题解 | 获得积分最多的人(一)

第一次写的代码使用了嵌套子查询，并在嵌套查询里使用了聚合函数，不够简洁，原因是对join (inner join)的理解不够深。

select 
    u.name,
    t.grade_num
from user u
join (
    select 
        user_id,
        sum(grade_num) as grade_num
    from grade_info
    where type = 'add'
    group by user_id
) t
on u.id = t.user_id
order by t.grade_num desc
limit 1

更加简洁的代码，直接inner join。

SELECT u.name, SUM(g.grade_num) AS grade_sum
FROM user u
JOIN grade_info g ON u.id = g.user_id
GROUP BY u.id, u.name
ORDER BY grade_sum DESC
LIMIT 1;