with
-- 每天的访问人数(存在一个用户在一天内多次访问的情况,需去重))
    t1 as (
        select
            date(visit_time) v_time,
            count(distinct user_id) v_num
        from
            visit_tb
        group by
            v_time
    ),
-- 每天的下单人数(存在一个用户在一天内多次下单的情况,需去重)
    t2 as (
        select
            date(order_time) o_time,
            count(distinct user_id) o_num
        from
            order_tb
        group by
            o_time
    )
select
    o_time date,
    concat(round((o_num / v_num) * 100, 1), '%') cr
from
    t1
    join t2 on t1.v_time = t2.o_time
order by
    date