题解 | #国庆期间近7日日均取消订单量#

select * from (select
    order_time,
    round(
        sum(finish_order_num) over (
            order by
                order_time rows 6 preceding
        ) / 7,
        2
    ) as finish_rate,
    round(
        sum(unfinish_num) over (
            order by
                order_time rows 6 preceding
        ) / 7,
        2
    ) as unfinish_rate
from
    (
        select
            date (order_time) as order_time,
            sum(if (fare is not null, 1, 0)) as finish_order_num,
            sum(if (fare is null, 1, 0)) as unfinish_num
        from
            tb_get_car_order as tgco
            left join tb_get_car_record as tgcr on tgcr.order_id = tgco.order_id
        group by
            date (order_time)
    ) as t1) as t2
where
    order_time between '2021-10-1' and '2021-10-3'

先算完成订单数，未完成订单数，并按照天来分类

计算每天的完成率，未完成率，使用窗口函数计算preceding获取当前行前n行

sum(unfinish_num) over (
            order by
                order_time rows 6 preceding
        )

选择1号到3号