263 hashmap集合练习之：键是Student
【需求】
创建一个HashMap集合，键是Student对象，值是String住址
存储多个键值对，并遍历
要求：如果学生对象的变量值相同，视为同一个对象
【思路】
1. 定义Student类，注意！要在Student类里重写2个方法（可快捷生成），保证学生的唯一性
---1-equals方法
---2-HashMap方法
2. 创建HashMap集合
3. 创建学生对象
4. 添加学生到集合的值
5. 遍历集合
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package e263;
public class Student263 {
private String name;
private int age;
public Student263(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 boolean equals(Object o) {
if (this == o) return true;
if (o == null || getclass() != o.getclass()) return false;
Student263 that = (Student263) o;
if (age != that.age) return false;
return name.equals(that.name);
}
@Override
public int hashCode() {
int result = name.hashCode();
result = 31 * result + age;
return result;
}
}
--------------------------------------------------------------
package e263;
import java.util.HashMap;
import java.util.Set;
public class HashMapDemo {
public static void main(String[] args) {
HashMap<Student263,String> jihe263 = new HashMap<Student263,String>();
Student263 ss1 = new Student263("TRACY", 33);
Student263 ss2 = new Student263("BEN", 70);
Student263 ss3 = new Student263("CECILIA", 23);
Student263 ss4 = new Student263("CECILIA",23);
jihe263.put(ss1,"LONDON");
jihe263.put(ss2,"PARIS");
jihe263.put(ss3,"PEKING");
jihe263.put(ss4,"TOKYO");
//Student类里不重写equals and HashMap方法，则对象相同时，后加的对象叠加到集合里
//Student类里重写equals and HashMap方法，则对象相同时，后加的对象替换先加的对象
System.out.println("\t21.原始数据");
System.out.println(jihe263);
System.out.println("\t24.遍历并输出");
Set<Student263> jianjihe = jihe263.keySet();
for (Student263 jian : jianjihe){
String zhi = jihe263.get(jian);
System.out.println(jian.getName()+","+jian.getAge()+","+zhi);
}
}
}
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21.原始数据
{e263.Student263@93f92e08=LONDON, e263.Student263@e879c27b=TOKYO, e263.Student263@1f0d3b=PARIS}
24.遍历并输出
TRACY,33,LONDON
CECILIA,23,TOKYO
BEN,70,PARIS