如何对特定时间段进行求和和分组

Question

我有30分钟的数据，想要从当前日期的02:00 AM到下一个日期的02:00 AM的白天和小组，我该怎么办？：

create table test (code varchar2(50), hour number, daytime date); 

insert into test (code, hour, daytime) values ('K11','0.5','23-Feb-17')
insert into test (code, hour, daytime) values ('K11','0.5','23-02-17 00:30')
insert into test (code, hour, daytime) values ('K11','0.5','23-02-17 01:00')
insert into test (code, hour, daytime) values ('K11','0.5','23-02-17 01:30')
insert into test (code, hour, daytime) values ('K11','0.5','23-02-17 02:00')
insert into test (code, hour, daytime) values ('K11','0.5','23-02-17 02:30')
insert into test (code, hour, daytime) values ('K11','0.5','24-Feb-17')
insert into test (code, hour, daytime) values ('K11','0.5','24-02-17 00:30')
insert into test (code, hour, daytime) values ('K11','0.5','24-02-17 01:00')
insert into test (code, hour, daytime) values ('K11','0.5','24-02-17 01:30')
insert into test (code, hour, daytime) values ('K11','0.5','24-02-17 02:00')
insert into test (code, hour, daytime) values ('K11','0.5','24-02-17 02:30')
insert into test (code, hour, daytime) values ('K12','0.5','23-Feb-17')
insert into test (code, hour, daytime) values ('K12','0.5','23-02-17 00:30')
insert into test (code, hour, daytime) values ('K12','0.5','23-02-17 01:00')
insert into test (code, hour, daytime) values ('K12','0.5','23-02-17 01:30')
insert into test (code, hour, daytime) values ('K12','0.5','23-02-17 02:00')
insert into test (code, hour, daytime) values ('K12','0.5','23-02-17 02:30')
insert into test (code, hour, daytime) values ('K12','0.5','24-Feb-17')
insert into test (code, hour, daytime) values ('K12','0.5','24-02-17 00:30')
insert into test (code, hour, daytime) values ('K12','0.5','24-02-17 01:00')
insert into test (code, hour, daytime) values ('K12','0.5','24-02-17 01:30')
insert into test (code, hour, daytime) values ('K12','0.5','24-02-17 02:00')
insert into test (code, hour, daytime) values ('K12','0.5','24-02-17 02:30')

最后我想得到以下数据：

'K11',3,'23-02-17'
'K12',3,'23-02-17'

因此，例如K11从（'K11'，'0.5'，'23-02-17 02:00'）到（'K11'，'0.5'，'24-02-17 01： 00'）

Answer 1

您可以通过减去两个小时来完成此操作。这是一种方法：

select trunc(date - 2/24) as dte, sum(hours)
from t
group by trunc(date - 2/24)
order by dte;

Oracle现在支持interval语法，因此您也可以使用d：

select trunc(date - interval 2 hour) as dte, sum(hours)
from t
group by trunc(date - interval 2 hour)
order by dte;