想要按时间段分组查询,首先要了解level,connect by,oracle时间的加减.
关于level这里不多说,我只写出一个查询语句:
- ---level 是一个伪例
- select level from dual connect by level <=10
- ---结果:1
- 2
- 3
- 4
- 5
- 6
- 7
- 8
- 9
- 10
关于connect by可以看
http://www.cnblogs.com/johnnyking39/articles/1155497.html
oracle时间的加减看看试一下以下sql语句就会知道:
- select sysdate -1 from dual
- ----结果减一天,也就24小时
- select sysdate-(1/2) from dual
- -----结果减去半天,也就12小时
- select sysdate-(1/24) from dual
- -----结果减去1 小时
- select sysdate-((1/24)/12) from dual
- ----结果减去5分钟
- select sydate-(level-1) from dual connect by level<=10
- ---结果是10间隔1天的时间
下面是本次例子:
- select dt, count(satisfy_degree) as num from T_DEMO i ,
- (select sysdate - (level-1) * 2 dt
- from dual connect by level <= 10) d
- where i.satisfy_degree='satisfy_1' and
- i.insert_time<dt and i.insert_time> d.dt-2
- group by d.dt
例子中的sysdate - (level-1) * 2得到的是一个间隔是2天的时间
group by d.dt 也就是两天的时间间隔分组查询
自己实现例子:
create table A_HY_LOCATE1
(
MOBILE_NO VARCHAR2(32),
LOCATE_TYPE NUMBER(4),
AREA_NO VARCHAR2(32),
CREATED_TIME DATE,
AREA_NAME VARCHAR2(512),
);
select (sysdate-13)-(level-1)/4 from dual connect by level<=34 --从第一条时间记录开始(sysdate-13)为表中的最早的日期,“34”出现的分组数(一天按每六个小时分组 就应该为4)
一下是按照每6个小时分组
select mobile_no,area_name,max(created_time ),dt, count(*) as num from a_hy_locate1 i ,
(select (sysdate-13)-(level-1)/4 dt
from dual connect by level <= 34) d
where i.locate_type = 1 and
i.created_time<dt and i.created_time> d.dt-1/4
group by mobile_no,area_name,d.dt
另外一个方法:
--按六小时分组
select trunc(to_number(to_char(created_time, 'hh24')) / 6),count(*)
from t_test
where created_time > trunc(sysdate - 40)
group by trunc(to_number(to_char(created_time, 'hh24')) / 6)
--按12小时分组
select trunc(to_number(to_char(created_time, 'hh24')) / 6),count(*)
from t_test
where created_time > trunc(sysdate - 40)
group by trunc(to_number(to_char(created_time, 'hh24')) / 6)