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Oracle数据库按时间进行分组统计数据的方法

时间:2015-11-24 20:43来源:www.chengxuyuans.com 点击:

Oracle按不同时间分组统计的sql
如下表table1:

日期(exportDate)        数量(amount) 
--------------          ----------- 
14-2月 -08            20 
10-3月 -08            2 
14-4月 -08            6 
14-6月 -08            75 
24-10月-09           23 
14-11月-09           45 
04-8月 -10            5 
04-9月 -10            44 
04-10月-10           88 

注意:为了显示更直观,如下查询已皆按相应分组排序

1.按年份分组

select to_char(exportDate,'yyyy'),sum(amount) from table1 group by to_char(exportDate,'yyyy'); 
年份   数量 
----------------------------- 
2009  68 
2010  137 
2008  103 

2.按月份分组

select to_char(exportDate,'yyyy-mm'),sum(amount) from table1 group by to_char(exportDate,'yyyy-mm')
order by to_char(exportDate,'yyyy-mm'); 
月份      数量 
----------------------------- 
2008-02  20 
2008-03  2 
2008-04  6 
2008-06  75 
2009-10  23 
2009-11  45 
2010-08  5 
2010-09  44 
2010-10  88 

3.按季度分组

select to_char(exportDate,'yyyy-Q'),sum(amount) from table1 group by to_char(exportDate,'yyyy-Q') 
order by to_char(exportDate,'yyyy-Q'); 
季度     数量 
------------------------------ 
2008-1  22 
2008-2  81 
2009-4  68 
2010-3  49 
2010-4  88 

4.按周分组

select to_char(exportDate,'yyyy-IW'),sum(amount) from table1 group by to_char(exportDate,'yyyy-IW')
order by to_char(exportDate,'yyyy-IW'); 
周       数量 
------------------------------ 
2008-07  20 
2008-11  2 
2008-16  6 
2008-24  75 
2009-43  23 
2009-46  45 
2010-31  5 
2010-35  44 
2010-40  88


PS: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 

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)

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