这也会处理重复的数据(为每个user_id返回一行):

SELECT *
FROM (
  SELECT u.*, FIRST_VALUE(u.rowid) OVER(PARTITION BY u.user_id ORDER BY u.date DESC) AS last_rowid
  FROM users u
) u2
WHERE u2.rowid = u2.last_rowid

(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

我想是这样的。(请原谅我的语法错误;在这一点上，我习惯使用HQL !)

编辑:也误解了问题!修正了查询…

SELECT UserId, Value
FROM Users AS user
WHERE Date = (
    SELECT MAX(Date)
    FROM Users AS maxtest
    WHERE maxtest.UserId = user.UserId
)

只是需要在工作中写一个“活”的例子:)

它支持在同一日期为UserId设置多个值。

列: 用户id，值，日期

SELECT
   DISTINCT UserId,
   MAX(Date) OVER (PARTITION BY UserId ORDER BY Date DESC),
   MAX(Values) OVER (PARTITION BY UserId ORDER BY Date DESC)
FROM
(
   SELECT UserId, Date, SUM(Value) As Values
   FROM <<table_name>>
   GROUP BY UserId, Date
)

您可以使用FIRST_VALUE而不是MAX，并在解释计划中查找它。我没有时间玩它。

当然，如果搜索巨大的表，在查询中使用FULL提示可能会更好。

如果(UserID, Date)是唯一的，即同一个用户没有出现两次日期，则:

select TheTable.UserID, TheTable.Value
from TheTable inner join (select UserID, max([Date]) MaxDate
                          from TheTable
                          group by UserID) UserMaxDate
     on TheTable.UserID = UserMaxDate.UserID
        TheTable.[Date] = UserMaxDate.MaxDate;

我知道你要求使用Oracle，但是在SQL 2005中我们现在使用这个:

-- Single Value
;WITH ByDate
AS (
SELECT UserId, Value, ROW_NUMBER() OVER (PARTITION BY UserId ORDER BY Date DESC) RowNum
FROM UserDates
)
SELECT UserId, Value
FROM ByDate
WHERE RowNum = 1

-- Multiple values where dates match
;WITH ByDate
AS (
SELECT UserId, Value, RANK() OVER (PARTITION BY UserId ORDER BY Date DESC) Rnk
FROM UserDates
)
SELECT UserId, Value
FROM ByDate
WHERE Rnk = 1