这是一种看起来应该适合您的方法。
一些假设:
- 根据您提供的 1 second 子句,我假设重复是不包括 ID 的实际行重复。如果不是这种情况...通过
row_number() 窗口函数的一部分删除分区,它将改变行为
- 这将删除递归重复。也就是说,如果 3,4 甚至 15 行彼此在一秒之内,则保持 1。
- 无论第一行或最后一行是否重复,这都应该有效
这是代码。取消注释表中的两行以查看更改
declare @table table(id int, var1 int, var2 int, var3 int, date datetime2)
insert into @table
values
--(0,1,2,3,'2001-01-01 00:01:01.456'),
(1,1,2,3,'2001-01-01 01:01:01.456'), --dupe of 1/2/3
(2,1,2,3,'2001-01-01 01:01:02.214'), --dupe of 1/2/3
(3,1,2,3,'2001-01-01 01:01:02.234'), --dupe of 1/2/3
(4,1,2,3,'2001-01-01 01:01:02.244'), --dupe of 1/2/3
(5,1,2,3,'2001-01-01 01:01:04.789'), --dupe of 4/5
(6,1,2,3,'2001-01-01 01:01:04.989'), --dupe of 4/5
--(7,1,2,3,'2001-01-01 01:01:06.789'), --dupe of 6/7
(8,1,2,3,'2001-01-01 01:01:06.799') --dupe of 6/7
--apply the sequence
;with cte as(
select
*,
ROW_NUMBER() over (partition by var1, var2, var3 order by date) as RN --just in case... change this to just order by id, date if need be and remove the partition
from
@table),
--get first / most of the batch to remove
cte2 as(
select
c1.*
,c2.RN as RowsToRemove
from cte c1
left join
cte c2 on c1.RN < c2.rn and
datediff(second,c1.date,c2.date) < 1),
--remove the rows identified in the above cte
cte3 as(
select distinct
ID,
var1,
var2,
var3,
date,
RN
from cte2
where
RN not in (select distinct isnull(RowsToRemove,0) from cte2)),
--add another sequence. This is necessary for first/last row check for duplicate
cte4 as(
select
f.*,
row_number() over (partition by var1, var2, var3 order by date) RN2
from
cte3 f)
--return the results
select
f.ID,
f.var1,
f.var2,
f.var3,
f.date
from
cte4 f
left join
cte4 d on d.RN = f.RN - 1
where isnull(datediff(second,d.date,f.date),500) > 1
返回
+----+------+------+------+-----------------------------+
| ID | var1 | var2 | var3 | date |
+----+------+------+------+-----------------------------+
| 1 | 1 | 2 | 3 | 2001-01-01 01:01:01.4560000 |
| 5 | 1 | 2 | 3 | 2001-01-01 01:01:04.7890000 |
| 8 | 1 | 2 | 3 | 2001-01-01 01:01:06.7990000 |
+----+------+------+------+-----------------------------+