--连续登录问题可以采用“连续值-连续值，结果一致”的原理 with temp as ( select t1.user_id,t1.reg_time,t1.reg_port,date(t2.log_time) as log_date,t2.log_port ,row_number() over(partition by t1.user_id order by date(t2.log_time) asc) as rn from register_tb t1 inner join login_tb t2 on t1.user_id=t2.user_id ) select user_id from ...
