需要刪除連續三天不包括週末和節假日在SQL

Question

我有一個要求從表中有連續三天刪除行（排除天，如果週末之間進來）

CREATE TABLE [dbo].[Test] 
(
    [Scanid] [bigint] NULL, 
    [Employeeid] [int] NULL, 
    [Datescanned] [datetime] NULL 
) 

INSERT INTO [dbo].[Test]([Scanid], [Employeeid], [Datescanned]) 
VALUES (108639, 3820, '2016-04-28 17:12:33.000'), 
     (108639, 3820, '2016-04-28 18:05:46.000'), 
     (108639, 3820, '2016-04-28 20:58:36.000'), 
     (999999, 3820, '2016-04-29 10:08:00.000'), 
     (999999, 3820, '2016-04-29 10:12:10.000'), 
     (777777, 3820, '2016-05-02 10:12:00.000'), 
     (111111, 3820, '2016-04-04 10:12:00.000'), 
     (33333, 3820, '2016-04-11 17:23:00.000'), 
     (987623, 3820, '2016-04-18 11:12:00.000'), 
     (1234, 3820, '2016-05-26 10:00:00.000'), 
     (5678, 3820, '2016-05-27 10:00:00.000'), 
     (8920, 3820, '2016-05-31 10:00:00.000') 

輸出：

Scanid Employeeid Datescanned 
---------------------------------------- 
108639 3820 2016-04-28 17:12:33.000 
108639 3820 2016-04-28 18:05:46.000 
108639 3820 2016-04-28 20:58:36.000 
999999 3820 2016-04-29 10:08:00.000 
999999 3820 2016-04-29 10:12:10.000 
777777 3820 2016-05-02 10:12:00.000 
111111 3820 2016-04-04 10:12:00.000 
33333 3820 2016-04-11 17:23:00.000 
987623 3820 2016-04-18 11:12:00.000 
1234 3820 2016-05-26 10:00:00.000 
5678 3820 2016-05-27 10:00:00.000 
8920 3820 2016-05-31 10:00:00.000 

我們只能從日期掃描的字段獲取日期，然後在上面的例子中，我們應該刪除從'2016-04-28'到'2016-05-02'（2016-04-30和31日是週末，所以我們可以忽略），並刪除連續3日從'2016-05-26'到'2016-05-31'的行'（2016-05-29和30日是週末，所以我們可以忽略）。所以只有結果應顯示排在2016-04-04,2016-04-11,2016-04-18的天數之前或之後沒有連續3天。

Answer 1

這是你想要的精確輸出..

我可以看到你的問題，[即一個錯誤。從「2016年5月26日」，以連續3日刪除行「2016年5月31日」（2016年5月29日和是週末，所以我們可以忽略）「]，那些週末天不正確..正確日期爲2016-05-28 and 2016-05-29。

DROP TABLE [TestDates] 
GO 
CREATE TABLE [dbo].[TestDates](
     [Scanid] [bigint] NULL, 
     [Employeeid] [int] NULL, 
     [Datescanned] [datetime] NULL 
     ) 

    INSERT INTO [dbo].[TestDates] ([Scanid] ,[Employeeid],[Datescanned]) 
    VALUES (108639,3820,'2016-04-28 17:12:33.000'),(108639,3820,'2016-04-28 18:05:46.000'), 
      (108639,3820,'2016-04-28 20:58:36.000'),(999999,3820,'2016-04-29 10:08:00.000'), 
      (999999,3820,'2016-04-29 10:12:10.000'),(777777,3820,'2016-05-02 10:12:00.000'), 
      (111111,3820,'2016-04-04 10:12:00.000'),(33333,3820,'2016-04-11 17:23:00.000'), 
      (987623,3820,'2016-04-18 11:12:00.000'),(1234,3820,'2016-05-26 10:00:00.000'), 
      (5678,3820,'2016-05-27 10:00:00.00'), (8920, 3820, '2016-05-30 10:00:00.000') 
GO 

DROP TABLE #t 
GO 
SELECT DISTINCT Employeeid,CONVERT(date,Datescanned) Datescanned INTO #T 
FROM [TestDates] 
GO 


;WITH cte_cnt 
AS 
(
SELECT Employeeid, MIN(Datescanned) AS FROM_DATE 
     ,MAX(Datescanned) AS TO_DATE 
     , COUNT('A') AS DayDiff 
FROM (
     SELECT Employeeid,Datescanned, 
      ROW_NUMBER() OVER(ORDER BY Datescanned) AS ROW_NUMBER, 
      DATEDIFF(D, ROW_NUMBER() OVER(ORDER BY Datescanned) 
      ,CASE WHEN DATENAME(dw, cast (Datescanned as datetime)-1) = 'Sunday' THEN DATEADD(DAY, -2, Datescanned) ELSE Datescanned END) AS Diff 
     FROM #t) AS dt 
GROUP BY Employeeid, Diff) 
DELETE t 
--SELECT * 
FROM cte_cnt c 
     JOIN [TestDates] t 
      ON c.Employeeid=t.Employeeid 
WHERE CAST(t.Datescanned as DATE) BETWEEN c.FROM_DATE AND c.TO_DATE and c.DayDiff=3 
GO 

SELECT * 
FROM [TestDates] 
GO 

Answer 2

也許這?:

delete from Test 
where not exists (
    select 1 
    from Test t2 
    where cast(t2.Datescanned as date) 
     between 
      dateadd(day, 
       case datepart(dayofweek, cast(Test.Datescanned as date)) 
        when 1 then -4 when 2 then -4 
        else -2 
       end, 
       cast(Test.Datescanned as date) 
      ) 
     and 
      dateadd(day, 
       case datepart(dayofweek, cast(Test.Datescanned as date)) 
        when 4 then 4 when 5 then 4 
        else 2 
       end, 
       cast(Test.Datescanned as date) 
      ) 
) 

Answer 3

不佔度假的解決辦法是

SELECT 
    t.* 
FROM (SELECT DISTINCT 
    CAST(t1.datescanned AS date) first_date, 
    CAST(t2.datescanned AS date) second_date, 
    CAST(t3.datescanned AS date) third_date 
FROM test t1 
JOIN test t2 --add a join condition for employeeid as well 
    ON DATEDIFF(dd, CAST(t1.datescanned AS date), CAST(t2.datescanned AS date)) = 1 
    OR (DATEPART(WEEKDAY, CAST(t2.datescanned AS date)) = 2 
    AND DATEDIFF(dd, CAST(t1.datescanned AS date), CAST(t2.datescanned AS date)) = 3) 
JOIN test t3 --add a join condition for employeeid as well 
    ON DATEDIFF(dd, CAST(t2.datescanned AS date), CAST(t3.datescanned AS date)) = 1 
    OR (DATEPART(WEEKDAY, CAST(t3.datescanned AS date)) = 2 
    AND DATEDIFF(dd, CAST(t2.datescanned AS date), CAST(t3.datescanned AS date)) = 3) 
) x 
JOIN test t 
    ON CAST(t.datescanned AS date) = x.first_date 
    OR CAST(t.datescanned AS date) = x.second_date 
    OR CAST(t.datescanned AS date) = x.third_date 

自聯接表的兩倍，在

• 日期每次差異1或
• 3當週末發生時檢查週一是週一一天（星期= 2）

Sample demo

結果給出需要被刪除的行。但是這裏要注意的是，如果沒有差距，這會給你連續3天以上的時間。在這種情況下，您需要解釋是否要在第3天停止刪除。

Answer 4

以前的腳本很好。我會改進和週末和節假日檢查，添加一個函數，並在選擇語句中調用它。 下面是一個簡單的函數，你可以使用（我假設你有一個名爲假日表，其中包含每個國家都放假日期）

Create FUNCTION [dbo].[IsHolidayOrWeekend] 
(
    @date DateTime, 
    @stateId int 
) 
RETURNS Bit 
AS 
BEGIN 
    declare @dayOfWeek VARCHAR(9); 
    set @dayOfWeek = DATEName(DW, @date); 

    IF(@dayOfWeek = 'Saturday' OR @dayOfWeek = 'Sunday') 
     RETURN 1; 
    ELSE 
    begin 
     set @date = cast(@date as date) -- Remove the time portion 
     RETURN IsNull((SELECT 1 from Holiday where StateId = @provinceId and HolidayDate = @date), 0) 
    end; 
END 

Answer 5

我有一個額外的要求igonre假期也隨之週末。我創建了Test_calendar表，刪除所有節假日和週末，並將行號分配給活動日期。然後這是我使用的代碼。它工作，但如果我們有數百萬行，可能不是最快的。對我而言，數據很小，因此可以完成工作。請讓我知道，如果你可以簡化刪除過程，使其更快。

SELECT distinct scanid as [badgeid] , Employeeid,CONVERT(date, Datescanned) as Datescanned,RN 
    into #test1 
    FROM [dbo].[Test] a 
    inner join Test_Calendar b 
    on CONVERT(date, a.Datescanned)=b.Cal_date 
    order by Datescanned asc 


    declare @min int 
    declare @max int 
    declare @i int 
    select @min=MIN(rn) from #test1 
    select @max=Max(rn) from #test1 
    while(@min<@max) 
    begin 
    select @i=COUNT(*) from #test1 where rn in(@min ,@min+1,@min+2) 
    if(@i=3) 
    select * from #test1 where rn in(@min ,@min+1,@min+2) 
    set @min=@min+1 
    end