1

目标是在 10 分钟间隔内组合具有相同 ID 的 Time_a。并将 ID 分组。

ID   Time_a        ->           ID    
------------                    ----------
1    12:10:00                   1
1    12:15:00                   2
1    12:20:00                   2
2    12:25:00
2    12:35:00
2    02:00:00

它变成了两个“2”，因为第 5 行和第 6 行之间的时间间隔超过 10 分钟。我能够在 10 分钟的差异内组合，但它不区分 ID。

select ID    
from(
    select id, Time_a, min(time) OVER (order by id, time rows between 1 preceding and 1 preceding) prev_t_stamp
    from dual
)
where abs(Time_a-prev_t_stamp)>10/1440
4

1 回答 1

0

如果 time_a 是日期，那么它的工作方式如下。如果它是时间戳，则需要将其转换为日期，否则它将无法正常工作。这是我的测试脚本

drop table test;
create table test (id number, time_a date);
insert into test values (1, to_date('12:10','HH24:MI'));
insert into test values (1, to_date('12:15','HH24:MI'));
insert into test values (1, to_date('12:20','HH24:MI'));
insert into test values (2, to_date('12:25','HH24:MI'));
insert into test values (2, to_date('12:35','HH24:MI'));
insert into test values (2, to_date('14:00','HH24:MI'));

select id from (
select distinct id, case from (
select ID,case when abs(Time_a-prev_t_stamp)>10/1440 then '>' end case
from(
    select id, Time_a, min(time_a) OVER (order by id, time_a rows between 1 preceding and 1 preceding) prev_t_stamp
    from test
)
)
) order by 1

这导致：

table TEST dropped.
table TEST created.
1 rows inserted.
1 rows inserted.
1 rows inserted.
1 rows inserted.
1 rows inserted.
1 rows inserted.

ID
--
 1 
 2 
 2
于 2012-08-01T08:24:49.560 回答