Sql服务器：任何方式根据其他记录的日期按日期分组记录？

Question

我有一个表中有事件，我需要找到重复的事件。问题是彼此之间发生1秒钟的事件被认为是重复的。所以，如果我的表有这些值

id | var1 | var2 | var3 | date 
1 | 1 | 2 | 3 | 2001-01-01 01:01:01.456 
2 | 1 | 2 | 3 | 2001-01-01 01:01:02.234 
3 | 1 | 2 | 3 | 2001-01-01 01:01:04.789 

记录1和2被视为重复，因为他们是一秒钟之内，但3是不是因为它的后2

超过一秒钟有什么办法写一个查询只选择一系列重复中的​​第一条记录？

编辑：也可能有行将不需要被重复捕获。 Id是表的主键，不在匹配标准中使用;这只是为了澄清。

Answer 1

这里，似乎像它应该为你工作的方式。

一些假设：

1. 我想重复的是不包括ID的实际行重复，根据您提供1秒条款。如果不是这种情况...删除分区的部分row_number()窗口函数，它将改变行为
2. 这将删除递归重复项。也就是说，如果3,4，甚至15排在彼此的一秒钟内，它保持1
3. 这应该不论工作，如果第一行或最后一行是重复

下面的代码。取消注释了两行在表中看到的变化

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 

退货

+----+------+------+------+-----------------------------+ 
| 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 | 
+----+------+------+------+-----------------------------+ 

Answer 2

滞后是一个可能的解决方案，这样的事情：

select * from (
select *, lag(date,1) over(order by date) previoustime from yourtable 
) x 
where datediff(second,previoustime,date)<1 

Answer 3

select T1.date,... from MyTable T1 
left outer join MyTable T2 on cast(T1.date as date) = cast(T2.date as date) and 
datediff(second,T1.date,T2.date)<=1 
group by cast(T1.date as date)