如何返回重複的行，如果日期差等於或超過4天或96小時

我要取重複記錄如果重複記錄之間的時間差是超過96小時或4天，否則忽略重複條目，並返回記錄與第一個入口或最早的日期。我的表是這樣的：如何返回重複的行，如果日期差等於或超過4天或96小時

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

預期結果：

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

我嘗試下面的查詢，但它無法正常工作。

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)

來源

2016-04-13 ravi

應該發生什麼，如果一個'id'有以下三個記錄？ '{'2016-04-01 06：00：00.000'，'2016-04-03 18：00：00.000'，'2016-04-06 06：00：00.000'}'？每個記錄相距不到4天，但第一個和最後一個記錄相隔5天。 – MatBailie

如果有3次重複的ID，第一日期，會發生什麼 - '1/4/2016'，第二日期 - '3/4/2016'和第三日期'4分之6/ 2016'，由你的定義，第二是一個重複的第一個和第三個是第二個副本。應該刪除哪個？ – sagi

@MatBailie哈，同樣的問題，相同的日期。 – sagi

下面的查詢返回預期的結果，增加了在線評論：

-- 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

來源

2016-04-13 10:58:54 Arulkumar

我不相信。如果你的日期分別是'1,4,7,10,13'，那麼怎麼辦？這個答案將返回'第1，第7，第10，第13'，即使最後三個結果每個都只有3天的差距。這並不是說這是錯誤的，只是OP對於這種情況的要求遠未明確。 – MatBailie

請嘗試以下查詢，我已經測試它工作正常。

在

忽略列，您可以更改時間（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)

來源

2016-04-13 11:04:15

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

來源

2016-04-13 12:10:44

如何返回重複的行，如果日期差等於或超過4天或96小時

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