大廠高頻面試題-連續登陸問題

背景web

對於數據開發人員來講，手寫sql是比較熟悉的了，就有這樣一道題，面試時須要手寫sql，這就是很是經典的連續登陸問題，大廠小廠都愛問，這種題說簡單也不簡單，說難也不難，關鍵是要有思路。
面試

真題sql

hql統計連續登錄的三天及以上的用戶ruby

這個問題能夠擴展到不少類似的問題：連續幾個月充值會員、連續天數有商品賣出、連續打車、連續逾期。bash

數據提供
微信

 用戶ID、登入日期 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

輸出字段
app

+---------+--------+-------------+-------------+--+|   uid   | times  | start_date  |  end_date   |+---------+--------+-------------+-------------+--+

羣內討論編輯器

這道題在羣裏發出後，你們就展開了激烈的討論：學習

解決方案大數據

能夠看出來，有不少種不一樣的解決方案。

這裏就爲你們提供一種比較常見的方案：

建表

create table wedw_dw.t_login_info( user_id string  COMMENT '用戶ID',login_date date COMMENT '登陸日期')row format delimitedfields terminated by ',';

導數據

hdfs dfs -put /test/login.txt /data/hive/test/wedw/dw/t_login_info/

驗證數據

select * from wedw_dw.t_login_info;+----------+-------------+--+| user_id  | 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  |+----------+-------------+--+

解決方案

select t2.user_id         as user_id,count(1)           as times,min(t2.login_date) as start_date,max(t2.login_date) as end_datefrom(    select     t1.user_id    ,t1.login_date    ,date_sub(t1.login_date,rn) as date_diff    from    (        select         user_id        ,login_date        ,row_number() over(partition by user_id order by login_date asc) as rn         from        wedw_dw.t_login_info    ) t1) t2group by  t2.user_id,t2.date_diffhaving times >= 3;

結果

+----------+--------+-------------+-------------+--+| 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  |+----------+--------+-------------+-------------+--+

思路

先把數據按照用戶id分組，根據登陸日期排序

select         user_id        ,login_date        ,row_number() over(partition by user_id order by login_date asc) as rn         from        wedw_dw.t_login_info

+----------+-------------+-----+--+| user_id  | login_date  | rn  |+----------+-------------+-----+--+| 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   |+----------+-------------+-----+--+

2.用登陸日期減去排序數字rn，獲得的差值日期若是是相等的，則說明這兩天確定是連續的

select     t1.user_id    ,t1.login_date    ,date_sub(t1.login_date,rn) as date_diff    from    (        select         user_id        ,login_date        ,row_number() over(partition by user_id order by login_date asc) as rn         from        wedw_dw.t_login_info    ) t1    ;

+----------+-------------+-------------+--+| 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  |+----------+-------------+-------------+--+

3.根據user_id和日期差date_diff 分組，最小登陸日期即爲這次連續登陸的開始日期start_date，最大登陸日期即爲結束日期end_date，登陸次數即爲分組後的count(1)

select t2.user_id         as user_id,count(1)           as times,min(t2.login_date) as start_date,max(t2.login_date) as end_datefrom(    select     t1.user_id    ,t1.login_date    ,date_sub(t1.login_date,rn) as date_diff    from    (        select         user_id        ,login_date        ,row_number() over(partition by user_id order by login_date asc) as rn         from        wedw_dw.t_login_info    ) t1) t2group by  t2.user_id,t2.date_diffhaving times >= 3;

+----------+--------+-------------+-------------+--+| 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  |+----------+--------+-------------+-------------+--+