将set集合转为list并对对象进行排序

之前用hibernate的时候，遇到一对多的表结构。比如班级(class),学生(student)时，hibernate都是将student作为一个set集合放在班级对象之中，这是无序的，而我们往往要根据student的id或者分数进行排序。以前采用的办法很笨，遍历set，再借助数组将其按照所要求的顺序加入到list中。这两天发现一种比较好的方法如下：

public class Stude implements Comparable<Object> {

	
	private String id;
	private String name;
	private String address;
	
	public Stude() {
	}
	public Stude(String id,String name,String address) {
		this.id=id;
		this.name=name;
		this.address=address;
	}
	public String getId() {
		return id;
	}
	public void setId(String id) {
		this.id = id;
	}
	public String getName() {
		return name;
	}
	public void setName(String name) {
		this.name = name;
	}
	public String getAddress() {
		return address;
	}
	public void setAddress(String address) {
		this.address = address;
	}

	@Override
	public int compareTo(Object arg0) {
		Stude stu1=(Stude)arg0;
		/*根据id排序
		 * int stu0Id=Integer.parseInt(id);
		int stu1Id=Integer.parseInt(stu1.id);
		return stu0Id>stu1Id?1:(stu0Id==stu1Id?0:-1);*/
		return name.compareTo(stu1.getName());//根据name排序
	}
	
	
}

import java.util.ArrayList;
import java.util.Collections;
import java.util.HashSet;
import java.util.List;
import java.util.Set;


public class Test {

	
	
	
	public static void main(String[] args) {
		Stude stu=new Stude("1","gao","hz");
		Stude stu2=new Stude("2","aao","hz");
		Stude stu3=new Stude("3","zao","hz");

		Set<Stude> studeset=new HashSet<Stude>();
		studeset.add(stu);
		studeset.add(stu2);
		studeset.add(stu3);
	   List<Stude> stuList=new ArrayList<Stude>();
	   stuList.addAll(studeset);
	   Collections.sort(stuList);
	    for (Stude stude : stuList) {
			System.out.println(stude.getName());
		}
	}
	
	
}