--Step1：筛选统计期内，用户首次登录的信息（user_id以及date）
select user_id,min(date) as date from login
group by user_id--Step2：步骤一得到的表a和login表连接，计算出在登录后的第二天又登录的用户数
----------- 满足首次登录日期+1day=第二天登录日期
----------- 如果不满足，b表里对应的数据为null
select count(distinct b.user_id) from
(select user_id,min(date) as date from login
group by user_id) a
left join login b
on a.user_id=b.user_id and  b.date=date(a.date,'+1 day')--Step3:计算次日留存 次日登录用户数/登录用户数
select round(count(distinct b.user_id )*1.0/count(distinct a.user_id),3) as p
from
(select user_id,min(date) as date from login
group by user_id)as a
left join login b
on b.user_id=a.user_id
and b.date=date(a.date,'+1 day')

--Step1：筛选出所有日期,并升序排列
select date from  login
group by date
order by date--Step2：筛选出每个用户首次登录的时间
select user_id,min(date) as date from login
group by user_id--Step3:将两张表连接
select * from
(select date from  login
GROUP BY date
order by date) a
left join
(select user_id,min(date) as date from login
group by user_id) b
on a.date=b.date--按照a.date分组，求出每组的人数
select a.date,count(b.user_id) as new from
(select date from  login
GROUP BY date
order by date) a
left join
(select user_id,min(date) as date from login
group by user_id) b
on a.date=b.date
group by a.date

--step1:筛选出每日新登录用户的user_id
SELECT a.date,b.user_id as 登陆新用户 from
(SELECT date from login
group by date
order by date) a
left join
(SELECT user_id,min(date) as date from login
group by user_id) b
on a.date=b.date--step2:在每日新登录用户信息后追加一列第二天仍继续登录的user_id
--------条件满足 首次登录日期+1 day=第二天登录日期
left join login c
on DATE_ADD(b.date,INTERVAL 1 day)=c.date
group by a.date
order by a.date--step3:计算次日留存率
--首日登录人数 count(distinct b.user_id) 次日登录数count(distinct c.user_id)
--将结果null替换成0，可以用case when 解决
SELECT a.date ,
case when count(b.user_id)=0 then 0.000
else round(count(DISTINCT c.user_id)*1.0/count(DISTINCT b.user_id),3)
end as p
from
(SELECT date from login
group by date) a
left join
(SELECT user_id,min(date) as date from login
group by user_id) b
on a.date=b.date
left join login c
on date(b.date,'+1 day')=c.date
group by a.date
order by a.date
;

就是考察表连接，同时注意累计求和
SELECT
c.name as u_n,d.name as c_n,b.date as date,
sum(number) over (partition by b.user_id order by b.date)
FROM
passing_number as b
left join user as c on b.user_id = c.id
left join login as a on b.user_id = a.user_id
and b.date = a.date
left join client as d on a.client_id = d.id
order by
b.date,c.name