题解 | #获得积分最多的人(二)#
获得积分最多的人(二)
http://www.nowcoder.com/practice/b6248d075d2d4213948b2e768080dc92
with re as(
select *,
sum(g.grade_num) as grade_sum
from user as u
join grade_info as g
on u.id = g.user_id
group by id
order by grade_sum desc)
select re.id,
re.name,
re.grade_sum
from re
where re.grade_sum in(
select max(grade_sum) as grade_max from re); 