Mysql 为什么当我使用 group by Id 时 sum 函数不计算列的总和

Question

我们公司有员工，员工从他们的工资中借了一些钱，这些借来的钱是要让我们知道的。我们还需要知道每笔交易的工资中剩余的钱，以及借来的钱。

我创建了包括薪水列的员工表

作为

CREATE TABLE `employees` (
 `Employee_Id` int(11) NOT NULL AUTO_INCREMENT,
 `Employee_Name` varchar(100) NOT NULL,
 `Salary` decimal(10,0) NOT NULL,
 PRIMARY KEY (`Employee_Id`)
) ENGINE=InnoDB AUTO_INCREMENT=2

和salary_transaction表把每笔借来的钱存到money_amount列

CREATE TABLE salary_transaction (
 Salary_Transaction_Id int(11) NOT NULL AUTO_INCREMENT,
 Employee_Id int(11) NOT NULL,
 money_amount decimal(10,0) NOT NULL,
 PRIMARY KEY (Salary_Transaction_Id)
 
) ENGINE=InnoDB AUTO_INCREMENT=3  

这是我的查询，问题是它没有计算特定员工借用的 money_amount 的累计总和。

SELECT t.Salary_Transaction_Id,
t.Employee_Id,t.money_amount,
sum(t.money_amount) as Total_borrowed,
e.salary-sum(t.money_amount) as remaining from salary_transaction t
JOIN
(SELECT salary,Employee_Id,Employee_Name from employees ) e 
ON
t.Employee_Id = e.Employee_Id GROUP by t.salary_transaction_id 

编辑

我用脚本提供了我的问题

All scripts here

编辑 2

预期的总借用值

Answer 1

select t.salary_transaction_id, 
       t.employee_id, 
       t.money_amount, 
       sum(t.money_amount) over (partition by employee_id order by t.salary_transaction_id) as total_borrowed, 
       e.salary - sum(t.money_amount) over (partition by e.employee_id order by t.salary_transaction_id) remaining
from employees e
inner join salary_transaction t on t.employee_id = e.employee_id
group by t.employee_id, t.salary_transaction_id, t.money_amount

你可以使用OVER()函数。

salary_transaction_id | employee_id | money_amount | total_borrowed | remaining
         1            |       1     |     3000     |       3000     |    4000
         2            |       1     |     1000     |       4000     |    3000
         3            |       1     |      500     |       4500     |    2500

Answer 2

您想从每个员工的工资中减去他们借入的总金额。我会推荐一个带有预聚合的left join：

select e.*,
    coalesce(st.total_borrowed, 0) as total_borrowed,
    e.salary - coalesce(st.total_borrowed, 0) as remaining
from employee e
left join (
    select employee_id, sum(money_amount) as total_borrowed
    from salary_transaction
    group by employee_id
) st on st.employee_id = e.employee_id

您也可以使用相关子查询。在最新版本的 MySQL 中，横向连接很方便：

select e.*, st.total_borrowed, e.salary - st.total_borrowed as remaining
from employee e
cross join lateral (
    select coalesce(sum(money_amount), 0) as total_borrowed
    from salary_transaction st
    where st.employee_id = e.employee_id
) st

---

编辑：这两个查询都为每个员工提供一行。如果你真的想要每笔交易一行，借来的钱和剩余的工资，那就不同了。

在 MySQL 8.0 中，您可以使用窗口函数：

select e.*,
    st.salary_transaction_id, st.money_amount,
    coalesce(sum(st.money_amount) over(partition by employee_id order by st.salary_transaction_id), 0) as total_borrowed,
    e.salary - coalesce(sum(st.money_amount) over(partition by employee_id order by st.salary_transaction_id), 0) as remaining
from employee e
left join salary_transaction st using(employee_id)

在早期版本中，另一种选择是相关子查询：

select t.*, salary - total_borrowed
from (
    select e.*,
        st.salary_transaction_id, st.money_amount,
        (
            select coalesce(sum(st.money_amount), 0) 
            from salary_transaction st1
            where st.employee_id = e.employee_id and st1.salary_transaction_id <= st.salary_transaction_id
        ) as total_borrowed
    from employee e
    left join salary_transaction using(employee_id)
) t