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# $1454. 活跃用户
# https://leetcode-cn.com/problems/active-users/
# SQL架构
Create table If Not Exists Accounts (id int, name varchar(10));
Create table If Not Exists Logins (id int, login_date date);
Truncate table Accounts;
insert into Accounts (id, name) values ('1', 'Winston');
insert into Accounts (id, name) values ('7', 'Jonathan');
Truncate table Logins;
insert into Logins (id, login_date) values ('7', '2020-05-30');
insert into Logins (id, login_date) values ('1', '2020-05-30');
insert into Logins (id, login_date) values ('7', '2020-05-31');
insert into Logins (id, login_date) values ('7', '2020-06-01');
insert into Logins (id, login_date) values ('7', '2020-06-02');
insert into Logins (id, login_date) values ('7', '2020-06-02');
insert into Logins (id, login_date) values ('7', '2020-06-03');
insert into Logins (id, login_date) values ('1', '2020-06-07');
insert into Logins (id, login_date) values ('7', '2020-06-10');
# Write your MySQL query statement below
select
t4.id,
a.name
from
(
select
distinct id
from
(
select
id,
subdate(login_date, rn) as diff
from
(
select
id,
login_date,
row_number() over(
partition by id
order by
login_date
) as rn
from
(
select
id,
login_date
from
Logins
group by
id,
login_date
) t1
) t2
group by
id,
diff
having
count(*) >= 5
) t3
) t4
join Accounts a on t4.id = a.id
order by
id;
# clean-up
drop table Accounts;
drop table Logins;
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