select 
    v.user_id,
    count(distinct v.visit_time) as visit_nums
from  order_tb o 
    left join visit_tb v on v.user_id = o.user_id 
where date(o.order_time) = '2022-09-02' and date(v.leave_time) = '2022-09-02'
group by o.user_id
order by visit_nums desc
#虽然题目不难，但是卡了很久，使用group by 一定要搭配聚合函数