今天，我們來(lái)繼續(xù)數(shù)據(jù)庫(kù)面試題的話題，這次給大家整理的是連續(xù)登陸的問(wèn)題，題目如下：

我們先根據(jù)題目中的字段自己創(chuàng)建了表，填充一下相關(guān)的數(shù)據(jù)

create table login(
user_id int comment '用戶id',
access_time datetime comment '訪問(wèn)時(shí)間',
page_id int comment '頁(yè)面id',
dt date comment '登陸日期'
);

insert into login values
(1, '2021-06-01 11:13:15', 10, '2021-06-01'),
(1, '2021-06-02 11:13:15', 10, '2021-06-02'),
(1, '2021-06-03 11:13:15', 10, '2021-06-03'),
(1, '2021-06-04 11:13:15', 10, '2021-06-04'),
(1, '2021-06-05 11:13:15', 10, '2021-06-05'),
(1, '2021-06-06 11:13:15', 10, '2021-06-06'),
(1, '2021-06-07 11:13:15', 10, '2021-06-07'),
(2, '2021-06-01 11:13:15', 10, '2021-06-01'),
(2, '2021-06-03 11:13:15', 10, '2021-06-03'),
(2, '2021-06-04 11:13:15', 10, '2021-06-04'),
(2, '2021-06-05 11:13:15', 10, '2021-06-05'),
(3, '2021-06-01 11:13:15', 10, '2021-06-01'),
(3, '2021-06-07 11:13:15', 10, '2021-06-07'),
(3, '2021-06-08 11:13:15', 10, '2021-06-08'),
(3, '2021-06-09 11:13:15', 10, '2021-06-09'),
(3, '2021-06-10 11:13:15', 10, '2021-06-10'),
(3, '2021-06-11 11:13:15', 10, '2021-06-11'),
(3, '2021-06-12 11:13:15', 10, '2021-06-12'),
(3, '2021-06-13 11:13:15', 10, '2021-06-13'),
(4, '2021-06-01 11:13:15', 10, '2021-06-01'),
(4, '2021-06-03 11:13:15', 10, '2021-06-03'),
(4, '2021-06-05 11:13:15', 10, '2021-06-05'),
(4, '2021-06-07 11:13:15', 10, '2021-06-07'),
(4, '2021-06-09 11:13:15', 10, '2021-06-09'),
(4, '2021-06-11 11:13:15', 10, '2021-06-11'),
(5, '2021-06-01 11:13:15', 10, '2021-06-01'),
(5, '2021-06-07 11:13:15', 10, '2021-06-07'),
(5, '2021-06-08 11:13:15', 10, '2021-06-08'),
(5, '2021-06-09 11:13:15', 10, '2021-06-09'),
(5, '2021-06-11 11:13:15', 10, '2021-06-11'),
(5, '2021-06-12 11:13:15', 10, '2021-06-12'),
(5, '2021-06-13 11:13:15', 10, '2021-06-13');

接下來(lái)我們根據(jù)需求來(lái)分析完成，整理思路順序如下：

/*
思路：
將同一用戶的登陸時(shí)間歸納在一起  設(shè)置個(gè)排名  如果時(shí)間是連續(xù)的  那么 時(shí)間和排名的差值就是相等的
*/
-- 先看一下設(shè)置排名
select *, row_number() over(partition by user_id order by dt) ranking from login where month(dt) = 6;
/*
結(jié)果展示：
user_id access_time page_id dt ranking
1 2021-06-01 11:13:15 10 2021-06-01 1
1 2021-06-02 11:13:15 10 2021-06-02 2
1 2021-06-03 11:13:15 10 2021-06-03 3
1 2021-06-04 11:13:15 10 2021-06-04 4
1 2021-06-05 11:13:15 10 2021-06-05 5
1 2021-06-06 11:13:15 10 2021-06-06 6
1 2021-06-07 11:13:15 10 2021-06-07 7
2 2021-06-01 11:13:15 10 2021-06-01 1
2 2021-06-03 11:13:15 10 2021-06-03 2
2 2021-06-04 11:13:15 10 2021-06-04 3
2 2021-06-05 11:13:15 10 2021-06-05 4
3 2021-06-01 11:13:15 10 2021-06-01 1
3 2021-06-07 11:13:15 10 2021-06-07 2
3 2021-06-08 11:13:15 10 2021-06-08 3
3 2021-06-09 11:13:15 10 2021-06-09 4
3 2021-06-10 11:13:15 10 2021-06-10 5
3 2021-06-11 11:13:15 10 2021-06-11 6
3 2021-06-12 11:13:15 10 2021-06-12 7
3 2021-06-13 11:13:15 10 2021-06-13 8
4 2021-06-01 11:13:15 10 2021-06-01 1
4 2021-06-03 11:13:15 10 2021-06-03 2
4 2021-06-05 11:13:15 10 2021-06-05 3
4 2021-06-07 11:13:15 10 2021-06-07 4
4 2021-06-09 11:13:15 10 2021-06-09 5
4 2021-06-11 11:13:15 10 2021-06-11 6
5 2021-06-01 11:13:15 10 2021-06-01 1
5 2021-06-07 11:13:15 10 2021-06-07 2
5 2021-06-08 11:13:15 10 2021-06-08 3
5 2021-06-09 11:13:15 10 2021-06-09 4
5 2021-06-11 11:13:15 10 2021-06-11 5
5 2021-06-12 11:13:15 10 2021-06-12 6
5 2021-06-13 11:13:15 10 2021-06-13 7

*/

-- 將日期與排名做差
select *, date_sub(dt, interval ranking day) diff from
(select *, row_number() over(partition by user_id order by dt) ranking from login where month(dt) = 6) as t;
/*
結(jié)果展示
user_id access_time page_id dt ranking diff
1 2021-06-01 11:13:15 10 2021-06-01 1 2021-05-31
1 2021-06-02 11:13:15 10 2021-06-02 2 2021-05-31
1 2021-06-03 11:13:15 10 2021-06-03 3 2021-05-31
1 2021-06-04 11:13:15 10 2021-06-04 4 2021-05-31
1 2021-06-05 11:13:15 10 2021-06-05 5 2021-05-31
1 2021-06-06 11:13:15 10 2021-06-06 6 2021-05-31
1 2021-06-07 11:13:15 10 2021-06-07 7 2021-05-31
2 2021-06-01 11:13:15 10 2021-06-01 1 2021-05-31
2 2021-06-03 11:13:15 10 2021-06-03 2 2021-06-01
2 2021-06-04 11:13:15 10 2021-06-04 3 2021-06-01
2 2021-06-05 11:13:15 10 2021-06-05 4 2021-06-01
3 2021-06-01 11:13:15 10 2021-06-01 1 2021-05-31
3 2021-06-07 11:13:15 10 2021-06-07 2 2021-06-05
3 2021-06-08 11:13:15 10 2021-06-08 3 2021-06-05
3 2021-06-09 11:13:15 10 2021-06-09 4 2021-06-05
3 2021-06-10 11:13:15 10 2021-06-10 5 2021-06-05
3 2021-06-11 11:13:15 10 2021-06-11 6 2021-06-05
3 2021-06-12 11:13:15 10 2021-06-12 7 2021-06-05
3 2021-06-13 11:13:15 10 2021-06-13 8 2021-06-05
4 2021-06-01 11:13:15 10 2021-06-01 1 2021-05-31
4 2021-06-03 11:13:15 10 2021-06-03 2 2021-06-01
4 2021-06-05 11:13:15 10 2021-06-05 3 2021-06-02
4 2021-06-07 11:13:15 10 2021-06-07 4 2021-06-03
4 2021-06-09 11:13:15 10 2021-06-09 5 2021-06-04
4 2021-06-11 11:13:15 10 2021-06-11 6 2021-06-05
5 2021-06-01 11:13:15 10 2021-06-01 1 2021-05-31
5 2021-06-07 11:13:15 10 2021-06-07 2 2021-06-05
5 2021-06-08 11:13:15 10 2021-06-08 3 2021-06-05
5 2021-06-09 11:13:15 10 2021-06-09 4 2021-06-05
5 2021-06-11 11:13:15 10 2021-06-11 5 2021-06-06
5 2021-06-12 11:13:15 10 2021-06-12 6 2021-06-06
5 2021-06-13 11:13:15 10 2021-06-13 7 2021-06-06
*/
-- 從數(shù)據(jù)中我們看出 如果用戶是連續(xù)登陸的，那么差值的日期結(jié)果是一樣的
-- 然后根據(jù)用戶 與 時(shí)間差 分類(lèi) 統(tǒng)計(jì)每個(gè)出現(xiàn)的次數(shù) 如果次數(shù)在7以上 表示連續(xù)7天登陸
select user_id ,count(*) from
(select *, date_sub(dt, interval ranking day) diff from
(select *, row_number() over(partition by user_id order by dt) ranking from login where month(dt)=6) as t) as t1
group by user_id, diff having count(*) >= 7;
/*
user_id count(*)
1 7
3 7
*/
-- 因此我們數(shù)據(jù)中只有1和3有連續(xù)登陸過(guò)

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