Retention rate of SQL required questions

Catalog

One 、 subject

Two 、 step

        1、 First find out the earliest login date of each user （ Auxiliary column ）

        2、 Even the table

        3、 Statistics

        4、 Source code

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One 、 subject

         Query the number of new users per day and their retention rate for the next day and 30 days

Two 、 step

        1、 First find out the earliest login date of each user （ Auxiliary column ）

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、 Even the table

         Associate the table with auxiliary columns with the original table , The associated condition is not only id equal , Another condition is the difference between the login time and the earliest login time

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、 Statistics

COUNT(DISTINCT t1.user_id) as  Number of new users ,
COUNT(DISTINCT t2.user_id) /COUNT(DISTINCT t1.user_id) as  The next day ,
COUNT(DISTINCT t3.user_id) /COUNT(DISTINCT t1.user_id) as  For 30 days  

        4、 Source code

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

#  View the date each user logged in  
SELECT 
first_login,
COUNT(DISTINCT t1.user_id) as  Number of new users ,
COUNT(DISTINCT t2.user_id) /COUNT(DISTINCT t1.user_id) as  The next day ,
COUNT(DISTINCT t3.user_id) /COUNT(DISTINCT t1.user_id) as  For 30 days  


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;