题解 | 获得积分最多的人(二)

SELECT 
  U.id AS user_id,  
  U.name AS user_name,
  T2.grade_num AS grade_num
FROM (
  SELECT 
    user_id,
    SUM(grade_num) AS grade_num,
    DENSE_RANK() OVER(ORDER BY SUM(grade_num) DESC) AS rank_num
  FROM grade_info G
  WHERE type = 'add'
  GROUP BY user_id
) T2
JOIN user U ON T2.user_id = U.id
WHERE T2.rank_num = 1  
ORDER BY U.id ASC;

总分第一的