Question

我想根据时间将单行分为多行。下面是示例。

SrNo    Notification    StartDate                  EndDate
---------------------------------------------------------------------------
1       001003741915    2018-08-20 07:27:00.000    2018-08-21 16:23:00.000
2       001003779670    2018-08-21 03:36:00.000    2018-08-21 04:36:00.000
3       001003779830    2018-08-21 04:36:00.000    2018-08-21 21:35:00.000

预期输出如下：

SrNo    Notification    StartDate                  EndDate
---------------------------------------------------------------------------
1       001003741915    2018-08-20 07:27:00.000    2018-08-21 05:59:00.000
1       001003741915    2018-08-21 06:00:00.000    2018-08-21 16:23:00.000
2       001003779670    2018-08-21 03:36:00.000    2018-08-21 04:36:00.000
3       001003779830    2018-08-21 04:36:00.000    2018-08-21 05:59:00.000
3       001003779830    2018-08-21 06:00:00.000    2018-08-21 21:35:00.000

每天从06:00 AM开始到第二天06:00 AM。如果EndDate时间是06:00 AM以后的时间，则将此日期分为两行。第一行结束日期是2018-08-21 05：59：00.000，下一行开始2018-08-21 06：00：00.000。

Answer 1

您可以通过使用递归CTE来实现

WITH CTE AS (
SELECT ID, Notification, StartDate, EndDate 
FROM TAB1
UNION ALL
SELECT ID, Notification, DATEADD(DD,1,StartDate), EndDate 
FROM CTE
WHERE cast(StartDate as date) < cast(EndDate as date)
)
SELECT * FROM CTE order by id

Answer 2

如果开始日期和结束日期之间只有一天或更短的时间

如果我们将您的表称为t1：

SELECT [SrNo]
      ,[Notification]
      ,[StartDate]
      ,[EndDate]

  FROM [t1]
  where DATEADD(MINUTE, 59,   DATEADD(HOUR, 5, CAST(CAST(enddate AS DATE) AS DATETIME))) > enddate
  union 
  SELECT [SrNo]
      ,[Notification]
      ,[StartDate]
      ,DATEADD(MINUTE, 59,   DATEADD(HOUR, 5, CAST(CAST(enddate AS DATE) AS DATETIME))) [EndDate]

  FROM [t1]
  where DATEADD(MINUTE, 59,   DATEADD(HOUR, 5, CAST(CAST(enddate AS DATE) AS DATETIME))) between startdate and enddate

  union 
  SELECT [SrNo]
      ,[Notification]
      ,DATEADD(MINUTE, 00,   DATEADD(HOUR, 6, CAST(CAST(enddate AS DATE) AS DATETIME))) [StartDate]
      , [EndDate]

  FROM [t1]
  where DATEADD(MINUTE, 59,   DATEADD(HOUR, 5, CAST(CAST(enddate AS DATE) AS DATETIME))) between startdate and enddate

  order by srno
  ,enddate

Answer 3

以下查询将为您提供帮助。

CREATE TABLE #test
(
    Notifications varchar(50)
    ,StartDate datetime
    ,EndDate Datetime
    ,Id int
)

INSERT into #test
select              '001003741915','2018-08-20 07:27:00.000','2018-08-21 16:23:00.000',1
UNION select        '001003779670','2018-08-21 03:36:00.000','2018-08-21 04:36:00.000',2
UNION select        '001003779830','2018-08-21 04:36:00.000','2018-08-21 21:35:00.000',3
UNION select        '001003779835','2018-08-21 04:36:00.000','2018-08-24 21:35:00.000',4

;with cte
As (  SELECT 
        ID,Notifications,StartDate,dateadd(d, datediff(d, 1, StartDate+1), '06:00') as StartOfDay, EndDate,dateadd(d, datediff(d, 1, EndDate+1), '06:00')  as EndDayOfDate
    FROM #test
)
, Result
AS (


    select Id
            ,Notifications
            ,StartDate 
            ,CASE WHEN StartOfDay BETWEEN StartDate AND EndDate THEN  StartOfDay
                  WHEN ENDDate <StartOfDay THEN ENDDate
                  WHEN ENDDate <EndDayOfDate THEN ENDDate
                ELSE  EndDayOfDate  END AS  EndDate 
    from cte

    union ALL
    Select T.Id
            ,T.Notifications
            ,R.EndDate As StartDate
            ,CASE WHEN R.EndDate+1 < T.EndDate THEN R.EndDate+1 ELSE  T.EndDate   END AS EndDate 
    from cte  T
    INNER JOIN Result R
    ON R.Notifications=T.Notifications
    WHERE  R.EndDate <T.EndDate

)

SELECT * FROM Result order by id

Answer 4

根据 Nitika 的回答，我能够在 MySQL 中执行此操作并解决类似问题：

WITH RECURSIVE CTE (id, name, start_dt, end_dt) AS (
    SELECT id, name, cast(start_date as date) as start_dt, cast(end_date as date) as end_dt FROM event e1
    UNION ALL
    SELECT e2.id, e2.name, DATE_ADD(e2.start_dt, INTERVAL 1 DAY) as start_dt, e2.end_dt
    FROM CTE e2
    WHERE e2.start_dt < e2.end_dt
)
SELECT * FROM CTE order by start_dt

根据SQL Server中的日期和时间将单行拆分为多行

4 个答案: