将 group by 与内部联接一起使用时得到错误的总和

Question

关于这个问题 (How to get the sum in a joined table when using group by - getting wrong results) 我有两个表 orders 和 order_items。我需要按天对结果进行分组。但我还需要从另一个表中获取每天的energy_used 的总和。当我尝试使用连接时，我每天都会出错order_sum（它们没有被总结）。不知道我做错了什么。

我想每天都得到

• 当天创建的所有订单的order_items.energy_used 总和
• 当天创建的所有订单的orders.order_sum 总和
• 对应于当天创建的最新order的created_at和order_sum

这是我的orders 表

+----+-----------+---------+---------------------+
| id | order_sum | user_id | created_at          |
+----+-----------+---------+---------------------+
| 1  | 25.13     | 7       | 2020-01-25 09:13:00 |
| 2  | 10.00     | 7       | 2020-01-25 15:23:00 |
| 3  | 14.00     | 5       | 2020-01-26 10:14:00 |
| 4  | 35.00     | 1       | 2020-01-27 11:13:00 |
+----+-----------+---------+---------------------+

这是我的order_items 表

+----+----------+-------------+---------------------+
| id | order_id | energy_used | created_at          |
+----+----------+-------------+---------------------+
| 1  | 1        | 65          | 2020-01-25 09:13:00 |
| 2  | 1        | 12          | 2020-01-25 09:13:00 |
| 3  | 2        | 70          | 2020-01-26 10:14:00 |
| 4  | 2        | 5           | 2020-01-26 10:14:00 |
| 5  | 3        | 0           | 2020-01-27 11:13:00 |
+----+----------+-------------+---------------------+

这是我想要达到的理想结果

+---------------+-----------------+-------------------+---------------------+----------------+
| date_of_month | total_order_sum | total_energy_used | last_order_date     | last_order_sum |
+---------------+-----------------+-------------------+---------------------+----------------+
| 2020-01-25    | 35.13           | 77                | 2020-01-25 09:13:00 | 25.13          |
| 2020-01-26    | 14.00           | 75                | 2020-01-26 10:14:00 | 14.00          |
| 2020-01-27    | 35.00           | 0                 | 2020-01-27 11:13:00 | 35.00          |
+---------------+-----------------+-------------------+---------------------+----------------+

这是我尝试过的查询，但我得到了错误的结果，order_sum 的计算不正确。它与last_order_sum显示相同

select 
    date(o.created_at) date_of_month,
    i.total_energy_used,
    o.created_at last_order_date,
    o.order_sum last_order_sum,
    sum(order_sum) as total_order_sum
from orders o
inner join (
    select date(o1.created_at) date_of_month, sum(i1.energy_used) total_energy_used
    from orders o1
    inner join order_items i1 on o1.id = i1.order_id
    group by date(o1.created_at)
) i on i.date_of_month = date(o.created_at)
where o.created_at = (
    select max(o1.created_at)
    from orders o1
    where date(o1.created_at) = date(o.created_at)
)

这是一个小提琴： https://dbfiddle.uk/?rdbms=mysql_5.6&fiddle=92b8cc2920ad9f7a7cdd56bded5a3bf2

Answer 1

始终根据它们的关系将表连接在一起（在本例中为 orders.id 和 order_items.order_id），然后进行分组。为避免在加入时为多个 order_items 重复 order_sums，首先按 order_id 对 order_items 进行分组。

select 
    date(o.created_at) date_of_month,
    sum(i.total_energy_used),
    max(o.created_at),
    sum(order_sum) as total_order_sum
from orders o
inner join (
    select order_id, sum(total_energy_used) total_energy_used
    from order_items i
    group by order_id
) i on o.id = i.order_id
group by date(o.created_at)

从此时起，您可以使用 max(o.created_at) 再次对订单进行连接，以获得最后一个订单的 order_sum。 故事的寓意：注意你的粒度。

Answer 2

您的问题是您从orders 中进行选择，实际上您希望按日期汇总orders。因此，请从您加入的两个聚合子查询中进行选择。唯一的问题是last_order_sum，一旦我们知道最后的订单日期，我们就可以在进一步的子查询中选择它。

select 
  order_date,
  o.total_order_sum,
  oi.total_energy_used,
  o.last_order_date,
  (
    select order_sum
    from orders last_order
    where lastorder.created_at = o.last_order_date
  ) as last_order_sum
from 
(
  select
    date(created_at) as order_date,
    sum(order_sum) as total_order_sum,
    max(created_date) as last_order_date
  from orders
  group by date(created_at)
) o
inner join 
(
  select
    date(created_at) as order_date,
    sum(energy_used) as total_energy_used
  from order_items
  group by date(created_at)
) oi using(order_date)
order by order_date;

Answer 3

您所询问的内容与您在输出中向我们展示的内容不相关。假设这是一个错字：

select so.dtDay as date_of_month, so.order_sum as total_order_sum,
   eu.energy_used as total_energy_used,
   o.created_at as last_order_date,
   o.order_sum as last_order_sum
from (
select left(created_at,10) as dtDay, sum(order_sum) as order_sum, max(id) as last_insert_id
from orders
group by left(created_at,10)
order by created_at
) so
inner join orders o on o.id = so.last_insert_id
left join (select left(created_at,10) as dtDay, sum(energy_used) as energy_used
from order_items
group by left(created_at,10)) eu on so.dtDay = eu.dtDay;

DBFiddle