Question

我有3个表格，我想退货每个存款和信用金额的客户名称和总总和。

deposit        customers     credit
id             id            id
d_amount                     c_amount
customer_id    name          customer_id
                             type(credit,etc)

我是通过此查询=＆gt;

来完成的

SELECT  customers.name ,
sum(deposit.d_amount) as total_depot,
sum(credit.c_amount) as total_credit
from customers
inner join  deposit on deposit.customer_id  = customers.id
inner join  credit on credit.customer_id  = customers.id
and credit.type='credit'
group by customers.id order by customers.name asc

不幸的是， total_depot 和 total_credit 的结果不正确 但是当我单独这样做时，就像这样=＆gt;

SELECT  customers.name , sum(deposit.d_amount) as total_depot
from customers
inner join  deposit on deposit.customer_id  = customers.id
group by customers.id order by customers.name asc


SELECT  customers.name , sum(credit.d_amount) as total_credit
from customers
inner join  credit on credit.customer_id  = customers.id
and credit.type='credit'
group by customers.id order by customers.name asc

total_depot 和 total_credit 的结果是正确的我不知道错误在哪里。

Answer 1

第一个查询完全错误，JOINS会在结果中乘以行。一个客户的例子，他的3个学分和5个他的存款：

从客户中选择返回1行

客户INNER JOIN积分返回3行

客户INNER JOIN积分INNER JOIN存款返回15行

这不是你想要的。执行没有SUM和GROUP BY的查询，你会看到它。

这就是你想要的（简化，未经测试）：

select customers.id, cr.amount, dep.amount
  from customers
  left join (select customer_id, sum(credit.d_amount) as amount from credits group by customer_id) cr on cr.customer_id=customers.id
  left join (select customer_id, sum(deposits.d_amount) as amount from deposits group by customer_id) dep on dep.customer_id=customers.id

顺便说一句。当客户没有BOTH存款和信用时，需要左联接

Answer 2

在加入表之前进行聚合：

select c.name, d.total_deposit, cr.total_credit from customers c join (select d.customer_id, sum(d.d_amount) as total_deposit from deposit d group by d.customer_id ) d on d.customer_id = c.id join (select c.customer_id, sum(c.c_amount) as total_credit from credit c where c.type = 'credit' group by c.customer_id ) cr on cr.customer_id = c.id order by c.name asc;

Answer 3

非常感谢每个人'

我尝试了所有这些考试，但它不起作用

我终于做到了这一点。 其工作。

select customers.name,  total_depot , total_credit
from customers 
left join (select customer_id , d_amount , sum(deposit.d_amount) as total_deposit from deposit group by customer_id)  d on d.customers_id = customers.id 
left join (select customer_id , c_amount , sum(credit.c_amount) as total_fact from credit where credit.type='credit' group by customer_id)  c on c.id = customers.id 
group by customers.name

为什么SQL＆＃34;总和＆＃34;返回false值

3 个答案: