SQL面试题练习 —— 查询每个用户最大连续登录天数

目录

• [1 题目](#1 题目)
• [2 建表语句](#2 建表语句)
• [3 题解](#3 题解)

1 题目

查询每个用户最大连续登录天数

样例数据如下 login_log：

2 建表语句

--建表语句
create table if not exists login_log
(
    user_id        int comment '用户id',
    login_time date comment '登录时间'
);
--数据插入
INSERT overwrite table login_log
VALUES (1, '2022-11-28'),
       (1, '2022-12-01'),
       (1, '2022-12-02'),
       (1, '2022-12-03'),
       (2, '2022-12-01'),
       (2, '2022-12-04');

3 题解

（1）开窗，排序

select user_id,
       login_time,
       row_number() over (partition by user_id order by login_time) rn
from login_log;

执行结果

（2）利用等差数列的特性，如果是连续登录，login_time - num 则相等。

select user_id,
       login_time,
       -- 在Hive中，DATE_SUB函数只能以天为单位进行操作。DATE_SUB用于从指定日期减去指定天数。
       date_sub(login_time, rn) as diff
from (select user_id,
             login_time,
             row_number() over (partition by user_id order by login_time) as rn
      from login_log) t

（3）按照 diff 分组，获取每个用户每次连续登录的天数

select user_id,
       diff,
       count(*) days
from (select user_id,
             login_time,
             -- 在Hive中，DATE_SUB函数只能以天为单位进行操作。DATE_SUB用于从指定日期减去指定天数。
             date_sub(login_time, rn) as diff
      from (select user_id,
                   login_time,
                   row_number() over (partition by user_id order by login_time) as rn
            from login_log) t) tt
group by user_id,diff;

执行结果

（4）得出每个用户最大连续登录天数

select user_id,
       max(days) as max_days
from
(select user_id,
       diff,
       count(*) days
from (select user_id,
             login_time,
             -- 在Hive中，DATE_SUB函数只能以天为单位进行操作。DATE_SUB用于从指定日期减去指定天数。
             date_sub(login_time, rn) as diff
      from (select user_id,
                   login_time,
                   row_number() over (partition by user_id order by login_time) as rn
            from login_log) t) tt
group by user_id,diff) ttt
group by user_id ;

执行结果