题解 | #获得积分最多的人(二)#

with ts as(
    select user_id, sum(grade_num) as s from grade_info group by user_id
)


select t.user_id, name, max(s) as grade_num from ts t inner join user u on t.user_id = id
group by t.user_id, name
having grade_num = (select max(s) from ts)
order by t.user_id asc

group by 和窗口函数的关系是 select a, b, c...d, 窗口函数 from t
这里面select了几个元素，group by 后面就要把这些都写上，我们常常只写一个，容易忽略多个的情况，需要注意