题解 | #连续签到领金币#

-- 1.筛选出 有效数据
with t1 as (
select * from tb_user_log 
where in_time > '2021-07-07 00:00:00'  
and in_time <'2021-11-01 00:00:00' 
and artical_id=0 and sign_in=1
),
-- 时间减去等差数列 算连续值
t2 as(
select uid,in_time,
date_sub(date(in_time),INTERVAL n day) gk
from (
select uid,in_time,
row_number() over(partition by uid order by in_time) n
from t1) t
),
-- 开窗 用连续值分区 打上连续天数的序号
t3 as(
select
uid,in_time,DATE_FORMAT(in_time, '%Y%m%d') `month`,
row_number() over(partition by uid,gk order by in_time) xh
from t2
),
-- 根据序号打上额外金币数标签
t4 as(
select 
uid,in_time,`month`,
case 
when xh%7=0 then 6
when xh%7=3 then 2
else 0 end as ew
from t3
)
-- 聚合每日金币数，和额外金币数
select 
uid,date_format(`month`,'%Y%m'),count(distinct `month`)+sum(ew)
from t4
group by 1,2