Mysql 使用具有内部联接的group by时获取错误的和
关于这个问题(),我有两个表Mysql 使用具有内部联接的group by时获取错误的和,mysql,sql,database,Mysql,Sql,Database,关于这个问题(),我有两个表orders和order\u items。我需要按天对结果进行分组。但是我还需要从另一个表中获得每天使用的能量的总和。当我尝试使用联接时,每天的顺序都会出错(它们没有被求和)。不知道我做错了什么 我希望每天都能得到 当天创建的所有订单使用的order\u项目的总和 订单的总和。当天创建的所有订单的订单总和 在创建的和订单金额,对应于当天创建的最新订单 这是我的订单表格 +----+-----------+---------+--------------------
orders
和order\u items
。我需要按天对结果进行分组。但是我还需要从另一个表中获得每天使用的能量的总和。当我尝试使用联接时,每天的顺序都会出错(它们没有被求和)。不知道我做错了什么
我希望每天都能得到
- 当天创建的所有订单使用的
order\u项目的总和
- 订单的总和。当天创建的所有订单的订单总和
- 在
创建的和
订单金额,对应于当天创建的最新订单
这是我的订单
表格
+----+-----------+---------+---------------------+
| 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 |
+----+-----------+---------+---------------------+
+----+----------+-------------+---------------------+
| 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 |
+----+----------+-------------+---------------------+
这是我的订单项目表
+----+-----------+---------+---------------------+
| 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 |
+----+-----------+---------+---------------------+
+----+----------+-------------+---------------------+
| 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,但是我得到了错误的结果,顺序和计算不正确。它显示的内容与上次订单和相同
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)
)
这是一把小提琴:
始终根据表之间的关系(在本例中,orders.id与order\u items.order\u id)将表连接在一起,然后分组。为避免加入时多个订单项目的订单金额重复,请首先按订单id对订单项目进行分组
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)
从这一点开始,您可以使用max(o.created_at)再次对订单执行联接,以获得上一个订单的订单和。
这个故事的寓意是:关注您的粒度。始终根据表之间的关系(在本例中,orders.id与order\u items.order\u id)将表连接在一起,然后分组。为避免加入时多个订单项目的订单金额重复,请首先按订单id对订单项目进行分组
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)
从这一点开始,您可以使用max(o.created_at)再次对订单执行联接,以获得上一个订单的订单和。
这个故事的寓意是:关注您的粒度。您的问题是您正在从订单
中进行选择,实际上您需要按日期聚合订单。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\u order\u sum
,一旦我们知道最后一个订单日期,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;
您的问题是,您正在从订单
中进行选择,实际上您希望按日期聚合订单
。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\u order\u sum
,一旦我们知道最后一个订单日期,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;
你们所要求的和你们在输出中向我们展示的是不相关的。假设是打字错误:
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;
您的要求与您在输出中向我们展示的内容不相关。假设是打字错误:
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;
那么,在
创建的被视为订单日期?这适用于订单。在创建和订单项目。在创建?@ThorstenKettner是的,这是正确的如果你想分组,为什么不使用分组依据
?你使用的是什么版本的MySQL?(获取最后的订单金额可能有点棘手,使用MySQL 8比以前的版本更容易解决。)@ThorstenKettner我使用的是5.5So,created_at
被认为是订单日期吗?这适用于订单。在创建和订单项目。在创建?@ThorstenKettner是的,这是正确的如果你想分组,为什么不使用分组依据
?你使用的是什么版本的MySQL?(获取最后的订单金额可能有点棘手,使用MySQL 8比以前的版本更容易解决。)@ThorstenKettner我正在使用5.5谢谢,我想这就是我想要的。只需要找出最后一个连接就可以获得订单数量到目前为止,运气不好,我尝试使用order by created_at并限制一个,但无法获得最后一个订单数量。整个过程将是一个子查询(比如X),然后您必须从X中选择*左连接订单o on max_created_at=o.created_at。再次注意复制,在订单中创建的_总是唯一的吗?谢谢,我想这就是我要找的。只需要找出最后一个连接就可以获得订单数量到目前为止,运气不好,我尝试使用order by created_at并限制一个,但无法获得最后一个订单数量。整个过程将是一个子查询(比如X),然后您必须从X中选择*左连接订单o on max_created_at=o.created_at。再次注意重复,在订单中创建的_是否总是唯一的?