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Mysql排序获取排名的实例代码

时间:2021-01-26 15:22:59 | 栏目:Mysql | 点击:

代码如下所示:

SELECT @i:=@i+1 rowNum,
if(@total=t.s_score,@rank,@rank:=@i) rank,@total:=t.s_score,
t.*
from(
select t1.* ,t2.s_score from student t1 LEFT JOIN score t2 on t1.s_id=t2.s_id and t2.c_id="01" ORDER BY t2.s_score desc
)t,(select @i:=0,@rank:=0,@total:=null) s ;
SELECT @i:=@i+1 rowNum,
if(@total=t.s_score,@rank,@rank:=@rank+1) rank,@total:=t.s_score,
t.*
from(
select t1.* ,t2.s_score from student t1 LEFT JOIN score t2 on t1.s_id=t2.s_id and t2.c_id="01" ORDER BY t2.s_score desc
)t,(select @i:=0,@rank:=0,@total:=null) s ; 

Mysql 获取成绩排序后的名次

其实就是输出mysql的排序后的行号

RT:获取单个用户的成绩在所有用户成绩中的排名

可以分两步:

1、查出所有用户和他们的成绩排名

select id,maxScore,(@rowNum:=@rowNum+1) as rowNo 
from t_user, 
(select (@rowNum :=0) ) b 
order by t_user.maxScore desc  

2、查出某个用户在所有用户成绩中的排名

select u.rowNo from ( 
select id,(@rowNum:=@rowNum+1) as rowNo 
from t_user, 
(select (@rowNum :=0) ) b 
order by t_user.maxScore desc ) u where u.id="2015091810371700001"; 

总结

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