Oracle按时间段分组统计

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)