解体思路

1. 先group by city,driver_id求出一个新表格t1，即相同城市相同司机的平均评分，平均接***均里程.
2. 在t1表格基础上，打印出所有的行，并且新添一列进行从大到小排序，又生成新的表格t2，利用窗口函数rank（）对平均评分从大到小排序，相同分数是同样的排名，即第一有很多也可以同时输出
3. 在t2基础上，打印出除排序列以外的所有列就好啦

select	
	city,
    driver_id,
    avg_grade,
    avg_order_num,
    avg_mileage
from
(select 
	*,
    rank() over (partition by city order by avg_grade desc) as paixu
from
(select 
	city,
    driver_id,
    round(sum(grade)/sum(if(grade>0,1,0)),1) as avg_grade,
    round(count(order_time)/count(distinct date(order_time)),1) as avg_order_num,
    round(sum(mileage)/count(distinct date(order_time)),3) as avg_mileage
from tb_get_car_order tgco
join tb_get_car_record tfcr
on tgco.order_id=tfcr.order_id and tgco.uid=tfcr.uid
group by city,driver_id
order by avg_grade desc) t1)t2
where paixu=1
order by avg_order_num