Question

我有4个表，相关列总结如下：

customers:
  id
  name

credits:
  id
  customer_id   # ie customers.id
  amount

sales:
  id
  customer_id  # ie customers.id

sales_items:
  id
  sale_id      # ie sales.id
  price
  discount

我的想法是customers列出了我们所有的客户，credits列出了每次付款给我们的信息，sales列出每次他们从我们这里购买的东西（但不是什么东西）他们购买了）sales_items列出了他们在每次销售中购买的所有商品。因此，您可以看到credits和sales都与customers有关，但sales_items仅与sales有关。

作为示例数据集，请考虑：

customers:
  id  |  name
  5   |  Carter

credits:
  id  |  customer_id  |  amount
  1   |  5            |  100

sales:
  id  |  customer_id 
  3   |  5

sales_items:
  id  |  sale_id  |  price  |  discount
  7   |  3        |  5      |  0
  8   |  3        |  0      |  0
  9   |  3        |  10     |  0

我在MySQL中试过这个：

SELECT c.*,
  SUM( cr.amount ) AS paid,
  SUM( i.price + i.discount ) AS bought 
FROM customers AS c
LEFT JOIN sales       AS s  ON s.customer_id = c.id
LEFT JOIN sales_items AS i  ON i.sale_id = s.id
LEFT JOIN credits     AS cr ON cr.customer_id = c.id
WHERE c.id = 5

但它返回：

  id  |  name    |  paid  |  bought
  5   |  Carter  |  300   |  15

如果省略SUM()函数，则返回：

  id  |  name    |  paid  |  bought
  5   |  Carter  |  100   |  5
  5   |  Carter  |  100   |  0
  5   |  Carter  |  100   |  15

因此，对于sales_items中匹配的每条记录，它看起来都会返回一行，但每次都会在amount列中填入credits的相同值。我知道这种情况正在发生，但我不明白为什么会发生这种情况。

所以，有两个问题： 1。正在发生的事情是它通过所有行涂抹了一个值？ 2。我可以在MySQL上使用什么SQL，以便我可以回复：

id  |  name    |  paid  |  bought
5   |  Carter  |  100   |  15

我知道我可以在子查询中将其全部拆分，但是只有连接才能实现它吗？当我解决这个问题时，我希望能够学习关于连接的一两件事。谢谢。

修改：我为此创建了一个SQL小提琴：http://sqlfiddle.com/#!2/0051b/1/0

Answer 1

请试试这个

SELECT c.*,( SELECT SUM( cr.amount ) FROM customer  c INNER JOIN credits cr ON  
 cr.customer_id = c.id WHERE c.id = 5 GROUP BY cr.id ) AS paid
,SUM( i.price + i.discount ) AS bought  
FROM customers AS c INNER JOIN sales s  ON s.customer_id = c.id 
INNER JOIN sales_items i  ON i.sale_id = s.id 
INNER JOIN credits cr ON cr.customer_id = c.id 
WHERE c.id = 5 GROUP BY s.id,cr.id

Answer 2

select distinct (c.id, c.name), sum(i.price+i.discount) AS bought, cr.amount AS paid
from customer c, credits cr, sales s, sales_items i
where s.customer_id = c.id
and i.sale_id = s.id
and cr.customer_id = c.id and c.id = 5
group by c.id, c.name;

我不太确定，但试试这个。使用group by;这肯定是解决方案。

在MySql中加入4个表时重复的数据

2 个答案: