mysql - 无法弄清楚自连接查询

Question

我有一个具有以下结构的表。

+-----------------------+--------------+------+-----+---------+-------+
| Field                 | Type         | Null | Key | Default | Extra |
+-----------------------+--------------+------+-----+---------+-------+
| linq_order_num        | char(32)     | NO   | PRI | NULL    |       |
| order_status_id       | int(11)      | YES  | MUL | NULL    |       |
| order_id              | varchar(100) | YES  |     | NULL    |       |
| item_name             | varchar(120) | YES  |     | NULL    |       |
| item_cost             | float        | YES  |     | NULL    |       |
| custmer_id            | int(11)      | YES  | MUL | NULL    |       |
| order_date_time       | datetime     | YES  |     | NULL    |       |
| order_category        | varchar(120) | YES  |     | NULL    |       |
| ordered_by            | int(11)      | YES  | MUL | NULL    |       |
| linq_shipping_cost    | float        | YES  |     | NULL    |       |
| website_shipping_cost | float        | YES  |     | NULL    |       |
| total_cost            | float        | YES  |     | NULL    |       |
| advance_amount        | float        | YES  |     | NULL    |       |
| website               | varchar(120) | YES  |     | NULL    |       |
| other                 | varchar(120) | YES  |     | NULL    |       |
| rvn                   | int(11)      | YES  |     | NULL    |       |
| received_date         | datetime     | YES  |     | NULL    |       |
| delivered_date        | datetime     | YES  |     | NULL    |       |
| store_id              | int(11)      | YES  | MUL | NULL    |       |
+-----------------------+--------------+------+-----+---------+-------+

所以我每天都需要找到总订单成本。我可以使用此查询来获取它。

select sum(total_cost), date_format(order_date_time,"%Y-%m-%d") from 
       order_item group by date_format(order_date_time,"%Y-%m-%d")

我还需要在交货日期支付的剩余总金额。

select sum(total_cost-advance_amount),date_format(delivered_date,"%Y-%m-%d")
       from order_item group by date_format(delivered_date,"%Y-%m-%d")

并非所有的日子都会发生订单，也不是所有的日子都会发生交付。如果有一天没有订单，那一天的总成本应显示为零，显示的总剩余金额应为(total_cost -advance_amount)当天。

有没有一种方法可以将上述两个查询组合在一个查询中并获得结果？

因此总结特定日期 d:我需要 sum(total_cost) where ordered_date_time = d ,我需要 sum(total_cost -advance_amount) where delivered_date = d本质上是在寻找这样的表格:

Date            Total Cost          Total Delivery Amounts
d                 500                     2000
d1                0                       900
d2                900                     0 

我尝试使用子查询。问题是它不显示 d1 的案例，其中当天的总成本为 0。

查询:

select
    date_format(order_date_time,"%Y-%m-%d") date,
    sum(total_cost) total,
    sum(advance_amount) advance_amount,
    IFNULL( (select sum(total_cost-advance_amount)
from order_item a 
where date_format(a.delivered_date,"%Y-%m-%d") = date_format(d.order_date_time,"%Y-%m-%d") ),0 ) delivery_amount
from order_item d
group by date_format(order_date_time,"%Y-%m-%d"), delivery_amount

Answer 1

您可以将两个查询用作派生表并按日期加入它们。问题是，您需要一个 FULL OUTER JOIN，MySQL 不支持它。所以你首先需要从两列中提取所有日期

select date(order_date_time) as d from order_item
union
select date(delivered_date)  as d from order_item

在你的查询中使用左连接

select
    dates.dt,
    coalesce(tc.total_cost, 0),
    coalesce(tm.total_remaining, 0)
from (
    select date(order_date_time) as dt from order_item
    union
    select date(delivered_date)  as dt from order_item
) dates
left join (
     select sum(total_cost) as total_cost, date(order_date_time) as dt
     from order_item
     group by dt
) tc using(dt)
left join (
     select sum(total_cost-advance_amount) as total_remaining, date(delivered_date)
     from order_item
     group by dt
) tm using(dt)

我还将 date_format(..) 替换为 date(..)。您可以在外部选择或您的应用程序中格式化日期。