MySQL如何计算连续登录天数

建表、insert数据

create table tmp_login (
 user_id int(11) ,
 login_date datetime
);
insert into tmp_login values(2,'2020-05-29 11:12:12');
insert into tmp_login values(2,'2020-05-29 15:12:12');
insert into tmp_login values(2,'2020-05-30 11:12:12');
insert into tmp_login values(2,'2020-05-31 11:12:12');
insert into tmp_login values(2,'2020-06-01 11:12:12');
insert into tmp_login values(2,'2020-06-02 11:12:12');
insert into tmp_login values(2,'2020-06-03 11:12:12');
insert into tmp_login values(2,'2020-06-04 11:12:12');
insert into tmp_login values(2,'2020-06-05 11:12:12');
insert into tmp_login values(2,'2020-06-06 11:12:12');
insert into tmp_login values(2,'2020-06-07 11:12:12');
insert into tmp_login values(7,'2020-06-01 11:12:12');
insert into tmp_login values(7,'2020-06-02 11:12:12');
insert into tmp_login values(7,'2020-06-03 11:12:12');
insert into tmp_login values(7,'2020-06-05 11:12:12');
insert into tmp_login values(7,'2020-06-06 11:12:12');
insert into tmp_login values(7,'2020-06-07 11:12:12');
insert into tmp_login values(7,'2020-06-08 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。

来源：https://blog.csdn.net/BurningSilence/article/details/121558764