统计用户连续登陆最大的天数

注意： 断一天 也算连续登陆

 

1、创建建表语句

create table demo(
    id string, 
    dt string
) row format delimited fields terminated by '\t';

 

2、数据准备

1001    2021-08-01
1001    2021-08-02
1001    2021-08-03
1001    2021-08-05
1001    2021-08-06
1001    2021-08-07
1001    2021-08-10
1001    2021-08-12

 

3、数据导入

load data local inpath '/home/hadoop/demo.txt' into table demo;

 

4、sql 语句

select
    id,
    flag,
    datediff(max(dt), min(dt)) + 1 as days
from(
    select
        id,
        dt,
        lag_dt,
        dt_diff,
        sum(if(dt_diff > 2, 1, 0)) over(partition by id order by dt) as flag
    from(
        select
            id,
            dt,
            lag_dt,
            datediff(dt, lag_dt) as dt_diff
        from(
            select 
                id,
                dt,
                lag(dt, 1, '1970-01-01') over(partition by id order by dt) as lag_dt
        from demo
        ) t1
    ) t2
) t3
group by id, flag
having datediff(max(dt), min(dt)) + 1 > 6

 

5、sql 解析

首先将日期下移一位，用于得到当前日期和上一个日期之间相差的天数

        select
            id,
            dt,
            lag_dt,
            datediff(dt, lag_dt) as dt_diff
        from(
            select 
                id,
                dt,
                lag(dt, 1, '1970-01-01') over(partition by id order by dt) as lag_dt
            from demo
        ) t1

 

 

针对相差的天数大于2的做等差数列，区别是否连续天数

    select
        id,
        dt,
        lag_dt,
        dt_diff,
        sum(if(dt_diff > 2, 1, 0)) over(partition by id order by dt) as flag  -- 根据连续的定义更改 dt_diff > 2 值就可以
    from(
        select
            id,
            dt,
            lag_dt,
            datediff(dt, lag_dt) as dt_diff
        from(
            select 
                id,
                dt,
                lag(dt, 1, '1970-01-01') over(partition by id order by dt) as lag_dt
            from demo
        ) t1
    ) t2

 

 

分组聚合，查找连续登陆天数的最大值

select
    id,
    flag,
    datediff(max(dt), min(dt)) + 1 as days
from(
    select
        id,
        dt,
        lag_dt,
        dt_diff,
        sum(if(dt_diff > 2, 1, 0)) over(partition by id order by dt) as flag
    from(
        select
            id,
            dt,
            lag_dt,
            datediff(dt, lag_dt) as dt_diff
        from(
            select 
                id,
                dt,
                lag(dt, 1, '1970-01-01') over(partition by id order by dt) as lag_dt
            from demo
        ) t1
    ) t2
) t3
group by id, flag