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 dtfrom dual connect by level <= 34) dwhere i.locate_type = 1 andi.created_time<dt and i.created_time> d.dt-1/4group 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)