在ArrayList
我有两个不同的对象,
Student
和Employee
。我想逐个遍历它们。我能够遍历列表并使用Employee
对象而不是Student
对象。
我有以下代码:
package javaCollections;
import java.util.ArrayList;
import java.util.Iterator;
class Employee {
@Override
public String toString() {
return "employee [name=" + name + ", age=" + age + "]";
}
public String name;
public int 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;
}
Employee(String name, int age) {
this.age = age;
this.name = name;
}
}
class Student {
@Override
public String toString() {
return "student [stud_name=" + stud_name + ", rollNumber=" + rollNumber
+ "]";
}
String stud_name;
int rollNumber;
public Student(String stud_name, int rollNumber) {
super();
this.stud_name = stud_name;
this.rollNumber = rollNumber;
}
public String getStud_name() {
return stud_name;
}
public void setStud_name(String stud_name) {
this.stud_name = stud_name;
}
public int getRollNumber() {
return rollNumber;
}
public void setRollNumber(int rollNumber) {
this.rollNumber = rollNumber;
}
}
public class Arraylist {
ArrayList<Object> emparray;
public void addemp() {
Employee emp = new Employee("abc", 12);
emparray = new ArrayList<Object>();
emparray.add(emp);
Employee emp1 = new Employee("def", 12);
emparray.add(emp1);
Student std = new Student("efg", 123);
Student std1 = new Student("xyz", 123);
emparray.add(std);
emparray.add(std1);
}
public void iterateemp() {
/*
* Iterator<Object> itr=emparray.iterator();
*
* while(itr.hasNext()) { System.out.println(itr.next()); }
*/
for (Object e : emparray) {
System.out.println(((Employee) e).getAge());
System.out.println(((Employee) e).getName());
}
}
public static void main(String[] args) {
Arraylist al = new arraylist();
al.addemp();
al.iterateemp();
}
}
有人可以帮我这个吗?
答案 0 :(得分:6)
您需要做的是检查对象的实例。
for (Object e : emparray) {
if(e instanceof employee) {
System.out.println(((employee) e).getAge());
System.out.println(((employee) e).getName());
} else if(e instanceof student) {
// do something else
}
}
}
IMO这是一个糟糕的设计。
答案 1 :(得分:4)
最佳做法是创建名为Person
的公共基础,其共享字段为name
。然后,您可以在循环中将Object
替换为Person
。
import java.util.ArrayList;
import java.util.Iterator;
interface Person{
public String getName();
public void setName(String name);
}
class employee implements Person{
@Override
public String toString() {
return "employee [name=" + name + ", age=" + age + "]";
}
public String name;
public int 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;
}
employee(String name, int age) {
this.age = age;
this.name = name;
}
}
class student implements Person{
@Override
public String toString() {
return "student [stud_name=" + name + ", rollNumber=" + rollNumber
+ "]";
}
String name;
int rollNumber;
public student(String stud_name, int rollNumber) {
super();
this.name = stud_name;
this.rollNumber = rollNumber;
}
public int getRollNumber() {
return rollNumber;
}
public void setRollNumber(int rollNumber) {
this.rollNumber = rollNumber;
}
@Override
public String getName() {
return name;
}
@Override
public void setName(String name) {
this.name=name;
}
}
public class arraylist {
ArrayList<Person> emparray;
public void addemp() {
employee emp = new employee("abc", 12);
emparray = new ArrayList<Person>();
emparray.add(emp);
employee emp1 = new employee("def", 12);
emparray.add(emp1);
student std = new student("efg", 123);
student std1 = new student("xyz", 123);
emparray.add(std);
emparray.add(std1);
}
public void iterateemp() {
for (Person e : emparray) {
if (e instanceof employee) {
System.out.println(((employee) e).getAge());
}else{
/// do for student
}
System.out.println(e.getName());
}
}
public static void main(String[] args) {
arraylist al = new arraylist();
al.addemp();
al.iterateemp();
}
}
答案 2 :(得分:3)
for (Object e : emparray) {
if(e instanceof employee) {
System.out.println(((employee) e).getAge());
System.out.println(((employee) e).getName());
} else if(e instanceof student) {
System.out.println(((student) e).getRollNumber());
System.out.println(((student) e).getStud_name());
}
}
}