php - SQL SUM() 不同表不同列

Question

分类表

mysql> SELECT * FROM cats;
+------+-----------+
| c_id | c_name    |
+------+-----------+
|    1 | cats 1    |
|    2 | cats 2    |
|    3 | cats 3    |
+------+-----------+

元表

mysql> SELECT * FROM meta;
+------+------+----------+-------------+-------+
| m_id | c_id | name     | description | costs |
+------+------+----------+-------------+-------+
|    1 |    1 | Abhijit1 | description | 100   |
|    2 |    1 | Abhijit2 | description | 200   |
|    3 |    2 | Abhijit3 | description | 500   |
|    4 |    3 | Abhijit4 | description | 800   |
+------+------+----------+-------------+-------+

meta和cats表共同的是c_id元表猫 c_id(1) 元表有 2 (Abhijit1,Abhijit2) 行，m_id(1,2)

交易表

mysql> SELECT * FROM transactions;
+------+------+------------+--------+
| t_id | m_id | date       | amount |
+------+------+------------+--------+
|    1 |    1 | 2016-02-01 | 50     |
|    2 |    1 | 2016-02-06 | 50     |
|    3 |    3 | 2016-02-15 | 400    |
|    4 |    4 | 2016-02-19 | 150    |
+------+------+------------+--------+

事务和元表共同的是m_idm_id 1 的事务有 2 行 t_id(1,2)这个表主要是支付金额和日期

我想对每个类别的所有成本(来自元表)和金额(来自交易表)求和()。

表与

cats.c_id
    |
    |-----> meta.c_id
    |-----> meta.m_id
                |-----> transactions.m_id

这是错误的。猫 id 1 的成本是 300，但在这里我得到了 400

I Want Get Return From Query Like This:
+------+-----------+--------------+---------------+
| c_id | c_name    | SUM(m.costs) | SUM(t.amount) |
+------+-----------+--------------+---------------+
|    1 | cats 1    |          300 |           100 |
|    2 | cats 2    |          500 |           400 |
|    3 | cats 3    |          800 |           150 |
+------+-----------+--------------+---------------+

这里 SUM(m.costs) 都是 Cost For a category 而 SUM(t.amount) 都是 paid for a category

请帮助我或任何更好的表管理想法。

Answer 1

问题出在您加入交易以获得SUM(t.amount)。因此，如果单个 meta 记录与两个 transactions 记录相关联，则 costs 值会被计算两次。

您可以使用相关子查询来计算 SUM(t.amount) 来解决这个问题:

SELECT c.c_id, 
       c.c_name, 
       SUM(m.costs),
       (SELECT SUM(t.amount)
        FROM transactions AS t
        WHERE m.m_id = t.m_id)
FROM cats AS c
LEFT JOIN meta AS m ON c.c_id = m.c_id
GROUP BY c.c_id, c.c_name

输出:

c_id c_name  SUM(m.costs)   SUM(t.amount) 
-----------------------------------------
1    cats 1  300            100
2    cats 2  500            400
3    cats 3  800            150

Demo here