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 ReportStatus c
inner join ( SELECT
MAX(Date) AS MaxDate
FROM ReportStatus ) m
on c.date = m.maxdate
其他回答
SELECT DISTINCT Username, Dates,value
FROM TableName
WHERE Dates IN (SELECT MAX(Dates) FROM TableName GROUP BY Username)
Username Dates value
bob 2010-02-02 1.2
brad 2010-01-02 1.1
fred 2010-01-03 1.0
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 *
FROM MyTable T1
WHERE date = (
SELECT max(date)
FROM MyTable T2
WHERE T1.username=T2.username
)
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。
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
