Can COUNT() OVER (PARTITION BY ...) Handle DISTINCT Counts in SQL Server?

Distinct Count with Partition Functions: COUNT() OVER vs. DISTINCT

This question explores the use of SQL's OVER clause in calculating distinct counts. The user attempts to use the COUNT() OVER (PARTITION BY ...) function with the DISTINCT keyword to obtain a running total of unique users within a given month. However, this approach is met with an error.

The error arises because SQL Server's OVER functions currently do not support the DISTINCT keyword within the partition clause. As a result, the user is unable to apply distinct counting during partitioning.

A solution to this issue is to utilize the dense_rank() function. The following expression achieves the desired output:

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

By applying dense_rank() twice and subtracting the results, the expression effectively counts the number of distinct values in the specified column within each month. This method provides the same functionality as the originally intended COUNT() OVER (PARTITION BY ...) with DISTINCT.

The above is the detailed content of Can COUNT() OVER (PARTITION BY ...) Handle DISTINCT Counts in SQL Server?. For more information, please follow other related articles on the PHP Chinese website!