HiveSQL题——用户连续登陆

一、连续登陆

1.1 连续登陆3天以上的用户

0 问题描述

查询连续登陆3天以上的用户（字节面试题）

1 数据准备

create table if not exists  table1  (id int comment '用户id', `date` string comment'用户登录时间');

insert overwrite table table1 values
(1,'2019-01-01 19:28:00'),
(1,'2019-01-02 19:53:00'),
(1,'2019-01-03 22:00:00'),
(1,'2019-01-05 20:55:00'),
(1,'2019-01-06 21:58:00'),
(2,'2019-02-01 19:25:00'),
(2,'2019-02-02 21:00:00'),
(2,'2019-02-04 22:05:00'),
(2,'2019-02-05 20:59:00'),
(2,'2019-02-06 19:05:00'),
(3,'2019-03-04 21:05:00'),
(3,'2019-03-05 19:10:00'),
(3,'2019-03-06 19:55:00'),
(3,'2019-03-07 21:05:00');

2 数据分析

select
    distinct id
from (select
          id,
          diff
      from (
               select
                   id,
                   date_sub(dt, row_number()over (partition by id order by dt)) diff
               from (  --- 同一个用户一天可能登陆多次,所以,先去重
                        select
                            id,
                            date_format(`date`,'yyyy-MM-dd') as dt
                        from table1
                        -- current_date() 获取当前的年月日
                        where date_format(`date`,'yyyy-MM-dd') between date_sub(current_date(),7) and current_date()
                        group by id, date_format(`date`,'yyyy-MM-dd')
                    ) tmp1
           ) tmp2
      group by id, diff
      having count(1) >= 3) tmp3;

3 小结

“连续登陆”类型的解题思路：

（1）计算 date_sub(login_date,row_number() over (partition by user_id oder by login_date)) as diff;

（2）group by user_id,diff 分组；

（3）count(1) >= 3的用户 user_id 就是连续登陆3天及以上的用户

1.2 每个用户历史至今连续登录的最大天数

0 问题描述

查询每个用户历史至今连续登录的最大天数

1 数据准备

create table if not exists  table1  (id int comment '用户id', `date` string comment'用户登录时间');

insert overwrite table table1 values
(1,'2019-01-01 19:28:00'),
(1,'2019-01-02 19:53:00'),
(1,'2019-01-03 22:00:00'),
(1,'2019-01-05 20:55:00'),
(1,'2019-01-06 21:58:00'),
(2,'2019-02-01 19:25:00'),
(2,'2019-02-02 21:00:00'),
(2,'2019-02-04 22:05:00'),
(2,'2019-02-05 20:59:00'),
(2,'2019-02-06 19:05:00'),
(3,'2019-03-04 21:05:00'),
(3,'2019-03-05 19:10:00'),
(3,'2019-03-06 19:55:00'),
(3,'2019-03-07 21:05:00');

2 数据分析

select
    id,
    max(cnt) as days
from (
         select
             id,
             count(1) as cnt
         from (
                  select
                      id,
                      `date`,
                      date_sub(`date`, row_number() over (partition by id order by `date`)) diff
                  from (--用户在同一天可能登录多次，需要去重
                           select
                               id,
                               date_format(`date`, 'yyyy-MM-dd') as `date`
                           from table1
                           group by id, date_format(`date`, 'yyyy-MM-dd')
                       ) tmp1
              ) tmp2
         group by id, diff
     ) tmp3
group by id;

3 小结

“连续登陆”类型的解题思路：

（1）计算 date_sub(login_date,row_number() over (partition by user_id oder by login_date)) diff;

（2）group by user_id,diff 分组；

（3）max(cnt)得到就是每个用户历史至今连续登陆的 最大天数。

1.3 每个用户连续登录的最大天数(间断也算)

0 问题描述

统计各用户最长的连续登录天数，间断一天也算作连续。例如：一个用户在1,3,5,6号登录，则视为连续6天登录。

1 数据准备

create table if not exists  table1  (id int comment '用户id', `date` string comment'用户登录时间');

insert overwrite table table1 values
(1,'2019-01-01 19:28:00'),
(1,'2019-01-02 19:53:00'),
(1,'2019-01-03 22:00:00'),
(1,'2019-01-05 20:55:00'),
(1,'2019-01-06 21:58:00'),
(2,'2019-02-01 19:25:00'),
(2,'2019-02-02 21:00:00'),
(2,'2019-02-04 22:05:00'),
(2,'2019-02-05 20:59:00'),
(2,'2019-02-06 19:05:00'),
(3,'2019-03-04 21:05:00'),
(3,'2019-03-05 19:10:00'),
(3,'2019-03-06 19:55:00'),
(3,'2019-03-07 21:05:00');

2 数据分析

方式一：间断的那一天，构造array数组，利用炸裂函数进行补全，然后按照“用户连续登陆”的思路来做。

select
    id,
    max(cnt) as days
from (
         select
             id,
             diff,
             count(1) as cnt
         from (
                  select
                      id,
                      login_date,
                      next_login_date,
                      arr,
                      new_login_date,
                      date_sub(new_login_date, row_number() over (partition by id order by new_login_date)) diff
                  from (
                           select
                               id,
                               login_date,
                               next_login_date,
                               arr,
                               new_login_date
                           from (
                                    select
                                        id,
                                        login_date,
                                        next_login_date,
                                     --间断的那一天，构造array数组，利用炸裂函数进行补全
                                        if(
                                                datediff(next_login_date, login_date) = 2,
                                                array(login_date, date_add(login_date, 1)),
                                                array(login_date)
                                            ) as arr

                                    from (
                                             select
                                                 id,
                                                 login_date,
                                                 --窗口函数 lead(向后取n行)
                                                 --lead(column1,n,val)over(partition by column2 order by column3) 查询当前行的后边第n行数据，如果没有就为null
                                                 lead(login_date, 1, '9999-12-31')
                                                      over (partition by id order by login_date) next_login_date
                                             from (--用户在同一天可能登录多次，需要去重
                                                      select
                                                          id,
                                                          date_format(`date`, 'yyyy-MM-dd') as login_date
                                                      from table1
                                                      group by id, date_format(`date`, 'yyyy-MM-dd')
                                                  ) tmp1
                                         ) tmp2
                                ) tmp3
                                    lateral view explode(arr) tmp as new_login_date
                       ) tmp4
              ) tmp5
         group by id, diff
     ) tmp6
group by id;

方式二：对用户多段stage的连续登陆进行划分,思路类似:会话划分

select
    id,
    max(diff) as days
from (
         select
             id,
             stage,
             datediff(max(login_date), min(login_date)) + 1 as diff
         from (
                  select
                      id,
                      login_date,
                      -- 思路类似：会话划分，字符串拼接得到stage
                      concat(id, '-', sum(start_point)
                                          over (partition by id order by login_date rows between unbounded preceding and current row )) stage
                  from (
                           select
                               id,
                               login_date,
                                --间隔一天也算连续，所以差值大于2的数据打上标签
                               if(datediff(login_date, last_login_date) > 2, 1, 0) start_point
                           from (
                                    select
                                        id,
                                        login_date,
                                        --窗口函数 lag(向前取n行)
                                        --lag(column1,n,val)over(partition by column2 order by column3) 查询当前行的前边第n行数据，如果没有就为null
                                        lag(login_date, 1, '1970-01-01')
                                            over (partition by id order by login_date) as last_login_date
                                    from (
                                             select
                                                 id,
                                                 date_format(`date`, 'yyyy-MM-dd') as login_date
                                             from table1
                                             group by id, date_format(`date`, 'yyyy-MM-dd')
                                         ) tmp1
                                ) tmp2
                       ) tmp3
              ) tmp4
         group by id, stage
     ) tmp5
group by id;

3 小结

** “间断连续”类型的解题思路：**

（1）构造array数组；

（2）炸裂函数+ 侧写视图 ： lateral view +explode将一行变多行，补全间断的那几天

（3）补全后之后就按照“连续登陆”问题的解决思路进行处理

• 计算 date_sub(login_date,row_number() over (partition by user_id oder by login_date)) diff;
• group by user_id,diff 分组；
• max(cnt)得到就是每个用户历史至今连续登陆的最大天数。