Question

我正在尝试撰写一份声明，用于计算员工出勤率，并通过计算重复id来执行他们name，id以及他过去3个月工作的日期NewTimeAttendance第1，第2和第3个月。

我试过数：

Select COUNT(employeeid)
  from NewTimeAttendance
 where employeeid=1 and (month=1 or month = 2 or month = 3)

这绝对有效，但仅适用于一名员工。

第二次尝试：

SELECT COUNT(NewEmployee.EmployeeID)
  FROM NewEmployee INNER JOIN NewTimeAttendance 
    ON NewEmployee.EmployeeID = NewTimeAttendance.EmployeeID
   and (month=1 or month =2 or month = 3)

这是有效的，但它会计算所有员工，我希望它将每个EmployeeId，EmployeeName和number of days作为新记录执行。

最后一次尝试:(在看到代码之前......这是错误的..但我正在尝试）

for i in  0..27 loop
   SELECT COUNT(NewEmployee.EmployeeID),NewEmployee.EmployeeId,EmployeeName
     FROM NewEmployee INNER JOIN NewTimeAttendance 
       ON NewEmployee.EmployeeID(i) = NewTimeAttendance.EmployeeID
      and (month=1 or month =2 or month = 3)
end loop

我真的需要帮助......提前谢谢。

Answer 1

您需要GROUP BY功能，请尝试：

SELECT COUNT(NewEmployee.EmployeeID), NewEmployee.EmployeeId,EmployeeName
 FROM NewEmployee INNER JOIN NewTimeAttendance 
  ON NewEmployee.EmployeeID = NewTimeAttendance.EmployeeID
   and (month=1 or month =2 or month = 3)
GROUP BY NewEmployee.EmployeeID, EmployeeName

计算员工出勤率

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