MySql中带有分组语句的两个不同表的总和差异[重复]

Question

我正在创建一个带有销售/付款选项的客户门户，其中所有销售都在一个表中，所有付款都在另一个表中。对于客户状态表，我需要得到每个客户的所有付款和每个客户的所有销售加上它们之间的差额的总和。

示例表：

表clients

====================================
id  client_name
====================================
1   School "Mike Jordan"
2   Car wash "Blah blah"

表purchasing

===================================
id  client_id  amount
===================================
1   1          1000
2   1          500
3   2          800
4   2          800

表payments

===================================
id  client_id  amount
1   1          300
2   2          200
3   2          200

我尝试过内部连接和分组，但这并没有正确地得到总和，总是更大的，就像它在计算我不想要的东西（没有总数，只是试图得到总和：

select clients.`client_name `,
sum(purchasing.amount) as SOLD,
sum(payments.amount) as PAID 
from clients
inner join payments on payments.client_id=clients.id
inner join purchasing on purchasing.client_id=clients.id
group by clients.id

我得到不同结果的地方（都更大）。

client_name           SOLD  PAID
================================
School "Mike Jordan"  1500  600
Car wash "Blah blah"  3200  800

如果我尝试按 payment.id 和/或 purchase.id 添加分组，则得到两个以上的结果...我知道存在分组问题，但我真的不知道它在哪里。

我想添加新列，这只是 SOLD 和 PAID 之间的区别，但是服务器生成错误消息，表明 SOLD 和 PAID 不在字段列表中。

知道如何通过单个查询来完成此任务吗？

Answer 1

你可以试试这个。使用UNION ALL 组合purchasing 和payments 表然后join 到clients。

下一步做SUM和group by

create table clients(
   id int,
   client_name varchar(50)
);

insert into clients values (1,'School "Mike Jordan"');
insert into clients values (2,'Car wash "Blah blah"');


create table purchasing(
  id int,
   client_id int,
   amount int
);


insert into purchasing values (1,1,1000);
insert into purchasing values (2,1,500);
insert into purchasing values (3,2,800);
insert into purchasing values (4,2,800);

create table payments(
  id int,
   client_id int,
   amount int
);
insert into payments values (1,1,300);
insert into payments values (2,2,200);
insert into payments values (3,2,200);

查询 1：

select c.client_name,
       sum(SOLD) as SOLD,
        sum(PAID) as PAID
from (
    SELECT client_id,amount as SOLD,0 as PAID  
    FROM purchasing
    UNION ALL 
    SELECT client_id,0,amount   
    FROM payments
)t1 INNER JOIN clients c on t1.client_id = c.id
group by c.client_name

Results：

|          client_name | SOLD | PAID |
|----------------------|------|------|
| Car wash "Blah blah" | 1600 |  400 |
| School "Mike Jordan" | 1500 |  300 |

Answer 2

在进行联接之前，您需要汇总表。

SELECT clients.`client_name`
    ,purchasing_sum.SOLD
    ,payments_sum.PAID
FROM clients
INNER JOIN (
    SELECT client_id
        ,sum(amount) as PAID
    FROM payments
    GROUP BY client_id
    ) payments_sum ON payments_sum.client_id = clients.id
INNER JOIN (
    SELECT client_id
        ,sum(amount) as SOLD
    FROM purchasing
    GROUP BY client_id
    ) purchasing_sum ON purchasing_sum.client_id = clients.id
GROUP BY clients.id