select * from test.score where (grade,name) in (select max(grade) as grade,name from test.score group by name)

------解决方案--------------------
不好意思，上面的语句是在自己的机子上测试的。改一下
SQL code

select * from user where (grade,name) in (select max(grade) as grade,name from user group by name)

------解决方案--------------------
如果是用java代码的话，假设已经有user 实体，可以使用如下代码：
Java code

Comparator<User> comparator = new Comparator<User>(){
    public int compare(User u1,User u2){
         return u2.getGrade() - u1.getGrade();//排倒序
    }
};
Map<String,List<User>> users = new HashMap<String,List<User>>();
//key 为用户名，list 是每个相同用户名的n个实体
int index = 1;
int length = users.values().iterator().next().size();
for(;index<=length;index++){
    System.out.println("-------------第"+index+"高分----------");
    for(String key;users.keySet()){
        User user = users.get(key).get(index);
        System.out.println(user.getId()+"\t"+user.getName()+"\t"+user.getGrade());
    }

}

------解决方案--------------------
挺有意思的，需求明白了，你的要求是什么？用纯sql 还是 纯java？还是什么

------解决方案--------------------
order by grade desc

------解决方案--------------------
2个排序不就可以了

------解决方案--------------------
select *,(
select count(*) from Test001  
where Test001.name=x.name and  
Test001.id<>x.id and  
Test001.grade>=x.grade) as row  
from (select id,name,grade from Test001) as x  
order by row;


------解决方案--------------------
Test001是表名，若是sqlserver2005/2008 或oracle 9之后都有 rownumber 函数就好办多了，
要是想从低到高可是把 Test001.grade>=x.grade 变为 Test001.grade《=x.grade

------解决方案--------------------
Test001是表名，若是sqlserver2005/2008 或oracle 9之后都有 rownumber 函数就好办多了，
要是想从低到高可是把 Test001.grade>=x.grade 变为 Test001.grade《=x.grade