sql - postgresql 聚合的聚合(sum of sum)

Question

我的员工有很多销售业绩并且属于各个部门。我想了解一个部门每天的销售额是多少。

为简单起见，假设一名工作人员只属于一个部门。

示例:

部门:

| id | name            |
| 1  | Men's Fashion   |
| 2  | Women's Fashion |

worker :

| id  | name   |
| 1   | Timmy  |
| 2   | Sally  |
| 3   | Johnny |

销售:

| id  |  worker_id | datetime          | amount |
| 1   |  1         | 2013-1-1 08:00:00 |  1     |
| 2   |  1         | 2013-1-1 09:00:00 |  3     |
| 3   |  3         | 2013-1-2 08:00:00 |  8     |

部门员工

| id | worker_id | department_id |
| 1  |   1       |    1          |
| 2  |   2       |    1          |
| 3  |   3       |    2          |

我想得到

|  department     | amount |
| Men's Fashion   | 4      |
| Women's Fashion | 8      |

要获取单个 worker 的总销售额，我可以这样做

SELECT worker_id, SUM(amount) FROM sales
GROUP BY worker_id

如何获取这些金额(每个 worker 的销售总额)并按部门汇总？

Answer 1

不要对总和进行求和，而是通过 Department_employees 表从 sales 加入到部门:

select d.name, sum(s.amount)
from sales s
join department_employees de on de.worker_id = s.worker_id
join departments d on d.id = de.department_id
group by d.name