我试图弄清3个等级（普通，加班1.5和加班2.0）的值，然后确定'普通'是否等于8，如果是，则计算（SUM-0.4） ）/7.6。如果没有，那么它只是（SUM / 7.6）......这样我就可以计算出全职员工的人数。
所以这就是我现在所处的位置。这些值似乎并不是我手动计算的结果，我认为我的过程是错误的：
SELECT Supervisor,
SUM(CASE WHEN DAYOFWEEK(workdate) = 2 THEN IF(employeehours.paycode = '01 Ordinary' AND employeehours = 8,((employeehours-0.4)/7.6),(employeehours/7.6)) END) `Monday`,
SUM(CASE WHEN DAYOFWEEK(workdate) = 3 THEN IF(employeehours.paycode = '01 Ordinary' AND employeehours = 8,((employeehours-0.4)/7.6),(employeehours/7.6)) END) `Tuesday`,
SUM(CASE WHEN DAYOFWEEK(workdate) = 4 THEN IF(employeehours.paycode = '01 Ordinary' AND employeehours = 8,((employeehours-0.4)/7.6),(employeehours/7.6)) END) `Wednesday`,
SUM(CASE WHEN DAYOFWEEK(workdate) = 5 THEN IF(employeehours.paycode = '01 Ordinary' AND employeehours = 8,((employeehours-0.4)/7.6),(employeehours/7.6)) END) `Thursday`,
SUM(CASE WHEN DAYOFWEEK(workdate) = 6 THEN IF(employeehours.paycode = '01 Ordinary' AND employeehours = 8,((employeehours-0.4)/7.6),(employeehours/7.6)) END) `Friday`,
SUM(CASE WHEN DAYOFWEEK(workdate) = 7 THEN IF(employeehours.paycode = '01 Ordinary' AND employeehours = 8,((employeehours-0.4)/7.6),(employeehours/7.6)) END) `Saturday`
FROM payroll.employeedatanew_copy
JOIN payroll.employeehours ON employeedatanew_copy.`ID Number` = employeehours.employeeid
WHERE employeehours.workdate BETWEEN '2013-12-02' AND '2013-12-07'
GROUP BY supervisor