Question

我遇到一个问题，需要在一天内显示每个组，用户，状态，某个字段的最大值。

   groupno   user    status                      date  amount
        74  user1   status_1    2016-01-01 05:40:00.0   900
        74  user1   status_1    2016-01-01 05:45:00.0   1200
        79  user1   status_2    2016-01-01 05:45:31.0   31
        79  user1   status_2    2016-01-01 05:50:00.0   300
        79  user1   status_2    2016-01-01 05:55:00.0   600
        79  user1   status_2    2016-01-01 06:00:00.0   900
        79  user1   status_2    2016-01-01 06:05:00.0   1200
        90  user1   status_1    2016-01-01 06:07:52.0   172
        90  user1   status_1    2016-01-01 06:10:00.0   300
        90  user1   status_1    2016-01-01 06:15:00.0   600

我正在尝试使用内部联接来最大化每个组的最大数量，但是因为我再次检索所有值而无法正常工作：

select t.groupno, t.user, t.status, t.date, t.amount
from ( select groupno,user, status, date, max (amount) as maxact from table group by 1,2,3,4) as m
Inner
join table as t
on t.user= m.user
and t.amount= m.maxact
and t.status = m.status
and t.date = m.date

Answer 1

使用窗口功能：

select t.*,
       max(t.amount) over (partition by groupno, user, status, date_trunc('day', date)) as max_amount
from t;

我认为groupno / user / status /天是计算最大值的维度。

Answer 2

如果你只想要最大金额，就日期而言 - 你可以拿MAX并将日期时间作为日期：

DECLARE @temp TABLE (groupno int, [user] varchar(20), status varchar(20), date datetime, amount int)

INSERT INTO @temp
VALUES
 (74,'user1', 'status_1', '2016-01-01 05:40:00.0', 900)
,(74,'user1', 'status_1', '2016-01-01 05:45:00.0', 1200)
,(79,'user1', 'status_2', '2016-01-01 05:45:31.0', 31)
,(79,'user1', 'status_2', '2016-01-01 05:50:00.0', 300)
,(79,'user1', 'status_2', '2016-01-01 05:55:00.0', 600)
,(79,'user1', 'status_2', '2016-01-01 06:00:00.0', 900)
,(79,'user1', 'status_2', '2016-01-01 06:05:00.0', 1200)
,(90,'user1', 'status_1', '2016-01-01 06:07:52.0', 172)
,(90,'user1', 'status_1', '2016-01-01 06:10:00.0', 300)
,(90,'user1', 'status_1', '2016-01-01 06:15:00.0', 600)


SELECT groupno, [user], [status], CAST([date] as date) [Date], MAX(amount) as MAXamount
FROM @temp
GROUP BY groupno, [user], [status], CAST([date] as date)

Answer 3

不确定我是否完全按照这个问题查看，但是我希望你想要每组最大数量的行。为此，您可以尝试：

SELECT t.groupno, t.user, t.status, t.date, t.amount
FROM t
INNER JOIN
(SELECT groupno, MAX(amount) as mx
 FROM t
 GROUP BY groupno) sub
ON t.groupno = sub.groupno
WHERE t.amount = sub.mx

仅显示每个组的最大值

3 个答案: