一、题目描述
求解用户登陆信息表中，每个用户连续登陆平台的天数，连续登陆基础为汇总日期必须登陆，表中每天只有一条用户登陆数据（计算中不涉及天内去重）。
表描述：user_id：用户的id；
sigin_date：用户的登陆日期。
二、解法分析
注：求解过程有多种方式，下述求解解法为笔者思路，其他解法可在评论区交流。
思路：
该问题的突破的在于登陆时间，计算得到连续登陆标识，以标识分组为过滤条件，得到连续登陆的天数，最后以user_id分组，以count()函数求和得到每个用户的连续登陆天数。
连续登陆标识 =（当日登陆日期 - 用户的登陆日期）- 开窗排序的顺序号(倒序)
三、求解过程及结果展示
1.数据准备
-- 1.建表语句
drop table if exists test_sigindate_cnt;
create table test_sigindate_cnt(
user_id string
,sigin_date string
)
;
-- 2.测试数据插入语句
insert overwrite table test_sigindate_cnt
select 'uid_1' as user_id,'2021-08-03' as sigin_date
union all
select 'uid_1' as user_id,'2021-08-04' as sigin_date
union all
select 'uid_1' as user_id,'2021-08-01' as sigin_date
union all
select 'uid_1' as user_id,'2021-08-02' as sigin_date
union all
select 'uid_1' as user_id,'2021-08-05' as sigin_date
union all
select 'uid_1' as user_id,'2021-08-06' as sigin_date
union all
select 'uid_2' as user_id,'2021-08-01' as sigin_date
union all
select 'uid_2' as user_id,'2021-08-05' as sigin_date
union all
select 'uid_2' as user_id,'2021-08-02' as sigin_date
union all
select 'uid_2' as user_id,'2021-08-06' as sigin_date
union all
select 'uid_3' as user_id,'2021-08-04' as sigin_date
union all
select 'uid_3' as user_id,'2021-08-06' as sigin_date
union all
select 'uid_4' as user_id,'2021-08-03' as sigin_date
union all
select 'uid_4' as user_id,'2021-08-02' as sigin_date
;2.计算过程
select user_id
,count(1) as sigin_cnt
from (
select
user_id
,datediff('2021-08-06',sigin_date) as data_diff
,row_number() over (partition by user_id order by sigin_date desc) as row_num
from test_sigindate_cnt
) t
where data_diff - row_num = -1
group by
user_id
;3.计算结果及预期结果对比
3.1 预期结果
| 汇总日期 | 用户id | 登陆天数 |
| 2021-08-06 | uid_1 | 6 |
| 2021-08-06 | uid_2 | 2 |
| 2021-08-06 | uid_3 | 1 |
3.2 计算结果
以上就是SQL查询语句求出用户的连续登陆天数的详细内容，更多关于SQL语句求用户的连续登陆天数的资料请关注编程网其它相关文章！
