如果日期差等于或大于 4 天或 96 小时，如何返回重复行答案

【问题标题】：how to return duplicate rows if date difference is equal or more than 4 days or 96 hours如果日期差等于或大于 4 天或 96 小时，如何返回重复行
【发布时间】：2016-08-04 07:43:23
【问题描述】：

我必须获取重复记录如果重复记录之间的日期差异超过 96 小时或 4 天，否则忽略重复条目并返回第一个条目或最早日期的记录。我的桌子是这样的：

ID           SDATE
----------- -----------------------
1           2016-04-13 14:54:18.983
1           2016-04-08 12:55:47.907
2           2016-04-13 14:54:18.983
3           2016-04-13 14:54:18.983
4           2016-04-13 14:54:18.983
5           2016-04-13 14:54:18.983
5           2016-04-11 12:55:47.907
6           2016-04-13 14:54:18.983
6           2016-04-13 14:54:18.983

预期结果：

ID           SDATE
----------- -----------------------
1           2016-04-13 14:54:18.983
1           2016-04-08 12:55:47.907
2           2016-04-13 14:54:18.983
3           2016-04-13 14:54:18.983
4           2016-04-13 14:54:18.983    
5           2016-04-11 12:55:47.907
6           2016-04-13 14:54:18.983

我尝试了以下查询，但它不起作用。

WITH tt AS (
SELECT 1 as ID, GETDATE() as SDATE
UNION ALL
SELECT 1 as ID, '2016-04-09 12:55:47.907' as SDATE
UNION ALL
SELECT 2 as ID, GETDATE() as SDATE
UNION ALL
SELECT 3 as ID, GETDATE() as SDATE
UNION ALL
SELECT 4 as ID, GETDATE() as SDATE
UNION ALL
SELECT 5 as ID, GETDATE() as SDATE
UNION ALL
SELECT 5 as ID, '2016-04-11 12:55:47.907' as SDATE
UNION ALL
SELECT 6 as ID, GETDATE() as SDATE
UNION ALL
SELECT 6 as ID, GETDATE() as SDATE
)
SELECT MIN(SDATE) as SDATE, ID FROM tt as tbl
GROUP BY ID,  DATEADD(HH, DATEDIFF(HH,0,SDATE) + 96,0)

【问题讨论】：

如果一个id有以下3条记录怎么办？ {'2016-04-01 06:00:00.000', '2016-04-03 18:00:00.000', '2016-04-06 06:00:00.000'}?每条记录相隔不到 4 天，但第一条和最后一条记录相隔 5 天。
如果有 3 个重复 ID，第一个日期 - 1/4/2016 ，第二个日期 - 3/4/2016 和第三个日期 6/4/2016 会发生什么情况第二个的副本。应该删除哪个？
@MatBailie Hah，同样的问题，同样的日期。
stackoverflow.com/questions/13196190/…从这里开始
嗨@MatBailie，如果有超过 2 条记录而不是我们查看的第一个和最后一个日期。

标签： sql sql-server sql-server-2008 tsql group-by

【解决方案1】：

以下查询返回预期结果，添加了内联 cmets：

-- Simply grouping each ID and get unique row with minimum date
SELECT MIN(SDATE) [SDate], ID
FROM tt
GROUP BY ID

UNION 

-- Get the row with each ID's difference is more than 96 hours
SELECT D.MaxDate  [SDate], D.ID
FROM (
    SELECT MIN(SDATE) [MinDate], MAX(SDATE) [MaxDate], ID
    FROM tt
    GROUP BY ID
) D
WHERE DATEDIFF(HH, D.MinDate, D.MaxDate) >= 96

【讨论】：

我不相信。如果你有1st, 4th, 7th, 10th, 13th 的日期呢？这个答案将返回1st, 7th, 10th, 13th，即使最后三个结果都只有 3 天的差距。这并不是说这是错误的，只是 OP 对这种情况的要求还很不清楚。

【解决方案2】：

请尝试以下查询，我已经测试它工作正常。

在忽略列中，您可以更改时间段（HH 或 DAY）

-- drop table #temptbl

WITH tt AS (
SELECT 1 as ID, GETDATE() as SDATE
UNION ALL
SELECT 1 as ID, '2016-04-09 12:55:47.907' as SDATE
UNION ALL
SELECT 2 as ID, GETDATE() as SDATE
UNION ALL
SELECT 3 as ID, GETDATE() as SDATE
UNION ALL
SELECT 4 as ID, GETDATE() as SDATE
UNION ALL
SELECT 5 as ID, GETDATE() as SDATE
UNION ALL
SELECT 5 as ID, '2016-04-11 12:55:47.907' as SDATE
UNION ALL
SELECT 6 as ID, GETDATE() as SDATE
UNION ALL
SELECT 6 as ID, GETDATE() as SDATE
)
SELECT Id,SDATE,case when DATEDIFF(HH,SDATE,GETDATE()) >94 THEN 0 else 1 end AS ignore,
  ROW_NUMBER() OVER ( PARTITION BY tt.ID ORDER BY tt.SDATE desc ) as Rowid
INTO #temptbl
FROM tt 
SELECT Id, sdate   from #temptbl
WHERE (#temptbl.ignore = 0) or (#temptbl.Rowid = 1)

【讨论】：

【解决方案3】：

declare @table table
(
ID int,           SDATE datetime)
insert into @table
(
ID      ,SDATE )
values 
(1,'2016-04-13 14:54:18'),
(1,'2016-04-08 12:55:47'),
(2,'2016-04-13 14:54:18'),
(3,'2016-04-13 14:54:18'),
(4,'2016-04-13 14:54:18'),
(5,'2016-04-13 14:54:18'),
(5,'2016-04-11 12:55:47'),
(6,'2016-04-13 14:54:18'),
(6,'2016-04-13 14:54:18')


;with cte as
(
select  id,min(sdate) mindate,max(sdate) maxdate, datediff(dd,min(sdate),max(sdate)) daysdiff,count(*) as Dups
 from @table
 group by   id 
 )
select  cte.id, t.sdate
from cte
join    @table t on t.id = cte.id 
where   cte.dups > 1 and cte.daysdiff > 4
union all
select  cte.id, 
        mindate
from cte
where   (cte.dups > 1 and cte.daysdiff <= 4) or
        cte.dups = 1

【讨论】：