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 BY  Mth     Management studio doesn t seem too happy about this  The error disappears when I remove the DISTINCT keyword  but then it won t be a distinct count   DISTINCT does not appear to be possible within the partition functions   How do I go about finding the distinct count  Do I use a more traditional method such as a correlated subquery   Looking into this a bit further  maybe these OVER functions work differently to Oracle in the way that they cannot be used in SQL-Server to calculate running totals   I ve added a live example here on SQLfiddle where I attempt to use a partition function to calculate a running total

User · Answer

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.

User · Answer

Necromancing   It s relativiely simple to emulate a COUNT DISTINCT over PARTITION BY  with MAX via DENSE RANK    WITH baseTable AS       SELECT  RM1  AS RM   ADR1  AS ADR     UNION ALL SELECT  RM1  AS RM   ADR1  AS ADR     UNION ALL SELECT  RM2  AS RM   ADR1  AS ADR     UNION ALL SELECT  RM2  AS RM   ADR2  AS ADR     UNION ALL SELECT  RM2  AS RM   ADR2  AS ADR     UNION ALL SELECT  RM2  AS RM   ADR3  AS ADR     UNION ALL SELECT  RM3  AS RM   ADR1  AS ADR     UNION ALL SELECT  RM2  AS RM   ADR1  AS ADR     UNION ALL SELECT  RM3  AS RM   ADR1  AS ADR     UNION ALL SELECT  RM3  AS RM   ADR2  AS ADR    CTE AS       SELECT RM  ADR  DENSE RANK   OVER PARTITION BY RM ORDER BY ADR  AS dr      FROM baseTable   SELECT      RM      ADR       COUNT CTE ADR  OVER  PARTITION BY CTE RM ORDER BY ADR  AS cnt1       COUNT CTE ADR  OVER  PARTITION BY CTE RM  AS cnt2      -- Not supported     -- COUNT DISTINCT CTE ADR  OVER  PARTITION BY CTE RM ORDER BY CTE ADR  AS cntDist      MAX CTE dr  OVER  PARTITION BY CTE RM ORDER BY CTE RM  AS cntDistEmu  FROM CTE   Note  This assumes the fields in question are NON-nullable fields   If there is one or more NULL-entries in the fields  you need to subtract 1

User · Answer

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

User · Answer

I think the only way of doing this in SQL-Server 2008R2 is to use a correlated subquery  or an outer apply   SELECT  datekey          COALESCE RunningTotal  0  AS RunningTotal          COALESCE RunningCount  0  AS RunningCount          COALESCE RunningDistinctCount  0  AS RunningDistinctCount FROM    document         OUTER APPLY             SELECT  SUM Amount  AS RunningTotal                      COUNT 1  AS RunningCount                      COUNT DISTINCT d2 dateKey  AS RunningDistinctCount             FROM    Document d2             WHERE   d2 DateKey  lt   document DateKey           rt    This can be done in SQL-Server 2012 using the syntax you have suggested   SELECT  datekey          SUM Amount  OVER ORDER BY DateKey  AS RunningTotal FROM    document   However  use of DISTINCT is still not allowed  so if DISTINCT is required and or if upgrading isn t an option then I think OUTER APPLY is your best option

[sql] Partition Function COUNT() OVER possible using DISTINCT

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