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 TABEL1 WHERE DATE= (SELECT MAX(CREATED_DATE) FROM TABEL1)
其他回答
我看到大多数开发人员在使用内联查询时没有考虑它对大数据的影响。
简单地说，你可以通过:
SELECT a.username, a.date, a.value
FROM myTable a
LEFT OUTER JOIN myTable b
ON a.username = b.username
AND a.date < b.date
WHERE b.username IS NULL
ORDER BY a.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
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。
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
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));
