题解 | 获得积分最多的人(一)
获得积分最多的人(一)
https://www.nowcoder.com/practice/1bfe3870034e4efeb4b4aa6711316c3b
with t as (
select name, sum_grade
from (
select user_id, sum(grade_num) over (partition by user_id) as sum_grade
from grade_info
) a join user b on a.user_id=b.id
)
select distinct name, sum_grade as grade_num
from t
where sum_grade=(
select max(sum_grade) from t
)
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