Question

我搜索过但没找到任何对我有用的东西。我有3张桌子。一个是客户端ID名称查找，支付表和一个invloice表。每个客户都有1个或多个发票记录，其中包含欠款。客户在付款表中有0个或更多记录。付款可能针对多张发票，或者一张发票可能有多笔付款。我想得到每个客户的欠款。例如，为每个客户添加所有发票，并减去每个客户的所有付款。

Invoices               Payments                   Clients
--------------------   ------------------       -----------
|Id|ClientId|Amount|   |Id|ClientId|Paid|       |Id|  Name   |
|1 |   3    | 200  |   | 1|  2     | 10 |       | 1|   Bob   |
|2 |   2    |  10  |   | 2|  1     | 20 |       | 2|  Tom    |
|3 |   4    | 100  |   | 3|  2     | 100|       | 3| John    |
|4 |   2    | 240  |   | 4|  3     | 240|       | 4| Peter   |
| 5|   1    |  20  |   ------------------       --------------
--------------------

Expected Result
----------------------
|Name |Spent|Paid|Total|
|John | 200 | 240|  -40|
| Tom | 250 | 110| 140 |
|Peter| 100 |   0| 100 |
| Bob |  20 |  20|   0 |
------------------------

Answer 1

请考虑以下事项：

DROP TABLE IF EXISTS invoices;

CREATE TABLE invoices
(id INT NOT NULL AUTO_INCREMENT PRIMARY KEY
,client_id INT NOT NULL
,amount DECIMAL(5,2) NOT NULL
);

INSERT INTO invoices VALUES
(1,3,200),
(2,2, 10),
(3,4,100),
(4,2,240),
(5,1, 20);

DROP TABLE IF EXISTS payments;

CREATE TABLE payments
(id INT NOT NULL AUTO_INCREMENT PRIMARY KEY
,client_id INT NOT NULL
,paid DECIMAL(5,2)
);

INSERT INTO payments VALUES
(1,2,10 ), 
(2,1,20 ), 
(3,2,100), 
(4,3,240);

DROP TABLE IF EXISTS clients;

CREATE TABLE clients
(id INT NOT NULL AUTO_INCREMENT PRIMARY KEY
,name VARCHAR(12) NOT NULL
);

INSERT INTO clients VALUES
(1,'Bob'),
(2,'Tom'),
(3,'John'),
(4,'Peter');

SELECT c.name
     , SUM(CASE WHEN x.type = 'invoice' THEN x.amount ELSE 0 END) spent
     , SUM(CASE WHEN x.type = 'payment' THEN x.amount ELSE 0 END) paid
     , SUM(CASE WHEN x.type = 'invoice' THEN x.amount ELSE x.amount * -1 END) total
  FROM clients c
  JOIN 
     ( SELECT client_id
            , 'invoice' type
            , amount
         FROM invoices
        UNION
       SELECT client_id
            , 'payment'
            , paid
         FROM payments
     ) x
    ON x.client_id = c.id
 GROUP
    BY c.id;

+-------+--------+--------+--------+
| name  | spent  | paid   | total  |
+-------+--------+--------+--------+
| Bob   |  20.00 |  20.00 |   0.00 |
| Tom   | 250.00 | 110.00 | 140.00 |
| John  | 200.00 | 240.00 | -40.00 |
| Peter | 100.00 |   0.00 | 100.00 |
+-------+--------+--------+--------+

要了解其工作原理，请尝试仅执行子查询位SELECT client_id... FROM payments

Mysql从发票中减去付款

1 个答案: