Question

我需要根据参加的会议，扣除和奖金来计算员工的月薪; 根据工作岗位，员工每次会议的薪酬不同。

解决方案是： salary =（Pay_per_minute * meetings_attended）+奖金 - 扣除;

我有四张桌子：

工作： ID，title，pay_per_meeting

员工： Id，Name，job_id

奖金： ID，金额，employee_id，日期

扣减： ID，金额，employee_id，日期

会议： ID，employee_id，日期

SELECT
COUNT(meetings.employee_id) as meetings_attended, 
COUNT(deductions.amount) as debt, 
COUNT(bonuses.amount) bonus, 
(SELECT jobs.pay_per_attendance from jobs where jobs.id = (select job_id from employees where id=meetings.employee_id)) as pay,
((meetings_attended * pay) + bonus - debt) as salary 
FROM meetings 
JOIN deductions ON deductions.employee_id = meetings.employee_id 
JOIN bonuses ON bonuses.employee_id = meetings.employee_id 
WHERE meetings.employee_id = 1 
GROUP BY MONTH(meetings.date), MONTH(deductions.date), MONTH(bonuses.date)

每当我删除工资行时，上面的查询都会返回许多不正确的值，但会出现未知列薪，meeting_attended，债务和奖金的错误，我确定分组有问题，但我不能只看到它。

Answer 1

您无法在定义的同一选择列表中引用列别名，您需要引用基础列。子查询无法访问主查询中计算的聚合。您需要重复聚合表达式，或将所有内容移动到子查询中，并在外部查询中使用它进行计算。

此外，所有COUNT()表达式都将返回相同的内容，因为它们只是计算行数（我假设这些值都不是NULL）。您可能希望COUNT(DISTINCT <column>)获得不同的计数，并且您需要使用唯一的列，因此它们应该是主键列，例如COUNT(DISTINCT deductions.id)。

另一个问题是，当您尝试在多个连接时对值进行求和并计算时，最终得到的结果太高，因为行在所有表的叉积中都会重复。见Join tables with SUM issue in MYSQL。解决方案是计算子查询中每个表的总和。

SELECT m.month, m.meetings_attended, d.debt, b.bonus, 
    m.meetings_attended * j.pay_per_meeting + b.amount - d.amount AS salary
FROM (
    SELECT MONTH(date) AS month, COUNT(*) AS meetings_attended
    FROM meetings
    WHERE employee_id = 1
    GROUP BY month) AS m
JOIN (
    SELECT MONTH(date) AS month, COUNT(*) AS bonus, SUM(amount) AS amount
    FROM bonuses
    WHERE employee_id = 1
    GROUP BY month) AS b ON m.month = b.month
JOIN (
    SELECT MONTH(date) AS month, COUNT(*) AS debt, SUM(amount) AS amount
    FROM deductions
    WHERE employee_id = 1
    GROUP BY month) AS d ON m.month = d.month
CROSS JOIN employees AS e
JOIN jobs AS j ON j.id = e.job_id
WHERE e.employee_id = 1

对不同表上的列求和，并乘以另一个表上的列的值

1 个答案: