题解 | #获得积分最多的人(一)#

SELECT
    t1.name name, t1.grade_sum grade_sum
FROM
    (
        SELECT
            t.name as name,
            sum(t.gnum) as grade_sum
        FROM
            (
                SELECT
                    u.name name,
                    g.grade_num gnum
                FROM
                   `user` u
                   LEFT JOIN
                   `grade_info` g
                   ON u.id = g.user_id
            ) t
        GROUP BY
            t.name
    ) t1
ORDER BY
    t1.grade_sum DESC
LIMIT 0,1