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
)

检查这个链接，如果你的问题似乎类似的页面，然后我会建议你下面的查询，将给出该链接的解决方案

选择不同的sno,item_name,max(start_date) over(sno分区)，max(end_date) over(sno分区)，max(creation_date) over(sno分区)， Max (last_modified_date) over(sno分区) 从uniq_select_records 按sno,item_name asc排序;

是否会给出与该链接相关的准确结果

我不知道你的列的确切名称，但它应该是这样的:

SELECT userid, value
FROM users u1
WHERE date = (
    SELECT MAX(date)
    FROM users u2
    WHERE u1.userid = u2.userid
)

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 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;