题解 | 查询连续登陆的用户
查询连续登陆的用户
https://www.nowcoder.com/practice/9944210610ec417e94140ac09512a3f5
select user_id from (select user_id,ds,count(*) over(partition by user_id,ds) day from (select t1.user_id,log_time,row_number() over(partition by user_id order by log_time) rk, date_sub(date(log_time),interval row_number() over(partition by user_id order by log_time) day) ds from register_tb t1 join login_tb t2 on t1.user_id = t2.user_id) as t3) as t4 group by user_id having max(day) >=3 order by user_id asc
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