with total_grade as (
select
user_id,
sum(grade_num) as grade_num,
dense_rank() over (order by sum(grade_num) desc) as rn
from
grade_info
group by
user_id
)
select
u.id,
u.name,
t_g.grade_num
from
total_grade as t_g
left join
user as u
on
t_g.user_id = u.id
where
t_g.rn = 1
order by
u.id


京公网安备 11010502036488号