Oracle按不同时间分组统计的sql
如下表table1：
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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
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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
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select to_char(exportDate,'yyyy'),sum(amount) from table1 group by to_char(exportDate,'yyyy');
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年份 数量
-----------------------------
2009 68
2010 137
2008 103
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2.按月份分组
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select to_char(exportDate,'yyyy-mm'),sum(amount) from table1 group by to_char(exportDate,'yyyy-mm')
order by to_char(exportDate,'yyyy-mm');
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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
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3.按季度分组
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select to_char(exportDate,'yyyy-Q'),sum(amount) from table1 group by to_char(exportDate,'yyyy-Q')
order by to_char(exportDate,'yyyy-Q');
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季度 数量
------------------------------
2008-1 22
2008-2 81
2009-4 68
2010-3 49
2010-4 88
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4.按周分组
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select to_char(exportDate,'yyyy-IW'),sum(amount) from table1 group by to_char(exportDate,'yyyy-IW')
order by to_char(exportDate,'yyyy-IW');
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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
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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:
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----level 是一个伪例
select level from dual connect by level <=10
---结果：
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oracle时间的加减看看试一下以下sql语句就会知道:
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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
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下面是本次例子：
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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
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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：
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create table A_HY_LOCATE1
(
MOBILE_NO VARCHAR2(32),
LOCATE_TYPE NUMBER(4),
AREA_NO VARCHAR2(32),
CREATED_TIME DATE,
AREA_NAME VARCHAR2(512),
)；
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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
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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
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另外一个方法：
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--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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