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 | |
+-----------------------+--------------+------+-----+---------+-------+
因此,我需要每天查找总订单成本。我可以通过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")
此外,我还需要在交付日期支付剩余的总金额。
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")
并非所有日子都会发生订单而且并非所有交付日都会发生。如果有没有订单的日子,那天的总费用应该显示为零,显示的剩余总金额应该是(total_cost)的总和-advance_amount)当天。
有没有办法可以在一个查询中组合上述两个查询并得到结果?
总结一下特定的日子d: 我需要sum(total_cost)其中ordered_date_time = d, 我需要sum(total_cost -advance_amount),其中deliver_date = d 基本上寻找这样的表:
Date Total Cost Total Delivery Amounts
d 500 2000
d1 0 900
d2 900 0
我尝试使用子查询。问题是它没有显示d1的情况,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
答案 0 :(得分:1)
您可以将两个查询用作派生表,并在日期加入它们。问题是,你需要一个FULL OUTER JOIN,这是MySQL不支持的。因此,您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
对您的查询使用左连接
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应用程序中格式化日期。