MySQL排名 保留名次空缺 问题

select a.cid, a.sid, a.score , count(a.score<b.score)+1 as rank
from sc a left join sc b on a.cid=b.cid and a.score<b.score   #笛卡尔积连接, 然后筛选满足a.score<b.score的
group by a.cid, a.sid
order by a.cid, a.score desc;

select a.cid, a.sid, a.score , count(distinct b.score)+1 as rank  #这里使用的是distinct b.score
from sc a left join sc b on a.cid=b.cid and a.score<b.score   #笛卡尔积连接, 然后筛选满足a.score<b.score的
group by a.cid, a.sid
order by a.cid, a.score desc;

select a.cid,a.sid,a.score,b.cid,b.sid,b.score
from  sc a left join sc b on a.cid=b.cid and a.score< b.score
order by a.cid, a.score desc;