Question

我有两个表，如下所示：

mysql> select*from receipt;                                                                                
+----+------------+-------+---------------------+
| id | receipt_id | money | created_at          |
+----+------------+-------+---------------------+
|  1 | receipt_3  |   100 | 2018-10-30 00:00:00 |
|  2 | receipt_4  |   200 | 2018-10-29 00:00:00 |
|  3 | receipt_5  |   300 | 2018-10-31 00:00:00 |
+----+------------+-------+---------------------+
3 rows in set (0.00 sec)

mysql> select*from material;
+----+------------+---------------------+
| id | receipt_id | created_at          |
+----+------------+---------------------+
|  1 | receipt_3  | 2018-10-30 00:00:00 |
|  2 | receipt_3  | 2018-10-30 00:00:00 |
|  3 | receipt_5  | 2018-10-31 00:00:00 |
+----+------------+---------------------+
3 rows in set (0.00 sec)

我得到的结果是这样的：“从material的{{1}}内部联接created_at的{{1}}中选择{sum（money），receipt。material。 receipt = receipt_id。material按receipt_id分组。material；”

created_at

但是我想通过忽略相同的记录来获得它：

+------------+---------------------+
| sum(money) | created_at          |
+------------+---------------------+
|        200 | 2018-10-30 00:00:00 |
|        300 | 2018-10-31 00:00:00 |
+------------+---------------------+
2 rows in set (0.00 sec)

想了很久但没有答案，请帮助，非常感谢。

Answer 1

只需使用sum(distinct money)即可：

select sum(distinct money) as sum_money,
       m.created_at 
  from receipt r join material m 
    on r.receipt_id=m.receipt_id 
 group by m.created_at;

+------------+---------------------+
| sum_money  | created_at          |
+------------+---------------------+
|        100 | 2018-10-30 00:00:00 |
|        300 | 2018-10-31 00:00:00 |
+------------+---------------------+

DB-Fiddle Demo

Answer 2

您可以尝试在子查询中将DISTINCT用于material，而忽略与JOIN相同的记录

SELECT sum(r.money),t1.created_at
FROM (
    select DISTINCT receipt_id,created_at   
    from material
)t1 JOIN receipt r on t1.receipt_id = r.receipt_id  
GROUP BY t1.created_at

合并两个表的唯一列

2 个答案: