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