Sql 如何对特定时间段进行汇总和分组

我有30分钟的数据，我想计算从当前日期凌晨2:00到下一个日期凌晨2:00的白天小时数和分组数，我该怎么做

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'

---

例如，它将K11的小时数从（'K11'，'0.5'，'23-02-17 02:00'）加至（'K11'，'0.5'，'24-02-17 01:00'）

你可以通过减去两个小时来实现这一点。这里有一种方法：

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，因此您还可以：

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

---

不，这不起作用，它增加了两个小时，就是这样，但没有分组。@JoeGreen。我不明白你的评论。首先，这减去两个小时，但不加上它们。其次，

groupby

进行聚合。