你可以使用Count函数两次。

在这种情况下，它将是:

SELECT COUNT (DISTINCT DocumentId), COUNT (DISTINCT DocumentSessionId) 
FROM DocumentOutputItems

你的查询没有问题，但你也可以这样做:

WITH internalQuery (Amount)
AS
(
    SELECT (0)
      FROM DocumentOutputItems
  GROUP BY DocumentId, DocumentSessionId
)
SELECT COUNT(*) AS NumberOfDistinctRows
  FROM internalQuery

这对我很管用。在oracle中:

SELECT SUM(DECODE(COUNT(*),1,1,1))
FROM DocumentOutputItems GROUP BY DocumentId, DocumentSessionId;

在jpql:

SELECT SUM(CASE WHEN COUNT(i)=1 THEN 1 ELSE 1 END)
FROM DocumentOutputItems i GROUP BY i.DocumentId, i.DocumentSessionId;

希望这能起作用，我正在prima vista上写

SELECT COUNT(*) 
FROM DocumentOutputItems 
GROUP BY DocumentId, DocumentSessionId

如果您使用的是固定长度的数据类型，则可以将其转换为二进制，从而非常容易和快速地完成此操作。假设documententid和DocumentSessionId都是int，因此都是4字节长…

SELECT COUNT(DISTINCT CAST(DocumentId as binary(4)) + CAST(DocumentSessionId as binary(4)))
FROM DocumentOutputItems

My specific problem required me to divide a SUM by the COUNT of the distinct combination of various foreign keys and a date field, grouping by another foreign key and occasionally filtering by certain values or keys. The table is very large, and using a sub-query dramatically increased the query time. And due to the complexity, statistics simply wasn't a viable option. The CHECKSUM solution was also far too slow in its conversion, particularly as a result of the various data types, and I couldn't risk its unreliability.

然而，使用上述解决方案几乎没有增加查询时间(与简单使用SUM相比)，并且应该是完全可靠的!它应该能够帮助其他处于类似情况的人，所以我把它贴在这里。

这段代码使用distinct on 2参数，并提供特定于这些不同值的行数计数。它在MySQL中为我工作，就像一个魅力。

select DISTINCT DocumentId as i,  DocumentSessionId as s , count(*) 
from DocumentOutputItems   
group by i ,s;