sql面试题 连续签到领金币（连续登录问题）

牛客面试题

• 问题：计算每个用户2021年7月以来每月获得的金币数

问题一：连续登录问题

    select uid,date(in_time) as dt,
    row_number() over(partition by uid order by date(in_time)) 编号,
	date(in_time)-(row_number() over(partition by uid order by date(in_time))) dt2
	from tb_user_log
	where date(in_time) between '2021-07-07' and '2021-10-31'
					and artical_id=0 and sign_in=1

1. 利用每个登录日期减去生成的编号会得到一个数值，可知：如果数值一样，则相同的数值的行数记为连续登录天数。

连续登录天数满3得3，满7得7.

select *,row_number() over(partition by uid,dt2 order by dt) 连续登录天数,
		case
		when row_number() over(partition by uid,dt2 order by dt)%7=3 then 3
        when row_number() over(partition by uid,dt2 order by dt)%7=0 then 7
        else 1 end as 金币
        
from ( 	  --  生成编号，利用每个登陆日期减去编号得到一个数值，可利用该数值得到连续登录天数
    select uid,date(in_time) as dt,
    row_number() over(partition by uid order by date(in_time)) 编号,
	date(in_time)-(row_number() over(partition by uid order by date(in_time))) dt2
	from tb_user_log
	where date(in_time) between '2021-07-07' and '2021-10-31'
					and artical_id=0 and sign_in=1
)t

• 利用 连续登录天数 % 7 =3 输出3
• 连续登录天数 %7 =7 输出7 ，否则输出1.
• 连续登录天数的开窗函数（错误点）
• row_number() over(partition by uid,dt2 order by dt) 连续登录天数

完整代码

select uid,date_format(dt,"%Y%m") month,
	sum(金币) coin
from( -- 根据得到的数值，可知一旦数值不一样说明登录天数不连续。进而判断得到的每天得到的金币
select *,row_number() over(partition by uid,dt2 order by dt) 连续登录天数,
		case
		when row_number() over(partition by uid,dt2 order by dt)%7=3 then 3
        when row_number() over(partition by uid,dt2 order by dt)%7=0 then 7
        else 1 end as 金币
        
from ( 	  --  生成编号，利用每个登陆日期减去编号得到一个数值，可利用该数值得到连续登录天数
    select uid,date(in_time) as dt,
    row_number() over(partition by uid order by date(in_time)) 编号,
	date(in_time)-(row_number() over(partition by uid order by date(in_time))) dt2
	from tb_user_log
	where date(in_time) between '2021-07-07' and '2021-10-31'
					and artical_id=0 and sign_in=1
)t
)tt
group by uid,month
order by uid,month ;