Mysql排序获取排名的实例代码
时间:2020-11-03 13:35 作者:admin
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/' target='_blank'>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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(责任编辑:admin)