题解 | 牛客每个人最近的登录日期(六)

# 截至到某天总共刷了多少题
with
t1 as(
    select
        user.name as name,
        number as pass_num,
        login.date as time1
        # passing_number.date as time2
    from
        login
        left join
        passing_number on login.user_id=passing_number.user_id
        left join
        user on login.user_id=user.id
    where
        login.date=passing_number.date
),
t2 as(
    select
        name as u_n,
        time1 as date,
        sum(pass_num)over(partition by name order by time1) as ps_num
    from
        t1
)

select * from t2 order by date,u_n

partition by 后使用order by就可解决问题