问题

下面是某游戏公司记录的用户每日登录数据, 计算每个用户最大的连续登录天数，定义连续登录时可以间隔一天。举例：如果一个用户在 1,3,5,6,9 登录了游戏，则视为连续 6 天登录。

user_id          dt
1001    2021-12-12
1002    2021-12-12
1001    2021-12-13
1001    2021-12-14
1001    2021-12-16
1002    2021-12-16
1001    2021-12-19
1002    2021-12-17
1001    2021-12-20

解答

这是个连续问题的升级版，当满足某种要求时我们也是算作连续的，所以不能使用传统的连续编号，然后做差值的解法了。

核心思路解析如下：

登录日期	第一步：上一个日期	第二步：判断登录日期与上一个日期差值是否在2之内	第三步：然后根据标记开窗做sum
1	1	0	0
3	1	0	0
5	3	0	0
6	5	0	0
9	6	1	1
11	9	0	1

with temp as (       
select '1001' as user_id , '2021-12-12' as  dt
union all 
select '1002' as user_id , '2021-12-12' as  dt
union all
select '1001' as user_id , '2021-12-13' as  dt
union all
select '1001' as user_id , '2021-12-14' as  dt
union all
select '1001' as user_id , '2021-12-16' as  dt
union all
select '1002' as user_id , '2021-12-16' as  dt
union all
select '1001' as user_id , '2021-12-19' as  dt
union all
select '1002' as user_id , '2021-12-17' as  dt
union all
select '1001' as user_id , '2021-12-20' as  dt
)

select
user_id
,max(diff) as max_login_days
from
( select
    user_id 
    ,user_group
    ,datediff(max(dt),min(dt))+1 as diff  --拿到每个用户下，连续时间里面最大日期与最小日期的差值加1就得到来连续天数
    from
    (
select
        user_id
        ,dt
        -- 如果当前日期与上一个日期的差值在2之内，那么就给0，否则给1
        ,sum(if(datediff(dt,last_dt)<=2,0,1)) over(partition by user_id order by dt) as user_group
        from
        (
select
            user_id
            ,dt
            ,lag(dt,1,dt) over(partition by user_id order by dt) as last_dt --根据user_id分组，拿到当前行的上一个日期，没有上一个就给自己本身的值
            from temp
)t1
)t1
group by user_id ,user_group
)t1
group by user_id 
;