oracle查询连续n天登录的用户

-- 查询连续3天登录的用户

1 先创建一个表,如下:

 create table USER_DATA
 (
 USER_ID NUMBER,
 LOGIN_TIME DATE
 );

2 插入用户登录数据:

 insert into user_data (USER_ID, LOGIN_TIME) values ('', to_date('21-07-2014', 'dd-mm-yyyy'));
 insert into user_data (USER_ID, LOGIN_TIME) values ('', to_date('22-07-2014', 'dd-mm-yyyy'));
 insert into user_data (USER_ID, LOGIN_TIME) values ('', to_date('24-07-2014', 'dd-mm-yyyy'));
 insert into user_data (USER_ID, LOGIN_TIME) values ('', to_date('25-07-2014', 'dd-mm-yyyy'));
 insert into user_data (USER_ID, LOGIN_TIME) values ('', to_date('21-07-2014', 'dd-mm-yyyy'));
 insert into user_data (USER_ID, LOGIN_TIME) values ('', to_date('28-07-2014', 'dd-mm-yyyy'));
 insert into user_data (USER_ID, LOGIN_TIME) values ('', to_date('29-07-2014', 'dd-mm-yyyy'));
 insert into user_data (USER_ID, LOGIN_TIME) values ('', to_date('30-07-2014', 'dd-mm-yyyy'));
 insert into user_data (USER_ID, LOGIN_TIME) values ('', to_date('21-07-2014', 'dd-mm-yyyy'));
 insert into user_data (USER_ID, LOGIN_TIME) values ('', to_date('22-07-2014', 'dd-mm-yyyy'));
 insert into user_data (USER_ID, LOGIN_TIME) values ('', to_date('23-07-2014', 'dd-mm-yyyy'));
 insert into user_data (USER_ID, LOGIN_TIME) values ('', to_date('24-07-2014', 'dd-mm-yyyy'));
 insert into user_data (USER_ID, LOGIN_TIME) values ('', to_date('26-07-2014', 'dd-mm-yyyy'));

3 查询连续3天登录的用户

思路:现有的数据里面已经是经过处理的,每个用户每天只有一条数据,并且登录时间是每天的零点,如果是真实的数据则进行trunc后就行,然后选择的时候,需要依据user_id和login_time进行排序,这样在外层选择时,rownm在每个用户按日志排序时是连续的,然后查询时用日志减去rownum,这样如果是连续的日期,其相减的结果就是一样的,然后根据此结果来判断是否是连续登录就行了,查询连续3天登录的用户id的具体代码如下:

 select distinct user_id
 from
 (
     select b.user_id, b.d_temp, count(*)
     from
     (
         select a.user_id, a.login_time - rownum d_temp
         from
         (
             select t.*
             from user_data t
             order by t.user_id, t.login_time
         )a
     )b
     group by b.user_id, b.d_temp
     having count(*) >= 3
 );

以上查询出来的结果就是至少连续登录3天的用户id了.