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
这将为你编辑的问题提供正确的结果。
子查询确保只找到最近日期的行，而外部的GROUP BY将负责联系。当同一用户的同一日期有两个条目时，它将返回值最高的那个。
SELECT t.username, t.date, MAX( t.value ) value
FROM your_table t
JOIN (
SELECT username, MAX( date ) date
FROM your_table
GROUP BY username
) x ON ( x.username = t.username AND x.date = t.date )
GROUP BY t.username, t.date
其他回答
您可以使用聚合函数MAX和GROUP BY
SELECT username, MAX(date), value FROM tablename GROUP BY username, value
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
我为我的申请做了一些事情:
查询结果如下:
select distinct i.userId,i.statusCheck, l.userName from internetstatus
as i inner join login as l on i.userID=l.userID
where nowtime in((select max(nowtime) from InternetStatus group by userID));
你也可以使用分析秩函数
with temp as
(
select username, date, RANK() over (partition by username order by date desc) as rnk from t
)
select username, rnk from t where rnk = 1
SELECT Username, date, value
from MyTable mt
inner join (select username, max(date) date
from MyTable
group by username) sub
on sub.username = mt.username
and sub.date = mt.date
会解决最新的问题。在大型表上，即使有良好的索引，它也可能工作得不太好。
