

This will give you exactly what you were asking for: The number of distinct UserAccountKeys within each month.