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


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