类别表
mysql> SELECT * FROM cats;
+------+------+-----------+
| c_id | p_id | c_name |
+------+------+-----------+
| 1 | 1 | cats 1 |
| 2 | 1 | cats 2 |
| 3 | 1 | cats 3 |
+------+------+-----------+
元表
mysql> SELECT * FROM meta;
+------+------+------+---------+-------------+-------+
| m_id | p_id | c_id | name | description | costs |
+------+------+------+---------+-------------+-------+
| 1 | 1 | 1 | Abhijit | description | 100 |
| 2 | 1 | 1 | Abhijit | description | 200 |
| 3 | 1 | 2 | Abhiji2 | description | 500 |
+------+------+------+---------+-------------+-------+
交易表
mysql> SELECT * FROM transactions;
+------+------+------+---------------------+--------+
| t_id | p_id | m_id | date | amount |
+------+------+------+---------------------+--------+
| 1 | 1 | 1 | 2016-02-16 11:17:06 | 50 |
| 2 | 1 | 1 | 2016-02-16 11:17:06 | 50 |
| 3 | 1 | 2 | 2016-02-16 11:17:06 | 50 |
| 4 | 1 | 2 | 2016-02-16 11:17:06 | 150 |
+------+------+------+---------------------+--------+
我想对每个类别成本(来自元表)和金额(来自交易表)求和。
我使用:
mysql> SELECT c.*, SUM(t.amount), SUM(m.costs)
FROM cats c
LEFT JOIN meta m ON m.c_id=c.c_id
LEFT JOIN transactions t ON t.m_id=m.m_id
GROUP BY c.c_id;
+------+------+-----------+--------+---------------+--------------+
| c_id | p_id | c_name | add_by | SUM(t.amount) | SUM(m.costs) |
+------+------+-----------+--------+---------------+--------------+
| 1 | 1 | Abhijit | 1 | 100 | 400 |
| 2 | 1 | Abhiji2 | 1 | 200 | 500 |
+------+------+-----------+--------+---------------+--------------+
这是错误的。 ID 1 的猫的成本是 300,但这里我得到 400
我想从这样的查询中获得返回:
+------+------+-----------+--------+---------------+--------------+
| c_id | p_id | c_name | add_by | SUM(t.amount) | SUM(m.costs) |
+------+------+-----------+--------+---------------+--------------+
| 1 | 1 | Abhijit | 1 | 100 | 300 |
| 2 | 1 | Abhiji2 | 1 | 200 | 500 |
+------+------+-----------+--------+---------------+--------------+
最佳答案
我认为您的 JOIN
条件之一存在拼写错误(或错误)。我认为您最初的查询是这样的:
SELECT c.*, SUM(t.amount), SUM(m.costs)
FROM cats c
LEFT JOIN meta m ON m.c_id = c.c_id
LEFT JOIN transactions t ON t.m_id = m.c_id
GROUP BY c.c_id;
仔细注意ON t.m_id = m.c_id
,这与您的预期输出一致。无论如何,我对您的查询进行了如下修改:
SELECT c.c_id, c.p_id, c.c_name, t2.transactionCosts, t1.metaCosts
FROM cats c
LEFT JOIN
(
SELECT c_id, SUM(costs) AS metaCosts
FROM meta
GROUP BY c_id
) t1
ON c.c_id = t1.c_id
LEFT JOIN
(
SELECT m_id, SUM(amount) AS transactionCosts
FROM transactions
GROUP BY m_id
) t2
ON c.c_id = t2.m_id
WHERE t2.transactionCosts IS NOT NULL OR t1.metaCosts IS NOT NULL;
第一个子查询计算每个 c_id
的元总计,第二个子查询计算每个 m_id
的交易总计。然后将这些结果与 cats
表连接在一起以获得最终结果。
点击下面的链接即可运行演示:
关于php - Sql sum() 来自不同表的列,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/35424640/