Hive SQL经典面试题：统计连续登陆的三天及以上的用户_hive求连续3天的人数

Hive SQL经典面试题

最近发现一道大数据面试经常会问的SQL题目：统计连续登录的三天及以上的用户（或者类似的：连续3个月充值会员用户、连续N天购买商品的用户等），下面就来记录一下解题思路。
要求输出格式：

+---------+--------+-------------+-------------+--+
|   uid   | times  | start_date  |  end_date   |
+---------+--------+-------------+-------------+--+
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首先建表：

create table test.user_login_info(
 user_id string  COMMENT '用户ID'
,login_date date COMMENT '登录日期'
)
row format delimited
fields terminated by ',';
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原始数据：

导入原始数据：

hdfs dfs -put /root/user_login_info.txt  /user/hive/warehouse/test.db/user_login_info/
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查看：

select * from user_login_info;
+--------------------------+-----------------------------+
| user_login_info.user_id  | user_login_info.login_date  |
+--------------------------+-----------------------------+
| user01                   | 2018-02-28                  |
| user01                   | 2018-03-01                  |
| user01                   | 2018-03-02                  |
| user01                   | 2018-03-04                  |
| user01                   | 2018-03-05                  |
| user01                   | 2018-03-06                  |
| user01                   | 2018-03-07                  |
| user02                   | 2018-03-01                  |
| user02                   | 2018-03-02                  |
| user02                   | 2018-03-03                  |
| user02                   | 2018-03-06                  |
+--------------------------+-----------------------------+

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解题思路：
1.先把数据按照user_id分组，login_date升序

select
      user_id
      ,login_date
      ,row_number() over(partition by user_id order by login_date asc) as rank
  from user_login_info
;

结果：
+----------+-------------+-------+
| user_id  | login_date  | rank  |
+----------+-------------+-------+
| user01   | 2018-02-28  | 1     |
| user01   | 2018-03-01  | 2     |
| user01   | 2018-03-02  | 3     |
| user01   | 2018-03-04  | 4     |
| user01   | 2018-03-05  | 5     |
| user01   | 2018-03-06  | 6     |
| user01   | 2018-03-07  | 7     |
| user02   | 2018-03-01  | 1     |
| user02   | 2018-03-02  | 2     |
| user02   | 2018-03-03  | 3     |
| user02   | 2018-03-06  | 4     |
+----------+-------------+-------+

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2.用 login_date - rank 得到的差值日期如果是一样的，则说明是连续登录的

select
      t1.user_id as user_id
      ,t1.login_date as login_date
      ,date_sub(t1.login_date , t1.rank) as date_diff
  from
     (
         select
                user_id
               ,login_date
               ,row_number() over(partition by user_id order by login_date asc) as rank
           from user_login_info
     )t1
;

结果：
+----------+-------------+-------------+
| user_id  | login_date  |  date_diff  |
+----------+-------------+-------------+
| user01   | 2018-02-28  | 2018-02-27  |
| user01   | 2018-03-01  | 2018-02-27  |
| user01   | 2018-03-02  | 2018-02-27  |
| user01   | 2018-03-04  | 2018-02-28  |
| user01   | 2018-03-05  | 2018-02-28  |
| user01   | 2018-03-06  | 2018-02-28  |
| user01   | 2018-03-07  | 2018-02-28  |
| user02   | 2018-03-01  | 2018-02-28  |
| user02   | 2018-03-02  | 2018-02-28  |
| user02   | 2018-03-03  | 2018-02-28  |
| user02   | 2018-03-06  | 2018-03-02  |
+----------+-------------+-------------+
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3.根据 user_id 和 date_diff 分组，login_date 的最小时间即 start_date ，最大时间即 end_date，取分组后的count>=3的即为最终结果

select
      t2.user_id as user_id
      ,count(*) as times
      ,min(t2.login_date) as start_date
      ,max(t2.login_date) as end_date
  from 
      (
        select
              t1.user_id as user_id
              ,t1.login_date as login_date
              ,date_sub(t1.login_date , t1.rank) as date_diff
          from
             (
                 select
                        user_id
                       ,login_date
                       ,row_number() over(partition by user_id order by login_date asc) as rank
                   from user_login_info
             )t1
      )t2
group by t2.user_id,t2.date_diff
having count(*) >= 3
;

结果：
+----------+--------+-------------+-------------+
| user_id  | times  | start_date  |  end_date   |
+----------+--------+-------------+-------------+
| user01   | 3      | 2018-02-28  | 2018-03-02  |
| user01   | 4      | 2018-03-04  | 2018-03-07  |
| user02   | 3      | 2018-03-01  | 2018-03-03  |
+----------+--------+-------------+-------------+
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以上为一种解决方案，大佬们如果有更好的方案欢迎留言交流。