Oracle 按時間段分組統計 (使用LEVEL)
想要按時間段分組查詢,首先要了解level,connect by,oracle時間的加減.
關於level這裡不多說,我只寫出一個查詢語句:
- ---level 是一個偽例
- selectlevelfrom dual connectbylevel <=10
- ---結果:1
- 2
- 3
- 4
- 5
- 6
- 7
- 8
- 9
- 10
- select sysdate -1 from dual
- ----結果減一天,也就24小時
- select sysdate-(1/2) from dual
- -----結果減去半天,也就12小時
-
select sysdate-(1/24)
- -----結果減去1 小時
- select sysdate-((1/24)/12) from dual
- ----結果減去5分鐘
- select sydate-(level-1) from dual connectbylevel<=10
- ---結果是10間隔1天的時間
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 connectbylevel <= 10) d
-
where
- i.insert_time<dt and i.insert_time> d.dt-2
- groupby d.dt
例子中的sysdate - (level-1) * 2得到的是一個間隔是2天的時間
group by d.dt 也就是兩天的時間間隔分組查詢
自己實現例子:
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 --從第一條時間記錄開始(sysdate-13)為表中的最早的日期,“34”出現的分組數(一天按每六個小時分組 就應該為4)
一下是按照每6個小時分組
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
另外一個方法:
--按六小時分組
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小時分組
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)