题解 | 国庆期间近7日日均取消订单量

啊啊啊好长，但是这样的是一步一步想出来的，很清晰详细
select t2.date1 as dt,round(finished/7,2),round(canceled/7,2)
from
(    select date1,count(distinct order_id) finished
    from
    (    select c.order_id,date(b.finish_time) as date1,
        date(c.finish_time) as date2
        from tb_get_car_order b,tb_get_car_order c
        where datediff(b.finish_time,c.finish_time)>=0
        and datediff(b.finish_time,c.finish_time)<=6
        and c.start_time is not null
        order by date1,date2) t1
    where day(date1)=1
    or day(date1)=2
    or day(date1)=3
    group by date1) t2,
(    select date1,count(distinct order_id) as canceled
    from
    (    select c.order_id,date(b.finish_time) as date1,
        date(c.finish_time) as date2
        from tb_get_car_order b,tb_get_car_order c
        where datediff(b.finish_time,c.finish_time)>=0
        and datediff(b.finish_time,c.finish_time)<=6
        and c.start_time is  null
        order by date1,date2) t1
    where day(date1)=1
    or day(date1)=2
    or day(date1)=3
    group by date1) t3
where t2.date1=t3.date1