题解 | #获得积分最多的人(二)#

WITH t1 AS(
SELECT user_id, SUM(grade_num) AS grade_sum,
RANK() OVER(ORDER BY SUM(grade_num) DESC) AS r_number
FROM grade_info
GROUP BY user_id)
SELECT t1.user_id, user.name, t1.grade_sum
FROM t1
INNER JOIN user
ON t1.user_id = user.id
WHERE r_number = 1