Question

select sum(t.amount) user_amount_sum 
  from users u
  join investor_transactions t
    on u.id = t.investor_id
 group 
    by u.id;

为每个用户返回user_amount_sum值40。（正确）

从中选择总和（investor_transactions.amount）AS user_amount_sum {{1}上的users 内部加入merchant_user。merchant_user = user_id上的users。id 内部联接investor_transactions。users = id。investor_transactions按investor_id分组。users;

但是现在为每个用户返回user_amount_sum值200。（是其他用户的5倍，merchant_user表中有5行）

注意1：两个查询之间的唯一区别是

id

NB 2：我要加入merchant_user表以进行过滤。

解决此问题的最佳做法是什么？？

Answer 1

SUM和所有聚合函数对（FROM ... WHERE的）中间结果进行操作；任何时候加入多个1：N关系，您都会遇到此问题。由于您仅使用其他联接进行过滤，因此您应该能够使用以下替代方法：

SELECT SUM(investor_transactions.amount) AS user_amount_sum 
FROM users 
INNER JOIN investor_transactions ON users.id = investor_transactions.investor_id 
INNER JOIN (SELECT DISTINCT user_id FROM merchant_user) AS mu ON mu.user_id = users.id 
GROUP BY users.id;

或

SELECT SUM(investor_transactions.amount) AS user_amount_sum 
FROM users 
INNER JOIN investor_transactions ON users.id = investor_transactions.investor_id 
WHERE users.id IN (SELECT DISTINCT user_id FROM merchant_user) 
GROUP BY users.id;

或

SELECT SUM(investor_transactions.amount) AS user_amount_sum 
FROM users AS u
INNER JOIN investor_transactions ON u.id = investor_transactions.investor_id 
WHERE EXISTS (SELECT * FROM merchant_user AS mu WHERE mu.user_id = u.id) 
GROUP BY users.id;

编辑：尽管我现在看了一下，但是您可能会跳过users表，因为您只使用了它的id字段，就像这样...

SELECT SUM(t.amount) AS user_amount_sum 
FROM investor_transactions AS t
INNER JOIN (SELECT DISTINCT user_id FROM merchant_user) AS mu ON t.user_id = mu.user_id
GROUP BY t.user_id;

，与上述其他替代方法类似。

当主表与另一个manyToOne关系表联接时，该字段的SUM相乘

1 个答案: