SQL必会题之留存率

目录

一、题目

二、步骤

        1、先求出每个用户最早的登录日期（辅助列）

        2、连表

        3、统计

        4、源码

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一、题目

        查询每天新增用户数以及他们次日留存率和三十日留存率

二、步骤

        1、先求出每个用户最早的登录日期（辅助列）

SELECT user_id,login_date,MIN(login_date) over (PARTITION by user_id ORDER BY login_date ) as first_login from tb_user_login;

        2、连表

        将有辅助列的表与原表进行关联，关联的条件不仅为id相等，还有一个条件就是其登录的时间与最早的登录时间的差值

SELECT user_id,login_date,MIN(login_date) over (PARTITION by user_id ORDER BY login_date ) as first_login from tb_user_login)t1 
LEFT JOIN tb_user_login as t2 
on t1.user_id = t2.user_id and DATEDIFF(t2.login_date,first_login)=1
left JOIN tb_user_login as t3
on t1.user_id = t3.user_id and DATEDIFF(t3.login_date,first_login)=29
GROUP BY first_login,uesr_id;

        3、统计

COUNT(DISTINCT t1.user_id) as 新增用户数,
COUNT(DISTINCT t2.user_id) /COUNT(DISTINCT t1.user_id) as 次日留存,
COUNT(DISTINCT t3.user_id) /COUNT(DISTINCT t1.user_id) as 三十日留存 

        4、源码

SELECT user_id,login_date,MIN(login_date) over (PARTITION by user_id ORDER BY login_date ) as first_login from tb_user_login;

# 查看每个用户登录的日期 
SELECT 
first_login,
COUNT(DISTINCT t1.user_id) as 新增用户数,
COUNT(DISTINCT t2.user_id) /COUNT(DISTINCT t1.user_id) as 次日留存,
COUNT(DISTINCT t3.user_id) /COUNT(DISTINCT t1.user_id) as 三十日留存 


from (SELECT user_id,login_date,MIN(login_date) over (PARTITION by user_id ORDER BY login_date ) as first_login from tb_user_login)t1 
LEFT JOIN tb_user_login as t2 
on t1.user_id = t2.user_id and DATEDIFF(t2.login_date,first_login)=1
left JOIN tb_user_login as t3
on t1.user_id = t3.user_id and DATEDIFF(t3.login_date,first_login)=29
GROUP BY first_login;