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,'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。
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易知道(ezd.cc)。
