表:

UserId, Value, Date.

我想获得UserId,为每个UserId的最大值(日期)的值。也就是说,具有最新日期的每个UserId的值。有没有一种方法可以在SQL中简单地做到这一点?(最好是Oracle)

更新:为任何歧义道歉:我需要得到所有的用户id。但是对于每个UserId,只有该用户拥有最新日期的行。


当前回答

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

其他回答

(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

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; 
SELECT a.userid,a.values1,b.mm 
FROM table_name a,(SELECT userid,Max(date1)AS mm FROM table_name GROUP BY userid) b
WHERE a.userid=b.userid AND a.DATE1=b.mm;

假设Date对于给定的UserID是唯一的,下面是一些TSQL:

SELECT 
    UserTest.UserID, UserTest.Value
FROM UserTest
INNER JOIN
(
    SELECT UserID, MAX(Date) MaxDate
    FROM UserTest
    GROUP BY UserID
) Dates
ON UserTest.UserID = Dates.UserID
AND UserTest.Date = Dates.MaxDate 
select VALUE from TABLE1 where TIME = 
   (select max(TIME) from TABLE1 where DATE= 
   (select max(DATE) from TABLE1 where CRITERIA=CRITERIA))