Android 为什么使用SQL；sum"；返回一个假值

我有3张表格，我想返回 客户姓名和总额每个存款和信贷金额的总和

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

我是通过这个查询来完成的=>

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，总仓库和总信用的结果不正确 但当我像这样单独做时=>

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

总仓库和总信用的结果是正确的 我不知道错误在哪里。

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;

在加入表之前执行聚合：

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;

---

第一个查询完全错误，连接将在结果中增加行数。例如，一名客户、3名他的信用卡和5名他的存款：

从客户返回1行中选择

客户内部加入积分返回3行

客户内部加入积分内部加入存款返回15行

那不是你想要的。不使用SUM和groupby执行查询，您将看到它

这就是您想要的（简化，未经测试）：

---

顺便说一句，如果客户既没有存款也没有信用卡，则需要使用左联接

第一次查询完全错误，联接将在结果中增加行数。例如，一名客户、3名他的信用卡和5名他的存款：

从客户返回1行中选择

客户内部加入积分返回3行

客户内部加入积分内部加入存款返回15行

那不是你想要的。不使用SUM和groupby执行查询，您将看到它

这就是您想要的（简化，未经测试）：

---

顺便说一句，如果客户既没有存款也没有信用卡，则需要左键联接

谢谢大家'

我尝试了所有的考试，但它不起作用

我终于做到了它的工作。

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

---

非常感谢大家'

我尝试了所有的考试，但它不起作用

我终于做到了它的工作。

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

---

我看不出你的第一个查询不起作用的简单原因。我可以看到您使用的是

group by

customers.id而不是name。我想您执行此查询时应该会出错。我看不出第一个查询不起作用的简单原因。我可以看到您使用的是

groupby

customers.id而不是name。我想执行此操作时应该会出错query@Birante . . . 嗯，那个评论毫无用处，对解决你的问题毫无帮助。@Birante。好吧，那评论完全没有用，对解决你的问题没有任何帮助。