SQL-datediff（分鍾）不包括周末

Question

您能幫我嗎，我正在使用下面的sql視圖（然后在Crystal報表中使用了它）。 我需要有日期差異（以分鍾為單位），但是現在我需要排除周末。 請幫忙 ：）

SELECT intwc                             AS wc,
       Datediff(n, start_date, end_date) AS time,
       mh_start_date                     AS date,
       'Repair'                          AS type
FROM   dbo.xxxxxxx 

Answer 1

這是@bendataclear答案的修改版本。 它直接計算周末分鍾數，而不是計算天數並乘以24 * 60。 它還考慮了周六/周日開始/結束的所有4種組合

我正在使用CONVERT(date,@StartDate)來獲取@StartDate的日期，其時間為00:00:00 ，然后將其用於計算部分星期日和星期六。 有更好的方法可以做到這一點，但我最簡單。

DECLARE @StartDate DATETIME
DECLARE @EndDate DATETIME
SET @StartDate = '2013/03/15 23:30:00'
SET @EndDate = '2013/03/17 00:30:00'


SELECT
(   DATEDIFF(MINUTE, @StartDate, @EndDate)
    - ( DATEDIFF(wk, @StartDate,@EndDate)*(2*24*60)
        -- End on Sunday
        -(CASE WHEN DATEPART(dw, @EndDate)  = 1 THEN 24.0*60-DATEDIFF(minute,CONVERT(date,@EndDate),@EndDate) ELSE 0 END)
        -- Start on Saturday
        -(CASE WHEN DATEPART(dw, @StartDate) = 7 THEN DATEDIFF(minute,CONVERT(date,@StartDate),@StartDate) ELSE 0 END)
        -- End on Saturday
        +(CASE WHEN DATEPART(dw, @EndDate)  = 7 THEN DATEDIFF(minute,CONVERT(date,@EndDate),@EndDate) ELSE 0 END)
        -- Start on Saturday
        +(CASE WHEN DATEPART(dw, @StartDate) = 1 THEN 24.0*60-DATEDIFF(minute,CONVERT(date,@StartDate),@StartDate) ELSE 0 END)
    )
)

Answer 2

這個答案假設您想在一分鍾內排除周末，這也完全基於此問題的答案：

DECLARE @StartDate DATETIME
DECLARE @EndDate DATETIME
SET @StartDate = '2013/03/15 23:30:00'
SET @EndDate = '2013/03/18 00:30:00'


SELECT
   (DATEDIFF(MINUTE, @StartDate, @EndDate))
  -(DATEDIFF(wk, @StartDate, @EndDate) * (2*24*60))
  -(CASE WHEN DATENAME(dw, @StartDate) = 'Sunday' THEN (24*60) ELSE 0 END)
  -(CASE WHEN DATENAME(dw, @EndDate) = 'Saturday' THEN (24*60) ELSE 0 END)

Answer 3

SELECT intwc                             AS wc,
   Datediff(n, start_date, end_date) AS time,
   mh_start_date                     AS date,
   'Repair'                          AS type
FROM   dbo.xxxxxxx 
Where DATEPART(dw, start_date) NOT IN (1, 7) and DATEPART(dw, end_date) NOT IN (1, 7)

Answer 4

因此，您確實需要能夠處理一些CASE語句來處理所有的極端情況。 這是我整理的一個示例。 Numbers表只是一個計數表，在這種情況下為1到30。

CREATE TABLE #times (id INT IDENTITY(1,1), start_stamp DATETIME, end_stamp DATETIME)

INSERT INTO #times
        ( 
          start_stamp ,
          end_stamp
        )
SELECT DATEADD(DAY, -2*Number, CURRENT_TIMESTAMP), DATEADD(DAY, -1*Number, CURRENT_TIMESTAMP)
FROM Common.NUMBERS
WHERE Number < 31

SELECT id, start_stamp, end_stamp,
CASE WHEN DATEDIFF(DAY, start_stamp, end_stamp) < 7 THEN
    CASE WHEN DATEPART(weekday, start_stamp) < DATEPART(weekday, end_stamp)
        THEN DATEDIFF(MINUTE, start_stamp, DATEADD(HOUR, -48, end_stamp))
        ELSE DATEDIFF(MINUTE, start_stamp, end_stamp) END
    ELSE DATEDIFF(MINUTE, start_stamp, DATEADD(HOUR, -48*(DATEDIFF(WEEK, start_stamp, end_stamp)), end_stamp)) END
    + CASE WHEN DATENAME(weekday,start_stamp) IN ('Sunday', 'Saturday') THEN 1440 ELSE 0 END
    + CASE WHEN DATENAME(weekday,end_stamp) IN ('Sunday', 'Saturday') THEN 1440 ELSE 0 END
FROM #times

可能有一種更優雅的方法，但是代碼允許您針對整個結果集運行並按行計算。