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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)
出处:http://blog.csdn.net/wanglipo/article/details/6556665
Oracle按时间段分组统计,码迷,mamicode.com
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原文地址:http://www.cnblogs.com/mq0036/p/3701776.html