只是使用 count() . 根据示例数据，您可以分别查看每一列：

select user_name,
       count(submitted) as num_submitted,
       count(reviewed) as num_reviewed,
       count(returned) as num_returned
from t
group by user_name;

例如，没有任何例子 returned 是非- NULL 其他任何一列都是 NULL .
如果可能的话，您可以使用条件聚合：

select user_name,
       count(submitted) as num_submitted,
       sum(case when submitted is not null and reviewed is not null then 1 else 0 end) as num_reviewed,
       sum(case when submitted is not null and reviewed is not null and returned is not null then 1 else 0 end) as num_returned
from t
group by user_name;

你也可以用 count() 玩算术游戏：

select user_name,
       count(submitted) as num_submitted,
       count(day(submitted) + day(reviewed)) as num_reviewed,
       count(day(submitted) + day(reviewed) + day(returned)) as num_returned
from t
group by user_name;

这是因为 day() 退货 NULL 如果值为 NULL . 以及 + 退货 NULL 如果任何值 NULL .

试试这个：

DECLARE @DataSource TABLE
(
    [id] INT
   ,[user_name] NVARCHAR(128)
   ,[submitted] DATE 
   ,[reviewed] DATE 
   ,[returned] DATE
);

INSERT INTO @DataSource ([id], [user_name], [submitted], [reviewed], [returned])
VALUES (1, 'tom', '01-01-2020', '02-01-2020', NULL)
      ,(2, 'mary', '01-15-2020', NULL, NULL)
      ,(3, 'joe', '01-25-2020', '02-07-2020', '03-04-2020')
      ,(4, 'tom', '01-07-2020', NULL, NULL)
      ,(5, 'tom', '01-04-2020', NULL, NULL)
      ,(6, 'mary', '01-16-2020', NULL, NULL)
      ,(7, 'joe', '02-08-2020', '02-08-2020', '03-07-2020')
      ,(8, 'mary', '01-05-2020', '01-20-2020', '03-19-2020')
      ,(9, 'joe', '01-21-2020', '02-09-2020', NULL);

SELECT [user_name]
      ,SUM(IIF([returned] IS NULL AND [reviewed] IS NULL AND [submitted] IS NOT NULL, 1, 0)) AS [ # Submitted]
      ,SUM(IIF([returned] IS NULL AND [reviewed] IS NOT NULL AND [submitted] IS NOT NULL, 1, 0)) AS [# Reviewed]
      ,SUM(IIF([returned] IS NOT NULL AND [reviewed] IS NOT NULL AND [submitted] IS NOT NULL, 1, 0)) AS [# Returned]
FROM @DataSource
GROUP BY [user_name];