通过HIVE SQL获取每个用户的最大连续登录时常

永远在减肥永远110的的小潘2025-05-31 23:13

样本数据导入：

复制代码

drop table if exists user_login;
create table user_login
( `user_id` bigint 
,login_date string
)
;


insert into table user_login 
values 
(1,'2025-04-01')
,(1,'2025-04-02')
,(1,'2025-04-03')
,(1,'2025-04-05')
,(1,'2025-04-06')
,(2,'2025-04-01')
,(2,'2025-04-02')
;在这里插入代码片

id	login_date
1	2025-04-01
1	2025-04-02
1	2025-04-03
1	2025-04-05
1	2025-04-06
2	2025-04-01
2	2025-04-02

sql 复制代码

WITH ranked_logins AS (
    SELECT 
        user_id,
        login_date,
        ROW_NUMBER() OVER (PARTITION BY user_id ORDER BY login_date) AS rn
    FROM (
        SELECT DISTINCT user_id, login_date
        FROM user_login
    ) AS distinct_logins
),
grouped_logins AS (
    SELECT
        user_id,
        login_date,
        rn,
        DATE_SUB(login_date,rn) AS grp
    FROM ranked_logins
),
streaks AS (
    SELECT
        user_id,
        grp,
        COUNT(*) AS streak_length
    FROM grouped_logins
    GROUP BY user_id, grp
)
SELECT
    user_id,
    MAX(streak_length) AS max_streak
FROM streaks
GROUP BY user_id;