题解 | 每个城市中评分最高的司机信息

select city,driver_id,avg_grade,avg_order_num,avg_mileage
from (
select city,driver_id,round(avg(grade),1)  avg_grade,round(count(distinct order_id)/count(distinct date(order_time)),1)  avg_order_num,round(sum(mileage)/count(distinct date(order_time)),3)  avg_mileage,rank() over(partition by city order by round(avg(grade),1) desc) rk
from tb_get_car_order join tb_get_car_record using(order_id)
group by city,driver_id
) t
where rk = 1
order by  city desc,avg_mileage asc

# 不难，但要经常做

# 这次卡住我的点：1.窗口函数不熟悉 2.对date函数不熟悉 3.题目太绕了，读题读半天。