通过代码

with t as (select    
    driver_id ,
    city,
    count(distinct date(order_time)) days,
    count(*) orders,
    sum(mileage) ways,
    sum(grade) / count(grade) avgs
FROM
    tb_get_car_order
JOIN
    tb_get_car_record
ON
    tb_get_car_order.order_id = tb_get_car_record.order_id
GROUP BY
    driver_id,city)

SELECT
    t.city,
    driver_id,
    round(avgs,1) avg_grade,
    round(orders / days ,1) avg_order_num,
    round(ways / days ,3) avg_mileage
from
    t
join(
select
    city,
    max(avgs) maxs
from
    t
group by 
    city) t2
on
    t.city = t2.city
WHERE
    t.avgs = t2.maxs
ORDER BY
    avg_order_num

思路

请统计每个城市中评分最高的司机平均评分、日均接单量和日均行驶里程数。

注：有多个司机评分并列最高时，都输出。

平均评分和日均接单量保留1位小数，

日均行驶里程数保留3位小数，按日均接单数升序排序。

平均评分 avg(grade)

兼职天数 count(distinct date(order_time)) days,

总接单量 count(driver_id) orders(这里*也可以但是格式问题就id)

总行驶里程数 sum(mileage) ways,

然后就好说了两表连接查出以上数据再分别计算

with t as (select    
    driver_id ,
    city,
    count(distinct date(order_time)) days,
    count(*) orders,
    sum(mileage) ways,
    sum(grade) / count(grade) avgs
FROM
    tb_get_car_order
JOIN
    tb_get_car_record
ON
    tb_get_car_order.order_id = tb_get_car_record.order_id
GROUP BY
    driver_id,city)

有多个司机评分并列最高时，都输出

这个怎么办呢？ 我选择对计算结果出来的表进行外连接max

也就是先查出来每个城市最大平均数

select
    city,
    max(avgs) maxs
from
    t
group by 
    city

然后跟计算结果

SELECT
    t.city,
    driver_id,
    round(avgs,1) avg_grade,
    round(orders / days ,1) avg_order_num,
    round(ways / days ,3) avg_mileage
from
    t

连接，where查出来等于最大值的人 最后order by一下

上表
join(上上表) t2
on
    t.city = t2.city
WHERE
    t.avgs = t2.maxs
ORDER BY
    avg_order_num