SELECT userid, MAX(value) KEEP (DENSE_RANK FIRST ORDER BY date DESC)
  FROM table
  GROUP BY userid

我认为你应该对之前的查询进行修改:

SELECT UserId, Value FROM Users U1 WHERE 
Date = ( SELECT MAX(Date)    FROM Users where UserId = U1.UserId)

假设Date对于给定的UserID是唯一的，下面是一些TSQL:

SELECT 
    UserTest.UserID, UserTest.Value
FROM UserTest
INNER JOIN
(
    SELECT UserID, MAX(Date) MaxDate
    FROM UserTest
    GROUP BY UserID
) Dates
ON UserTest.UserID = Dates.UserID
AND UserTest.Date = Dates.MaxDate 

select VALUE from TABLE1 where TIME = 
   (select max(TIME) from TABLE1 where DATE= 
   (select max(DATE) from TABLE1 where CRITERIA=CRITERIA))

SELECT a.userid,a.values1,b.mm 
FROM table_name a,(SELECT userid,Max(date1)AS mm FROM table_name GROUP BY userid) b
WHERE a.userid=b.userid AND a.DATE1=b.mm;

(T-SQL)首先获取所有用户及其最大日期。与表连接以查找maxdates上用户的对应值。

create table users (userid int , value int , date datetime)
insert into users values (1, 1, '20010101')
insert into users values (1, 2, '20020101')
insert into users values (2, 1, '20010101')
insert into users values (2, 3, '20030101')

select T1.userid, T1.value, T1.date 
    from users T1,
    (select max(date) as maxdate, userid from users group by userid) T2    
    where T1.userid= T2.userid and T1.date = T2.maxdate

结果:

userid      value       date                                    
----------- ----------- -------------------------- 
2           3           2003-01-01 00:00:00.000
1           2           2002-01-01 00:00:00.000