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

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)

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
)

在Oracle 12c+中，你可以使用Top n查询和分析函数排名来实现这一点，而且不需要子查询:

select *
from your_table
order by rank() over (partition by user_id order by my_date desc)
fetch first 1 row with ties;

上面的代码返回每个用户my_date最大的所有行。

如果你只想要一个最大日期的行，那么用row_number替换秩:

select *
from your_table
order by row_number() over (partition by user_id order by my_date desc)
fetch first 1 row with ties; 

在PostgreSQL 8.4或更高版本中，你可以这样使用:

select user_id, user_value_1, user_value_2
  from (select user_id, user_value_1, user_value_2, row_number()
          over (partition by user_id order by user_date desc) 
        from users) as r
  where r.row_number=1

Select  
   UserID,  
   Value,  
   Date  
From  
   Table,  
   (  
      Select  
          UserID,  
          Max(Date) as MDate  
      From  
          Table  
      Group by  
          UserID  
    ) as subQuery  
Where  
   Table.UserID = subQuery.UserID and  
   Table.Date = subQuery.mDate