Sql server 计算T-SQL中两个日期之间的工作日（扩展）_Sql Server_Tsql_Date_Count

Sql server 计算T-SQL中两个日期之间的工作日（扩展）

sql-server tsql date

Sql server 计算T-SQL中两个日期之间的工作日（扩展）,sql-server,tsql,date,count,Sql Server,Tsql,Date,Count,扩大这一问题：我想在t-sql中计算两个日期之间的工作日，只是周末，而不是整个工作日（24小时），比如说每个工作日从上午9点到下午5:30 我如何使用例如CMS的方法做到这一点： DECLARE @StartDate DATETIME DECLARE @EndDate DATETIME SET @StartDate = '2008/10/01' SET @EndDate = '2008/10/31' SELECT (DATEDIFF(dd, @StartDate, @EndDate

扩大这一问题：

我想在t-sql中计算两个日期之间的工作日，只是周末，而不是整个工作日（24小时），比如说每个工作日从上午9点到下午5:30

我如何使用例如CMS的方法做到这一点：

 DECLARE @StartDate DATETIME
DECLARE @EndDate DATETIME
SET @StartDate = '2008/10/01'
SET @EndDate = '2008/10/31'


SELECT
   (DATEDIFF(dd, @StartDate, @EndDate) + 1)
  -(DATEDIFF(wk, @StartDate, @EndDate) * 2)
  -(CASE WHEN DATENAME(dw, @StartDate) = 'Sunday' THEN 1 ELSE 0 END)
  -(CASE WHEN DATENAME(dw, @EndDate) = 'Saturday' THEN 1 ELSE 0 END)

我想你已经很好地掌握了，只需要把THEN 1改成THEN-1

然而，我不完全确定你想对你问题的时间部分做什么

DECLARE @StartDate DATETIME
DECLARE @EndDate DATETIME
SET @StartDate = '2013/03/01'
SET @EndDate = '2013/03/31'


SELECT
   (DATEDIFF(dd, @StartDate, @EndDate) + 1)
  -(DATEDIFF(wk, @StartDate, @EndDate) * 2)
  -(CASE WHEN DATENAME(dw, @StartDate) = 'Sunday' THEN -1 ELSE 0 END)
  -(CASE WHEN DATENAME(dw, @EndDate) = 'Saturday' THEN -1 ELSE 0 END)

找到了这个

 CREATE FUNCTION dbo.GetWorkingDays
  ( @StartDate datetime,
    @EndDate datetime )
RETURNS INT
AS
BEGIN
  DECLARE @WorkDays int, @FirstPart int
  DECLARE @FirstNum int, @TotalDays int
  DECLARE @LastNum int, @LastPart int
  IF (DATEDIFF(day, @StartDate, @EndDate) < 2)
    BEGIN
      RETURN ( 0 )
    END
  SELECT
   @TotalDays = DATEDIFF(day, @StartDate, @EndDate) - 1,
   @FirstPart = CASE DATENAME(weekday, @StartDate)
                 WHEN 'Sunday' THEN 6
                 WHEN 'Monday' THEN 5
                 WHEN 'Tuesday' THEN 4
                 WHEN 'Wednesday' THEN 3
                 WHEN 'Thursday' THEN 2
                 WHEN 'Friday' THEN 1
                 WHEN 'Saturday' THEN 0
               END,
   @FirstNum = CASE DATENAME(weekday, @StartDate)
                 WHEN 'Sunday' THEN 5
                 WHEN 'Monday' THEN 4
                 WHEN 'Tuesday' THEN 3
                 WHEN 'Wednesday' THEN 2
                 WHEN 'Thursday' THEN 1
                 WHEN 'Friday' THEN 0
                 WHEN 'Saturday' THEN 0
               END
  IF (@TotalDays < @FirstPart)
     BEGIN
       SELECT @WorkDays = @TotalDays
     END
  ELSE
     BEGIN
       SELECT @WorkDays = (@TotalDays - @FirstPart) / 7
       SELECT @LastPart = (@TotalDays - @FirstPart) % 7
       SELECT @LastNum = CASE
         WHEN (@LastPart < 7) AND (@LastPart > 0) THEN @LastPart - 1
         ELSE 0
       END
       SELECT @WorkDays = @WorkDays * 5 + @FirstNum + @LastNum
     END
  RETURN ( @WorkDays )
END
GO

创建函数dbo.GetWorkingDays
（@StartDate datetime，
@结束日期（日期时间）
返回整数
作为
开始
声明@WorkDays int、@FirstPart int
声明@FirstNum int、@TotalDays int
声明@LastNum int、@LastPart int
IF（DATEDIFF（day，@StartDate，@EndDate）<2）
开始
返回（0）
结束
挑选
@TotalDays=DATEDIFF（天、@StartDate、@EndDate）-1，
@第一部分=案例日期名称（工作日@StartDate）
“星期天”是什么时候
“星期一”是什么时候
“星期二”是什么时候
“星期三”是什么时候
“星期四”是什么时候
“星期五”是什么时候
“星期六”是什么时候
完,，
@FirstNum=案例日期名称（工作日@StartDate）
“星期天”是什么时候
“星期一”是什么时候
“星期二”是什么时候
“星期三”是什么时候
“星期四”是什么时候
“星期五”是什么时候
“星期六”是什么时候
结束
如果（@TotalDays<@FirstPart）
开始
选择@WorkDays=@TotalDays
结束
其他的
开始
选择@WorkDays=（@TotalDays-@FirstPart）/7
选择@LastPart=（@TotalDays-@FirstPart）%7
选择@LastNum=CASE
当（@LastPart<7）和（@LastPart>0）时，则@LastPart-1
其他0
结束
选择@WorkDays=@WorkDays*5+@FirstNum+@LastNum
结束
返回时间（@工作日）
结束
去

这是我一直在寻找的，下面是代码，以防其他人需要它：

参数包括：

@D1 (StartDate)
@D2 (EndDate)
@T1 (Business Hours - Start)
@T2 (Business Hours - End)

该函数返回一个带有小时数的浮点值

创建函数[dbo].[WorkHoursDiff]（@D1作为datetime，@D2作为datetime，@T1作为int，@T2作为int）
返回浮点数作为
开始
如果@D1>@D2返回0
将@whd声明为浮动
将@H1声明为float
将@H2声明为float
如果不是（日期部分（yyyy，@D1）=日期部分（yyyy，@D2）和日期部分（mm，@D1）=日期部分（mm，@D2）和日期部分（dd，@D1）=日期部分（dd，@D2））
开始
将@DD声明为datetime
设置为@whd=0
如果日期部分（dw，@D1）1和日期部分（dw，@D1）7
如果dbo.GetDateHour（@D1）@T2 SET@whd=0
否则，如果dbo.GetDateHour（@D1）>=@T1和dbo.GetDateHour（@D1）=@T1和dbo.GetDateHour（@D2）试试这个
Declare @startdate datetime,
@enddate datetime,
@countSunday int,
@countTuesday int,
@countWednesday int,@countThursday int,@countFriday int,@countSaturday int

set @countSunday=0
set @countMonday=0
set @countTuesday =0
set @countWednesday =0
set @countThursday =0
set @countFriday =0
set @countSaturday=0
set @startdate='2018-10-01'
set @enddate='2018-10-20'

while @startdate<=@enddate
Begin
    IF DatePart(WEEKDAY,@startdate) = 1
    SET @countSunday=@countSunday+1
    else if  DatePart(WEEKDAY,@startdate) = 2
      SET @countMonday=@countMonday+1  
      else if  DatePart(WEEKDAY,@startdate) = 3
      SET @countTuesday=@countTuesday+1  
      else if  DatePart(WEEKDAY,@startdate) = 4
      SET @countWednesday=@countWednesday+1  
      else if  DatePart(WEEKDAY,@startdate) = 5
      SET @countThursday=@countThursday+1  
      else if  DatePart(WEEKDAY,@startdate) = 6
      SET @countFriday=@countFriday+1  
      else if  DatePart(WEEKDAY,@startdate) = 7
      SET @countSaturday=@countSaturday+1  
    SET @startdate=DateAdd(d,1,@startdate)
END


select @countSunday as Sunday ,@countMonday as Monday,@countTuesday as Tuesday,@countWednesday as Wednesday,@countThursday as Thursday,@countFriday as Friday,@countSaturday as Saturday

DECLARE @DateFrom DateTime ='20181001' , @DateTo DateTime = '20181031'  ;
WITH CTE(dt)
AS
(
SELECT @DateFrom
UNION ALL
SELECT DATEADD(d, 1, dt) FROM CTE
  WHERE dt < @DateTo
)
SELECT datename("dw", dt) as Days,count(datename("dw", dt)) as numberofDays FROM CTE 
group by datename ("dw", dt)

Declare@startdate datetime，
@enddate日期时间，
@countSunday int，
@星期二国际，
@计数星期三整数、@计数星期四整数、@计数星期五整数、@计数星期六整数
设置为@countSunday=0
设置@countMonday=0
设置为@countbeday=0
设置为@count星期三=0
设置为@count星期四=0
设置为@countFriday=0
设置为@countSaturday=0
设置为@startdate='2018-10-01'
设置为@enddate='2018-10-20'
而@startdate试试这个
Declare @startdate datetime,
@enddate datetime,
@countSunday int,
@countTuesday int,
@countWednesday int,@countThursday int,@countFriday int,@countSaturday int

set @countSunday=0
set @countMonday=0
set @countTuesday =0
set @countWednesday =0
set @countThursday =0
set @countFriday =0
set @countSaturday=0
set @startdate='2018-10-01'
set @enddate='2018-10-20'

while @startdate<=@enddate
Begin
    IF DatePart(WEEKDAY,@startdate) = 1
    SET @countSunday=@countSunday+1
    else if  DatePart(WEEKDAY,@startdate) = 2
      SET @countMonday=@countMonday+1  
      else if  DatePart(WEEKDAY,@startdate) = 3
      SET @countTuesday=@countTuesday+1  
      else if  DatePart(WEEKDAY,@startdate) = 4
      SET @countWednesday=@countWednesday+1  
      else if  DatePart(WEEKDAY,@startdate) = 5
      SET @countThursday=@countThursday+1  
      else if  DatePart(WEEKDAY,@startdate) = 6
      SET @countFriday=@countFriday+1  
      else if  DatePart(WEEKDAY,@startdate) = 7
      SET @countSaturday=@countSaturday+1  
    SET @startdate=DateAdd(d,1,@startdate)
END


select @countSunday as Sunday ,@countMonday as Monday,@countTuesday as Tuesday,@countWednesday as Wednesday,@countThursday as Thursday,@countFriday as Friday,@countSaturday as Saturday

DECLARE @DateFrom DateTime ='20181001' , @DateTo DateTime = '20181031'  ;
WITH CTE(dt)
AS
(
SELECT @DateFrom
UNION ALL
SELECT DATEADD(d, 1, dt) FROM CTE
  WHERE dt < @DateTo
)
SELECT datename("dw", dt) as Days,count(datename("dw", dt)) as numberofDays FROM CTE 
group by datename ("dw", dt)

声明@DateFrom DateTime='20181001'，@DateTo DateTime='20181031'；
带热膨胀系数（dt）
作为
(
选择@DateFrom
联合所有
从CTE中选择日期添加（d，1，dt）
其中dt<@DateTo
)
选择datename（“dw”，dt）作为天数，从CTE中计算（datename（“dw”，dt））作为天数
按日期名称分组（“dw”，dt）
工时部分是工作日的营业时间。在这种情况下，我必须将datediff改为小时而不是天。假设我想知道星期一和星期二两个日期之间的工作时间，使用上述解决方案，我将有48小时（2天），但如果只有工作时间（上午9点到下午5点），将是16小时，这就是我需要的。