統計用戶連續登陸最大的天數

注意： 斷一天 也算連續登陸

 

1、創建建表語句

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

 

2、數據准備

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、數據導入

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

 

4、sql 語句

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

首先將日期下移一位，用於得到當前日期和上一個日期之間相差的天數

        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

 

 

針對相差的天數大於2的做等差數列，區別是否連續天數

    select
        id,
        dt,
        lag_dt,
        dt_diff,
        sum(if(dt_diff > 2, 1, 0)) over(partition by id order by dt) as flag  -- 根據連續的定義更改 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

 

 

分組聚合，查找連續登陸天數的最大值

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