MySQL如何计算连续登录天数

目录

方法一 row_number()方法二lead() 或 lag()

建表、insert数据

create table tmp_login (

  user_id int(11) ,

  login_date datetime

);

insert into tmp_login values(2,’2022-05-29 11:12:12′);

insert into tmp_login values(2,’2022-05-29 15:12:12′);

insert into tmp_login values(2,’2022-05-30 11:12:12′);

insert into tmp_login values(2,’2022-05-31 11:12:12′);

insert into tmp_login values(2,’2022-06-01 11:12:12′);

insert into tmp_login values(2,’2022-06-02 11:12:12′);

insert into tmp_login values(2,’2022-06-03 11:12:12′);

insert into tmp_login values(2,’2022-06-04 11:12:12′);

insert into tmp_login values(2,’2022-06-05 11:12:12′);

insert into tmp_login values(2,’2022-06-06 11:12:12′);

insert into tmp_login values(2,’2022-06-07 11:12:12′);

insert into tmp_login values(7,’2022-06-01 11:12:12′);

insert into tmp_login values(7,’2022-06-02 11:12:12′);

insert into tmp_login values(7,’2022-06-03 11:12:12′);

insert into tmp_login values(7,’2022-06-05 11:12:12′);

insert into tmp_login values(7,’2022-06-06 11:12:12′);

insert into tmp_login values(7,’2022-06-07 11:12:12′);

insert into tmp_login values(7,’2022-06-08 11:12:12′);

方法一 row_number()

1.查询所有用户的每日登录记录

select distinct user_id, date(login_date) as days 

from tmp_login;

2.row_number()计算登录时间排序

select user_id, days, row_number() over(partition by user_id order by days) as rn

from (

        select distinct user_id, date(login_date) as days from tmp_login) t1;

3.用登录时间 – row_number()，如果得到的日期相同，则认为是连续登录日期

select *, date_sub(days, interval rn day) as  results

from(

        select user_id, days, row_number() over(partition by user_id order by days) as rn

        from (

                select distinct user_id, date(login_date) as days from tmp_login) t1

) t2;

4. 按user_id、results分组就可得出连续登录天数

select user_id, count(*) as num_days

from (

        select *, date_sub(days, interval rn day) as  results

        from(

                select user_id, days, row_number() over(partition by user_id order by days) as rn

                from (

                        select distinct user_id, date(login_date) as days from tmp_login) t1

        ) t2) t3

group by user_id , results;

直接用日期减去row_number()，不用date_sub的话，遇到登录日期跨月时会计算错误，

方法二lead() 或 lag()

这种情况适合的场景是，需要查找连续登录超过n天的用户，n为确定值

如果n为4，即计算连续登录超过4天的用户

— lead计算连续登录

select distinct user_id

from(

        select user_id, days, datediff(lead(days, 3, ‘1970-01-01’) over(partition by user_id order by days), days)as results

        from (

                select distinct user_id, date(login_date) as days from tmp_login) t1) t2

where results = 3;

连续登录4天，则日期差应该为3。

以上为个人经验，希望能给大家一个参考，也希望大家多多支持共生网络。