SQL Server：DATEDIFF在兩個日期內排除一定的時間

Question

我試圖在幾分鐘內得到兩個日期之間的差異，是否可能從這個計算中排除午餐時間（又名13：30-14：00）我的SQL語句。什麼我目前做如下：

例排

SetupStart     StartTime     SetupTime 
---------------------------------------------------------------- 
2017-01-23 12:56:42.000  2017-01-23 14:41:06.000  105 

我現在的語句：

DATEDIFF(MINUTE, UserData.Orders.SetupStart, UserData.Orders.StartTime) AS[SetupTime] 

編輯：--------------

這是我目前的Select語句

SELECT DISTINCT 
        TOP (100) PERCENT UserData.Resources.Name AS Resource, UserData.Orders.OrderNo, UserData.Orders.StringAttribute4 AS Customer, UserData.Orders.Product, 
        CEILING(UserData.Orders.OpNo10Quantity) AS OpNo10Quantity, UserData.Orders.NumericalAttribute12 AS Speed, UserData.Orders.SetupStart, UserData.Orders.StartTime, 
        UserData.Orders.EndTime, CASE ToggleAttribute1 WHEN 1 THEN 'Yes' ELSE 'No' END AS Packed, UserData.Orders.StringAttribute5 AS OrderInstructions, UserData.Orders.NextResource, 
        UserData.Orders.DatasetId, UserData.Orders_Dataset.name AS ScheduleName, UserData.Orders.ShowOnReport, dbo.tblPreactorExportFullv10.Length AS L, 
        dbo.tblPreactorExportFullv10.Thickness AS T, dbo.tblPreactorExportFullv10.fWidth AS W, 
DATEDIFF(MINUTE, UserData.Orders.SetupStart, UserData.Orders.StartTime) AS [SAM SetupTime] 
FROM   UserData.Orders INNER JOIN 
        UserData.Resources ON UserData.Orders.Resource = UserData.Resources.ResourcesId AND UserData.Orders.Resource = UserData.Resources.ResourcesId INNER JOIN 
        UserData.Orders_Dataset ON UserData.Orders.DatasetId = UserData.Orders_Dataset.DatasetId AND UserData.Orders.DatasetId = UserData.Orders_Dataset.DatasetId INNER JOIN 
        dbo.tblPreactorExportFullv10 ON UserData.Orders.PartNo = dbo.tblPreactorExportFullv10.ProductCode 
WHERE  (UserData.Orders.DatasetId = 15) AND (UserData.Resources.Name = 'Moulder 6') AND (UserData.Orders.ShowOnReport = 1) 
      AND (UserData.Orders.OperationProgress <> 5) 
ORDER BY UserData.Orders.SetupStart 

Answer 1

我相信戈登的回答會更快，但另一個選項如下：

（編輯：只是爲了好玩，跑這20,000記錄，並以238毫秒返回）

Declare @YourTable table (SetupStart datetime, StartTime datetime) 
Insert Into @YourTable values 
('2017-01-23 12:56:42.000','2017-01-23 14:41:06.000'), 
('2017-01-23 15:00:00.000','2017-01-23 18:30:00.000'), -- No Lunch 
('2017-01-23 23:51:00.000','2017-01-23 23:53:46.000'), -- Anomoly mentioned 
('2017-01-23 13:15:00.000','2017-01-23 13:45:00.000') -- Started After Lunch 

Select A.* 
     ,B.* 
From (... your complex query here ...) A 
Cross Apply (
       Select SetupTime = count(*) 
       From (Select Top (DateDiff(MINUTE,A.SetupStart,A.StartTime)) T=cast(DateAdd(MINUTE,Row_Number() Over (Order By Number)-1,A.SetupStart) as time) 
         From master..spt_values) S 
       Where (cast(A.SetupStart as time)<'13:30' and cast(A.StartTime as time)>'14:00' and T not between '13:30' and '14:00') 
        or (cast(A.SetupStart as time) > '13:30' ) 
        or (cast(A.StartTime as time) < '14:00') 
     ) B 

返回從安裝時間

SetupStart    StartTime     SetupTime 
2017-01-23 12:56:42.000 2017-01-23 14:41:06.000 75 
2017-01-23 15:00:00.000 2017-01-23 18:30:00.000 210 
2017-01-23 23:51:00.000 2017-01-23 23:53:46.000 2 
2017-01-23 13:15:00.000 2017-01-23 13:45:00.000 30 

Answer 2

這是SQL Server中的痛苦，但你可以做到這一點：

(CASE WHEN CAST(o.SetupStart as time) > '14:00:00' OR 
      CAST(o.StartTime as time) < '13:30:00'   
     THEN DATEDIFF(MINUTE, o.SetupStart, o.StartTime) 
     WHEN CAST(o.SetupStart as time) >= '13:30:00' AND 
      CAST(o.StartTime as time) <= '14:00:00' 
     THEN 0 
     WHEN CAST(o.SetupStart as time) >= '13:30:00' 
     THEN DATEDIFF(MINUTE, 
        CAST(CAST(o.SetupStart as DATE) as DATETIME) + CAST('14:00:00' as TIME), 
        o.StartTime 
        ) 
     WHEN CAST(o.StartTime as Time) <= '14:00:00' 
     THEN DATEDIFF(MINUTE, 
        o.SetStart, 
        CAST(CAST(o.StartTime as DATE) as DATETIME) + CAST('13:30:00' as TIME) 
        ) 
     ELSE DATEDIFF(MINUTE, o.SetupStart, o.StartTime) - 30 
    END) AS [SetupTime] 

此處理以下情況：

• 時報兩個前或午飯後都。
• 時間都在午餐時間。
• SetupStart正在吃午飯。
• StartTime正在吃午飯。
• SetupStart是午餐前和StartTime是午飯後

Answer 3

最簡單的方法是什麼@Shakeer侯賽因建議

DATEDIFF(MINUTE, UserData.Orders.SetupStart, UserData.Orders.StartTime) AS[SetupTime] 

BECOMES

​​

此外，你可以做兩個DATEDIFF功能並將它們加在一起：

DATEDIFF(MINUTE, UserData.Orders.SetupStart, UserData.Orders.LunchStartTime) 
+ DATEDIFF(MINUTE, UserData.Orders.LunchEndTime, UserData.Orders.StartTime) AS[SetupTime] 

如果這些數據點不存在，您可以創建一個臨時表。

Answer 4

declare 
    @ldt_from time = '13:00' 
    ,@ldt_to time = '14:00'; 

-- dummy data prepare 
;with [my_setups] as 
(
    select [SetupStart] = cast('2017-01-23 13:56:42.000' as datetime), [StartTime] = cast('2017-01-23 14:41:06.000' as datetime) 
    union all 
    select [SetupStart] = cast('2017-01-23 12:45:00.000' as datetime), [StartTime] = cast('2017-01-23 12:46:00.000' as datetime) 
) 
-- end dummy data prepare 
select [minutes] = 
     datediff(mi, [SetupStart], [StartTime]) 
    - datediff(day, [SetupStart], [StartTime]) * (datediff(mi, @ldt_from, @ldt_to)) 
    + case when cast([SetupStart] as time) between @ldt_from and @ldt_to or cast([StartTime] as time) between @ldt_from and @ldt_to then 
     datediff 
     (
     mi 
     ,case when cast([SetupStart] as time) > @ldt_from then cast([SetupStart] as time) else @ldt_from end 
     ,case when cast([StartTime] as time) < @ldt_to  then cast([StartTime] as time) else @ldt_to end  
    ) 
     else 0 end   
from 
    [my_setups]