题解 | 国庆期间近7日日均取消订单量（最傻瓜解决）

select
'2021-10-01',
round(count(mileage) / 7,2),
round((count(*)-count(mileage)) / 7,2)
from tb_get_car_record a left join tb_get_car_order b using(order_id)
where date(finish_time) > date_sub("2021-10-01",interval 7 day)
and date(finish_time) <= "2021-10-01"

union

select
'2021-10-02',
round(count(mileage) / 7,2),
round((count(*)-count(mileage)) / 7,2)
from tb_get_car_record a left join tb_get_car_order b using(order_id)
where date(finish_time) > date_sub("2021-10-02",interval 7 day)
and date(finish_time) <= "2021-10-02"

union

select
'2021-10-03',
round(count(mileage) / 7,2),
round((count(*)-count(mileage)) / 7,2)
from tb_get_car_record a left join tb_get_car_order b using(order_id)
where date(finish_time) > date_sub("2021-10-03",interval 7 day)
and date(finish_time) <= "2021-10-03"

当然了，不建议大家用这个方法做