我有一个表,它是一个关于用户何时登录的集合条目。

username, date,      value
--------------------------
brad,     1/2/2010,  1.1
fred,     1/3/2010,  1.0
bob,      8/4/2009,  1.5
brad,     2/2/2010,  1.2
fred,     12/2/2009, 1.3

etc..

我如何创建一个查询,将给我每个用户的最新日期?

更新:我忘记了我需要有一个值与最近的日期。


当前回答

SELECT *     
FROM MyTable T1    
WHERE date = (
   SELECT max(date)
   FROM MyTable T2
   WHERE T1.username=T2.username
)

其他回答

SELECT MAX(DATE) AS dates 
FROM assignment  
JOIN paper_submission_detail ON  assignment.PAPER_SUB_ID = 
     paper_submission_detail.PAPER_SUB_ID 

这是一个简单的老派方法,适用于几乎所有的db引擎,但你必须小心重复:

select t.username, t.date, t.value
from MyTable t
inner join (
    select username, max(date) as MaxDate
    from MyTable
    group by username
) tm on t.username = tm.username and t.date = tm.MaxDate

使用窗口函数将避免由于重复的日期值而导致的任何可能的重复记录问题,所以如果你的db引擎允许它,你可以这样做:

select x.username, x.date, x.value 
from (
    select username, date, value,
        row_number() over (partition by username order by date desc) as _rn
    from MyTable 
) x
where x._rn = 1
SELECT t1.username, t1.date, value
FROM MyTable as t1
INNER JOIN (SELECT username, MAX(date)
            FROM MyTable
            GROUP BY username) as t2 ON  t2.username = t1.username AND t2.date = t1.date

对于Oracle,将结果集按降序排序,并获取第一个记录,因此您将获得最新的记录:

select * from mytable
where rownum = 1
order by date desc

您可以使用聚合函数MAX和GROUP BY

SELECT username, MAX(date), value FROM tablename GROUP BY username, value