经典SQL之连续3天登陆

经典SQL之连续3天登陆

经典SQL1——连续3天登陆

无论大厂还是小厂，在SQL题面试中，面试官都很喜欢问连续3天登陆，那么今天就来手撕一下

建表

在Hive中简单建表

drop table if exists user_login_details_last_3_day;
create table user_login_details_last_3_day 
(Userid int,login_date date
)

插入临时数据

简单插入几条测试数据

insert overwrite table user_login_details_last_3_day
select 1010123,'2021-08-02'
union all 
select 1010123,'2021-08-03'
union all 
select 1010123,'2021-08-04'
union all 
select 1010122,'2021-11-02'
union all 
select 1010122,'2021-12-09'
union all 
select 1010199,'2021-01-01'
union all 
select 1010199,'2021-04-23'
union all 
select 1010199,'2021-09-10'
union all 
select 1010199,'2021-09-11'
union all 
select 1010199,'2021-09-12'
union all 
select 1010155,'2021-08-02'
union all 
select 1010155,'2021-08-04'
union all 
select 1010155,'2021-08-05'
union all 
select 1010155,'2021-08-07';

思路

判断一个用户是否连续3天登陆的关键点是如何判断连续

• 通过用户ID分组升序排序得出排序号rk
• 然后用登陆日期减去这个排序号得出一个日期dt
• 如果是连续登陆，那么这个dt是相同的

实现

select Userid,count(1) as cntfrom (select Userid,login_date,row_number()over(partition by Userid order by login_date) as rk   from user_login_details_last_3_day) tgroup by Userid,date_sub(login_date,t.rk)
having count(1) >=3

结果如下：