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)

首先，我看错了问题，下面是一个完整的例子，结果是正确的:

CREATE TABLE table_name (id int, the_value varchar(2), the_date datetime);

INSERT INTO table_name (id,the_value,the_date) VALUES(1 ,'a','1/1/2000');
INSERT INTO table_name (id,the_value,the_date) VALUES(1 ,'b','2/2/2002');
INSERT INTO table_name (id,the_value,the_date) VALUES(2 ,'c','1/1/2000');
INSERT INTO table_name (id,the_value,the_date) VALUES(2 ,'d','3/3/2003');
INSERT INTO table_name (id,the_value,the_date) VALUES(2 ,'e','3/3/2003');

--

  select id, the_value
      from table_name u1
      where the_date = (select max(the_date)
                     from table_name u2
                     where u1.id = u2.id)

--

id          the_value
----------- ---------
2           d
2           e
1           b

(3 row(s) affected)

SELECT a.* 
FROM user a INNER JOIN (SELECT userid,Max(date) AS date12 FROM user1 GROUP BY userid) b  
ON a.date=b.date12 AND a.userid=b.userid ORDER BY a.userid;

如果你在使用Postgres，你可以使用array_agg像

SELECT userid,MAX(adate),(array_agg(value ORDER BY adate DESC))[1] as value
FROM YOURTABLE
GROUP BY userid

我不熟悉甲骨文。这是我想到的

SELECT 
  userid,
  MAX(adate),
  SUBSTR(
    (LISTAGG(value, ',') WITHIN GROUP (ORDER BY adate DESC)),
    0,
    INSTR((LISTAGG(value, ',') WITHIN GROUP (ORDER BY adate DESC)), ',')-1
  ) as value 
FROM YOURTABLE
GROUP BY userid 

两个查询返回的结果都与接受的答案相同。看到SQLFiddles:

接受的答案 我对Postgres的解决方案 我对甲骨文的解决方案

以下查询可以工作:

SELECT user_id, value, date , row_number() OVER (PARTITION BY user_id ORDER BY date desc) AS rn
FROM table_name
WHERE rn= 1

使用ROW_NUMBER()为每个UserId按递减日期分配唯一的排名，然后为每个UserId过滤到第一行(即ROW_NUMBER = 1)。

SELECT UserId, Value, Date
FROM (SELECT UserId, Value, Date,
        ROW_NUMBER() OVER (PARTITION BY UserId ORDER BY Date DESC) rn
      FROM users) u
WHERE rn = 1;