Oracle数据库按时间进行分组统计数据的方法
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
-----------------------------
200968
2010137
2008103
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
-----------------------------
2008-0220
2008-032
2008-046
2008-0675
2009-1023
2009-1145
2010-085
2010-0944
2010-1088
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
------------------------------
2008-122
2008-281
2009-468
2010-349
2010-488
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
------------------------------
2008-0720
2008-112
2008-166
2008-2475
2009-4323
2009-4645
2010-315
2010-3544
2010-4088
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)
https://www.uoften.com/dbs/oracle/20180415/73358.html
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