如何计算小时数，计算日，加时间和进出时间 [英] How to hours, calculate day , overtime from in and OUT time

问题描述

我有IN时间和外出时间，其中我从IN和Out时间计算小时数，现在我想要当小时= 8然后Day将是1并且如果小时多于8小时（例如：12小时减去从8小时= 4小时然后4小时乘以100分钟）



i想要商店程序...



我尝试过：

 SELECT INTime，Outtime，DATEDIFF（Hour，INTime，Outtime）AS Hours 
如果
 Hours = 8 
 FROM EmployeeAttedance 

解决方案

我将修复你拥有的SQL语句

  DECLARE   @ punches   TABLE （PunchIn  time ，PunchOut  time ）
  INSERT   @ punches   VALUES （'  07:58'，'  16:15'）
 
  SELECT  PunchIn，PunchOut 
，[Hours] = DateDiff（小时，PunchIn，PunchOut）
，[天数] =  CASE   WHEN  DateDiff（小时，PunchIn） ，PunchOut）> =  8  那么  1   ELSE   0   END  
  FROM   @ punches  

我必须让你知道有一些潜在的缺陷在你的逻辑中。

1-如果有人在16小时内进行双班制，那么这将是2天吗？

2-如何在夜班工作？

3-如果有人在1天工作，你是否希望减少工作时间8？

i have IN Time and Out Time ,in which i am calculating Hours from IN and Out Time,Now i want that when hours =8 then Day will be 1 and if Hours is more then to 8hours(for example :12hours minus from 8hours =4hours then 4hours multiply to 100rate)

i want store procedure ...

What I have tried:

SELECT     INTime, Outtime, DATEDIFF(Hour, INTime, Outtime) AS Hours
if
Hours = 8
FROM         EmployeeAttedance

解决方案

I will fix the SQL statement that you have

DECLARE   @punches TABLE ( PunchIn time, PunchOut time )
INSERT @punches VALUES ('07:58', '16:15')

SELECT	PunchIn, PunchOut
	,	[Hours] = DateDiff(Hour, PunchIn, PunchOut) 
	,	[Days] = CASE WHEN  DateDiff(Hour, PunchIn, PunchOut) >= 8 THEN 1 ELSE 0 END
FROM @punches

I must let you know that there are a few potential flaws in your logic.
1- If someone works a "double-shift" for 16 hours, would this be 2 days?
2- What about someone working an overnight shift?
3- If someone did work 1 day, did you want the hours reduced by 8?

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