目录
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 (user_id oder by login_date)) diff;
（2）group by user_id,diff 分组；
（3）count(1) >= 3天的用户就是连续登陆3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 (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”的情景进行处理
- 计算 date_sub(login_date,row_number() over (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。
