#order_tb: order_id
#visit_tb: user_id+visit_time

#需求 访问-下单

-- 这种登录问题，一定要去重，某个用户一天内多次访问丨登录的情况

select
date(visit_time) as date
,concat(round(count(distinct o.user_id,date(order_time))
/count(distinct v.user_id,date(visit_time))*100,1),'%') as cr
from visit_tb v
left join
order_tb o
on v.user_id = o.user_id
and date(o.order_time) = date(v.visit_time)
group by 1
order by date