MySQL之活跃用户 连续出现n次

Logins table:
+---------+------------+
| user_id | login_date |
+---------+------------+
| 1       | 2019-07-20 |
| 1       | 2019-07-20 |
| 1       | 2019-07-20 |
| 2       | 2019-07-20 |
| 2       | 2019-07-21 |
| 2       | 2019-07-21 |
| 3       | 2019-07-21 |
| 3       | 2019-07-21 |
| 3       | 2019-07-21 |
| 4       | 2019-06-25 |
| 4       | 2019-06-25 |
+---------+------------+

select
distinct a1.id, a1.name
from `Accounts` as a1
inner join (
    select
    id,
    @cnt:=if(@id=id and @pre_date=date_sub(login_date, interval 1 day), @cnt+1, 1) as cnt,
    @id:=id,
    @pre_date:=login_date
    from
    (select * from `Logins` group by id, login_date order by id, login_date) as a,
    (select @id:=null, @pre_date:=null, @cnt:=0) as b
) as b1
on a1.id = b1.id
where b1.cnt >= 5
order by a1.id

• 求n次，那就将5改成你要你次数