难道一个qualified子句不是既简单又最好吗?

select userid, my_date, ...
from users
qualify rank() over (partition by userid order by my_date desc) = 1

对于上下文，在Teradata这里一个像样的大小测试运行在17秒与这个合格版本和在23秒与“内联视图”/Aldridge解决方案#1。

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;

这也会处理重复的数据(为每个user_id返回一行):

SELECT *
FROM (
  SELECT u.*, FIRST_VALUE(u.rowid) OVER(PARTITION BY u.user_id ORDER BY u.date DESC) AS last_rowid
  FROM users u
) u2
WHERE u2.rowid = u2.last_rowid

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; 

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;

select   UserId,max(Date) over (partition by UserId) value from users;