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
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;
其他回答
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 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。
根据我的经验，最快的方法是取表中没有新行的每一行。
另一个优点是所使用的语法非常简单，而且查询的含义相当容易掌握(取所有行，确保所考虑的用户名不存在更新的行)。
不存在
SELECT username, value
FROM t
WHERE NOT EXISTS (
SELECT *
FROM t AS witness
WHERE witness.username = t.username AND witness.date > t.date
);
ROW_NUMBER
SELECT username, value
FROM (
SELECT username, value, row_number() OVER (PARTITION BY username ORDER BY date DESC) AS rn
FROM t
) t2
WHERE rn = 1
内连接
SELECT t.username, t.value
FROM t
INNER JOIN (
SELECT username, MAX(date) AS date
FROM t
GROUP BY username
) tm ON t.username = tm.username AND t.date = tm.date;
左外连接
SELECT username, value
FROM t
LEFT OUTER JOIN t AS w ON t.username = w.username AND t.date < w.date
WHERE w.username IS NULL
SELECT *
FROM MyTable T1
WHERE date = (
SELECT max(date)
FROM MyTable T2
WHERE T1.username=T2.username
)
SELECT *
FROM ReportStatus c
inner join ( SELECT
MAX(Date) AS MaxDate
FROM ReportStatus ) m
on c.date = m.maxdate
