题解 | #获得积分最多的人(三)#

select b.user_id user_id

,u.name name

,b.grade_num grade_num

from

(

select user_id

,sum(if(type = 'add',grade_num,grade_num*(-1))) grade_num

,dense_rank() over(order by sum(if(type = 'add',grade_num,grade_num*(-1))) desc) dr

from grade_info

group by 1 #求出用户当前总积分和其排名

)b

left join user u on b.user_id = u.id

where b.dr = 1

order by 1