java:Map借口及其子类HashMap四

java:Map借口及其子类HashMap四

使用非系统对象作为key,使用匿名对象获取数据

在Map中可以使用匿名对象找到一个key对应的value.

person:

public class HaspMapPerson {

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

	@Override
	public String toString() {
		return "姓名：" + name + ", 年龄：" + age ;
	}
	
	
}

　　

main:

Map<String, HaspMapPerson> allSet = new HashMap<String, HaspMapPerson>();
		allSet.put("zhangsan", new HaspMapPerson("zhangsan",30));
		allSet.put("lisi", new HaspMapPerson("lisi",33));
		
		//获取value值
		System.out.println( allSet.get(new String("zhangsan")) );

结果：姓名：zhangsan, 年龄：30

 

另外一种情况：

key:是对象， value是string

则无法通过key找到value,为什么之前的string可以？这里需要实现equals()和hashCode来区分是否是同一个对象。

 

//通过key找到value
		Map<HaspMapPerson, String> map = new HashMap<HaspMapPerson, String>();
		map.put(new HaspMapPerson("zhangsan",30), "zhangsan");
		map.put(new HaspMapPerson("lisi",33), "lis");
		
		System.out.println( map.get(new HaspMapPerson("zhangsan",30)) );

　　

结果为：null

需要修改Person中的 equals()和hashCode()方法：

增加：

public int hashCode()
	{
		return this.name.hashCode() * this.age;
	}
	
	public boolean equals(Object o)
	{
		
		if(this == o)
		{
			return true;
		}
		if( !(o instanceof HaspMapPerson)  )
		{
			return false;
		}
		HaspMapPerson p = (HaspMapPerson) o;
		if( this.name.equals(p.getName()) && this.age == p.getAge() )
		{
			return true;
		}else {
			return false;
		}
	}

　　

Person:

public class HaspMapPerson {

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

	@Override
	public String toString() {
		return "姓名：" + name + ", 年龄：" + age ;
	}
	
	
	public int hashCode()
	{
		return this.name.hashCode() * this.age;
	}
	
	public boolean equals(Object o)
	{
		
		if(this == o)
		{
			return true;
		}
		if( !(o instanceof HaspMapPerson)  )
		{
			return false;
		}
		HaspMapPerson p = (HaspMapPerson) o;
		if( this.name.equals(p.getName()) && this.age == p.getAge() )
		{
			return true;
		}else {
			return false;
		}
	}
	
	
	
	
	
	
	
}

　　

执行结果：

zhangsan