如何每周增加总数以得到余数？ SQL服务器

Question

I have a table that stores timesheet data of hours worked and for each day, there is an Hours column which has the number of hours worked for that day, and so based on a 5 day work week, on the fifth day, there will be a total from those 5 days in a column next to it called Week Total, so as a sample below for 1 week:

    TimeIn                TimeOut                Hours          Week Total
---------------------------------------------------------------------------
2019-02-18 07:10:00     2019-02-18 15:50:00      8.17
2019-02-19 07:20:00     2019-02-19 15:41:00      7.85
2019-02-20 06:57:00     2019-02-20 15:50:00      8.38
2019-02-21 07:12:00     2019-02-21 16:03:00      8.35
2019-02-22 07:09:00     2019-02-22 16:05:00      8.43           41.18  

The "Week Total" column value is nowhere to be found within any table within the database, and so I am needing to write a query that will calculate it for each week and if it totals to over 40 for any week, then take that remainder that is over 40 and add it together with other weeks' remainders.

Like for example, in the table shown above, the total is 41.18, and say for another week the total is 42.07. Since these weeks are over 40, it will be 1.18 + 2.07 for the total of anything extra that is over 40. So this needs to be calculated for each week which is 5 work days.

So far I have:

SELECT
    employeeID,
    SUM(DATEDIFF([second], [TimeIn], [TimeOut]) / 60.0  / 60.0) AS Hours
FROM 
    Time
WHERE 
    TimeIn BETWEEN DATEADD(YEAR, -1, DATEADD(MONTH, DATEDIFF(MONTH, 0, GETDATE()), 0)) 
               AND DATEADD(MONTH, DATEDIFF(MONTH, 0, GETDATE()), -1)
GROUP BY 
    employeeID

But this is just returning their EmployeeID along with their entire total hours worked within the date ranges I am filtering by.

How would I narrow it down to totals per each week of 5 work days in order to get total hours worked per week and if any are over 40, then take that remainder and add?

Answer 1

Does this help?

SELECT
employeeID,
SUM(DATEDIFF([second], [TimeIn], [TimeOut]) / 60.0  / 60.0) AS Hours,
CASE WHEN SUM(DATEDIFF([second], [TimeIn], [TimeOut]) / 60.0  / 60.0) > 40
     THEN SUM(DATEDIFF([second], [TimeIn], [TimeOut]) / 60.0  / 60.0) - 40
     ELSE 0 END OvertimeHours
FROM  Time
WHERE 
    TimeIn BETWEEN DATEADD(YEAR, -1, DATEADD(MONTH, DATEDIFF(MONTH, 0, GETDATE()), 0)) 
           AND DATEADD(MONTH, DATEDIFF(MONTH, 0, GETDATE()), -1)
GROUP BY 
    employeeID, DATEPART(WEEK,[TimeIn])

Answer 2

Add datepart(weekday, TimeIn) to the group by clause:

select employeeID, sum(HoursReminder)
from(
    SELECT
        employeeID,
        SUM(DATEDIFF([second], [TimeIn], [TimeOut]) / 60.0  / 60.0) - 40.0 AS HoursReminder
    FROM 
        Time
    WHERE 
        TimeIn BETWEEN DATEADD(YEAR, -1, DATEADD(MONTH, DATEDIFF(MONTH, 0, GETDATE()), 0)) 
                   AND DATEADD(MONTH, DATEDIFF(MONTH, 0, GETDATE()), -1)
    GROUP BY 
        employeeID
        , datepart(weekday, TimeIn)
    ) as fr
group by employeeID

This query will give you total reminder per employee.