多个mysql表列的Sum，Subtract和Join

Posted 2021-03-30

我有四个mysql表client，transaction，other_loan和payment。我想得到sum和load_amount的additional从transaction和sum的amount表other_loan + subtract到表sum的payment_amount的payment。我怎样才能实现它？

结果我想：

>  ID | Name     |  Amount
>  1  | Robin    | 8718  
>  2  | Reynaldo | 21
>  3  | Leomar   | 0

我的表：交易

>  tid | id| date       | load_amount | additional
>  1   | 1 | 2018-12-01 |    90       | 0  
>  2   | 1 | 2018-12-07 |    90       | 0
>  3   | 2 | 2018-12-08 |    49       | 2

表：other_loan

> oid | id| amount   |  date
> 1   | 1 | 7928     | 2018-12-10
> 2   | 1 | 750      | 2018-12-10

表：付款

> pid |id | payment_amount  | date
> 1   | 1 |      50         | 2015-12-10
> 2   | 1 |      90         | 2015-12-10
> 3   | 2 |      30         | 2015-12-10

表：客户

> id | Name       |  
> 1  | Robin      | 
> 2  | Cinderella |
> 3  | Leomar     | 

答案

由于您有多个交易，其他贷款金额和每个客户的付款，您不能互相执行表格的直接JOIN，因为它会导致行的复制，从而导致错误的值。相反，在执行SUM之前，我们在客户端基础上JOIN每个表中的所有值。此外，由于某些客户端在每个表中都没有条目，因此必须在结果上使用LEFT JOINs和COALESCE，以便空行不会导致SUM变为NULL。此查询应该为您提供所需的结果：

SELECT c.id, c.name,
       COALESCE(t.transactions, 0) + COALESCE(o.amounts, 0) - COALESCE(p.payments, 0) AS amount
FROM client c
LEFT JOIN (SELECT id, SUM(load_amount) + SUM(additional) AS transactions
           FROM transaction
           GROUP BY id) t on t.id = c.id
LEFT JOIN (SELECT id, SUM(amount) AS amounts
           FROM other_loan
           GROUP BY id) o ON o.id = c.id
LEFT JOIN (SELECT id, SUM(payment_amount) AS payments
           FROM payment
           GROUP BY id) p ON p.id = c.id
GROUP BY c.id

输出（适用于您的样本数据）：

id  name        amount
1   Robin       8718
2   Cinderella  21
3   Leomar      0

Demo on SQLFiddle