这个怎么样，

Select DocumentId, DocumentSessionId, count(*) as c 
from DocumentOutputItems 
group by DocumentId, DocumentSessionId;

这将得到documententid和DocumentSessionId的所有可能组合的计数

比如:

select count(*)
from
  (select count(*) cnt
   from DocumentOutputItems
   group by DocumentId, DocumentSessionId) t1

可能只是做了和你已经做的一样的事情，但是它避免了DISTINCT。

我用过这种方法，对我很有效。

SELECT COUNT(DISTINCT DocumentID || DocumentSessionId) 
FROM  DocumentOutputItems

对于我的案例，它提供了正确的结果。

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;

下面是不带subselect的简短版本:

SELECT COUNT(DISTINCT DocumentId, DocumentSessionId) FROM DocumentOutputItems

它在MySQL中工作得很好，我认为优化器更容易理解这一点。

编辑:显然我误解了MSSQL和MySQL -对不起，但也许它有帮助。