How to Calculate Distinct Counts Within Partitions in SQL Server Without Using DISTINCT?

Calculating Distinct Counts Using Partition Functions

In SQL Server, the COUNT() OVER function allows you to aggregate data within specified partitions. However, using the DISTINCT keyword within the partition functions is not supported. This leaves developers wondering how to obtain distinct counts within partitions.

Using DENSE_RANK

A simple solution for calculating distinct counts is to utilize the DENSE_RANK function. The following query demonstrates how:

DENSE_RANK() OVER (PARTITION BY [Mth] ORDER BY [UserAccountKey])
+ DENSE_RANK() OVER (PARTITION BY [Mth] ORDER BY [UserAccountKey] DESC)
- 1

This expression calculates the number of distinct UserAccountKeys within each month. It does so by first ranking the UserAccountKeys in ascending and descending order within each month. These rankings are then added together and 1 is subtracted to give the desired count.

Example

Consider the following data:

| Mth | UserAccountKey |
|---|---|
| 1 | A |
| 1 | B |
| 1 | C |
| 2 | D |
| 2 | E |
| 2 | F |

Using the above expression, the query will return the following results:

| Mth | NumUsers |
|---|---|
| 1 | 3 |
| 2 | 3 |

Conclusion

The DENSE_RANK function provides a concise and efficient method for calculating distinct counts within partitions in SQL Server, even though the DISTINCT keyword is not directly supported in thePARTITION BY clause.

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