选择时间差小于单列2小时的行

Question

表：TEST

选择当天时间差小于2小时的行（按日期分组）。

这里输出应该是前两行，因为前两行的时差（18-JAN-15 01.08.40.000000000 PM-18-JAN-15 11.21.28.000000000 AM＆lt; 2小时）

NB: compare rows of same date.

输出：

  CREATE TABLE TEST
  ( "ID" VARCHAR2(20 BYTE), 
    "CAM_TIME" TIMESTAMP (6)
  ) 

Insert into TEST (ID,CAM_TIME) values ('1',to_timestamp('18-JAN-15 11.21.28.000000000 AM','DD-MON-RR HH.MI.SSXFF AM'));
Insert into TEST (ID,CAM_TIME) values ('2',to_timestamp('18-JAN-15 01.08.40.000000000 PM','DD-MON-RR HH.MI.SSXFF AM'));

Insert into TEST (ID,CAM_TIME) values ('3',to_timestamp('23-JAN-15 09.18.40.000000000 AM','DD-MON-RR HH.MI.SSXFF AM'));
Insert into TEST (ID,CAM_TIME) values ('4',to_timestamp('23-JAN-15 04.22.22.000000000 PM','DD-MON-RR HH.MI.SSXFF AM'));

Answer 1

我采用了稍微不同的方法并使用LAG()和LEAD()分析函数：

WITH mydata AS (
  SELECT 1 AS id, timestamp '2015-01-15 11:21:28.000' AS cam_time
    FROM dual
   UNION ALL
  SELECT 2 AS id, timestamp '2015-01-15 13:08:40.000' AS cam_time
    FROM dual
   UNION ALL
  SELECT 3 AS id, timestamp '2015-01-23 09:18:40.000' AS cam_time
    FROM dual
   UNION ALL
  SELECT 4 AS id, timestamp '2015-01-23 16:22:22.000' AS cam_time
    FROM dual
)
SELECT id, cam_time FROM (
    SELECT id, cam_time
         , LAG(cam_time) OVER ( PARTITION BY TRUNC(cam_time) ORDER BY cam_time ) AS lag_time
         , LEAD(cam_time) OVER ( PARTITION BY TRUNC(cam_time) ORDER BY cam_time ) AS lead_time
      FROM mydata
) WHERE CAST(lead_time AS DATE) - CAST(cam_time AS DATE) < 1/12
     OR CAST(cam_time AS DATE) - CAST(lag_time AS DATE) < 1/12

Answer 2

此自连接查询可以完成工作：

SQL Fiddle

select distinct t1.id, t1.cam_time 
  from test t1 join test t2 on t1.rowid <> t2.rowid  
    and trunc(t1.cam_time) = trunc(t2.cam_time)
  where abs(t1.cam_time-t2.cam_time) <= 2/24
  order by t1.id

编辑：

如果cam_time是time_stamp类型，则条件应为：

where t1.cam_time between t2.cam_time - interval '2' Hour 
                      and t2.cam_time + interval '2' Hour

Answer 3

这应该这样做。使用“存在”，您可以检查子查询是否返回任何结果。在这种情况下，子查询返回符合2个条件的行：

1. 它们具有相同的日期（trunc(CAM_TIME)仅返回日期，然后进行比较）。
2. 他们的时差少于两个小时。您可以减去两次以获得天数差异。将该差值乘以24，即可得到小时数。

结果：

select
  t1.*
from 
  TEST t1
where
  exists
    ( select 'x' 
      from   TEST t2 
      where  t1.id <> t2.id                                -- Not the same row
             trunc(t2.CAM_TIME) = trunc(t1.CAM_TIME)       -- Same date
             and abs(t1.CAM_TIME - t2.CAM_TIME) * 24 < 2   -- < 2 hours