Collections.sort() 对 List 排序

    今天在项目中遇到一个collections的排序问题，到网上搜了下，还蛮好用的，先记下来，同时保存自己原有的思想：在jsp中调用collections.sort(list)-->userModel中实现comparator接口，然后实现相应的方法
class User {
String name;
String age;

public User(String name,String age){
  this.name=name;
  this.age=age;
}
public String getAge() {
  return age;
}
public void setAge(String age) {
  this.age = age;
}
public String getName() {
  return name;
}
public void setName(String name) {
  this.name = name;
}
}

//具体的比较类，实现Comparator接口

import java.util.Comparator;
import java.util.List;
import java.util.ArrayList;
import java.util.Collections;

public class ComparatorUser implements Comparator{

public int compare(Object arg0, Object arg1) {
  User user0=(User)arg0;
  User user1=(User)arg1;

   //首先比较年龄，如果年龄相同，则比较名字

  int flag=user0.getAge().compareTo(user1.getAge());
  if(flag==0){
   return user0.getName().compareTo(user1.getName());
  }else{
   return flag;
  } 
}

}




//测试类
public class SortTest {


public static void main(String[] args){
  List userlist=new ArrayList();
  userlist.add(new User("dd","4"));
  userlist.add(new User("aa","1"));
  userlist.add(new User("ee","5"));
  userlist.add(new User("bb","2")); 
  userlist.add(new User("ff","5"));
  userlist.add(new User("cc","3"));
  userlist.add(new User("gg","6"));
 
  ComparatorUser comparator=new ComparatorUser();
  Collections.sort(userlist, comparator);
  
  for (int i=0;i<userlist.size();i++){
   User user_temp=(User)userlist.get(i);
      System.out.println(user_temp.getAge()+","+user_temp.getName());
  }
 
}
}
/*
public int compareTo(Object o) {
UserModel um = (UserModel) o;
        //sort the string ignore capital and lowercase
        return this.primaryEmail.toUpperCase().compareTo(um.getPrimaryEmail().toUpperCase());
}
*/
//首先年龄排序，如果年龄相同，则按名字排序

结果:
   1, aa
   2, bb
   3, cc
   4, dd
   5, ee                    //注意:同样是5岁的人，则比较名字(ee,ff)，然后排序
   5, ff
   6, gg