Question

我要从这些日期中获取两个日期，分别为2019/07/26和2019/08/08

Year  Month  ActualDays  WorkingDays
------------------------------------
2019    07       06           04
2019    08       08           06


DECLARE @start DATETIME, @end DATETIME

SET @start = '2019-07-26'
SET @end = '2019-08-08'

;WITH c(d) AS
(
    SELECT TOP (DATEDIFF(DAY, @start, @end) + 1 )
        DATEADD(DAY, ROW_NUMBER() OVER (ORDER BY name)-1, @start)
    FROM 
        sys.all_columns
) 
SELECT 
    YEAR(t.[date]), MONTH(t.[date]), t.[days]
FROM 
    (SELECT 
         [date] = DATEADD(MONTH, DATEDIFF(MONTH, 0, d), 0), 
         [days] = COUNT(*)
     FROM 
         c 
     GROUP BY 
         DATEDIFF(MONTH, 0, d)
     UNION ALL 
     SELECT 
         d, NULL           
     FROM 
         c) AS t
WHERE 
    t.[days] IS NOT NULL

由此，我将按月和按年分组，需要工作日

Answer 1

您可以试试这个-

DECLARE @startdate DATE = '20190701'
      , @enddate DATE   = '20190716'
DECLARE @temp TABLE(thedate DATE, DoCount INT);

WITH CTE (thedate) AS 
(
  SELECT @startdate
  UNION ALL
  SELECT DATEADD(day, 1, thedate)
    FROM CTE
   WHERE thedate < @enddate
)     
INSERT INTO @temp
SELECT thedate,CASE WHEN DATEPART(dw, [thedate]) IN (1,7) THEN 0 ELSE 1 END
FROM CTE

SELECT YEAR(thedate),MONTH(thedate),
COUNT(DoCount) AS ActualDays,
SUM(DoCount)    WorkingDays
FROM @temp
GROUP BY YEAR(thedate),
MONTH(thedate)

SQL Server从两个日期开始按月和年分组工作（不包括周六和周日）

1 个答案: