Question

我有这两张桌子：

Table members
| id | name    |
----------------
| 01 | John    |
| 02 | Michael |
| 03 | Richard |

和

Table payments
| id  | eur  | name    | date       |
-------------------------------------
| 238 | 9.95 | John    | 1323377751 |
| 233 | 9.95 | Michael | 1397864609 |
| 220 | 9.95 | Michael | 1397852739 |
| 215 | 9.95 | John    | 1397852719 |
|   2 | 9.95 | Richard | 1323377751 | // This payment was made more than 24h ago

贝娄是我的疑问：

select id, m.name, sum(eur) as total 
from mambers as m
left join payments as p
on m.name = p.name
where p.date >= unix_timestamp(current_timestamp - interval 24 hour)
group by m.name

并且返回：

| John    | 19.90 |
| Michael | 19.90 |

但我需要在过去的24小时内同时包含没有付款的会员：

| John    | 19.90 |
| Michael | 19.90 |
| Richard | 00.00 |

Answer 1

尝试查询时发生了两次错误。

表名mambers不存在。
不明确的列名id。

请改为尝试：

select m.id, m.name, IFNULL(sum(eur),'00.00') as total 
from members as m
left join payments as p
on m.name = p.name and p.date >= unix_timestamp(current_timestamp - interval 24 hour)
group by m.name

结果：

ID  NAME      TOTAL
1   John      10
2   Michael   20
3   Richard   00.00

请参阅SQL Fiddle中的结果。

MySql group by和include

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