-- 统计用户的总积分
WITH t AS (
SELECT user_id,SUM(grade_num) AS grade_sum
FROM grade_info
GROUP BY user_id
)
-- 与用户表关联查询积分最高的用户
SELECT id,name,grade_sum
FROM t
JOIN user u on u.id = t.user_id
WHERE grade_sum = (SELECT MAX(grade_sum) FROM t)