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
)

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

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

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  

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

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;