java : 解析一个 ArrayList 并将它们设置为一个对象

Question

我有这两个类：

class Student
{
String name;
String age ;
}

class Person
{

String name;
String age ;
String grade ;
}

在下面的代码中，我创建了Student 对象并将它们设置在ArrayList 中。在ArrayList 中设置数据后，我需要解析ArrayList 并在另一个对象中设置值：

public class Work {

    public static void main(String args[]) {

        List StudentItems = new ArrayList();

        Student stud1 = new Student();
        Student stud2 = new Student();

        stud1.name = "ABC";
        stud1.age = "28";
        stud2.name = "XYZ";
        stud2.age = "38";

        StudentItems.add(stud1);
        StudentItems.add(stud2);

        Person[] pers = new Person[StudentItems.size()];

        for (int i = 0; i < StudentItems.size(); i++) {
            pers[i] = new Person();

// I am confused here , could anyone please help 

        }

    }
}

Answer 1

鉴于您的课程布局确实正确（请参阅我的评论），您可以为 person 编写一个构造函数，将 Student 作为输入：

        pers[i] = new Person (StudentItems[i]);

注意，我会重命名变量：

        persons [i] = new Person (students[i]);

您与新 CTor 的人员将如下所示：

class Person
{
  String name;
  String age ;
  String grade ;

  public Person () {}
  public Person (s Student) {
    name = s.name;
    age = s.age;
  }
}

更可能的是，您想更改 Student 和 Person 的名称，并从 Person 派生 Student。然后，每个学生都是一个人，在你的循环中，它就是：

        persons [i] = students[i];

Answer 2

试试看。这将完成工作

你的 Person 类应该是这样的：

package com.student.person.work;

/**
 *
 * @author sarath_sivan
 */
public class Person {

    private String name;
    private int age;
    private String grade;

    public String getName() {
       return this.name;
    }

    public void setName(String name) {
        this.name = name;
    }

    public int getAge() {
        return this.age;
    }

    public void setAge(int age) {
        this.age = age;
    }

    public String getGrade() {
        return this.grade;
    }

    public void setGrade(String grade) {
        this.grade = grade;
    }

}

你的学生类应该是这样的：

package com.student.person.work;

/**
 *
 * @author sarath_sivan
 */
public class Student {

    private String name;
    private int age;

    public String getName() {
       return this.name;
    }

    public void setName(String name) {
        this.name = name;
    }

    public int getAge() {
        return this.age;
    }

    public void setAge(int age) {
        this.age = age;
    }

}

最后是 Work 类：

package com.student.person.work;

import java.util.ArrayList;
import java.util.List;

/**
 *
 * @author sarath_sivan
 */
public class Work {

    public static String calculateGrade() {
        String grade = "";
        // Your code to find the grade.
        //............
        return grade;
    }

    public static void doWork() {
        List<Student> studentList = new ArrayList<Student>();

        Student student = new Student();
        student.setName("ABC");
        student.setAge(24);
        studentList.add(student);

        student = new Student();
        student.setName("DEF");
        student.setAge(28);
        studentList.add(student);

        student = new Student();
        student.setName("GHI");
        student.setAge(21);
        studentList.add(student);

        List<Person> personList = new ArrayList<Person>();

        for(Student students : studentList) {
            Person person = new Person();
            person.setName(students.getName());
            person.setAge(students.getAge());
            person.setGrade(Work.calculateGrade());// Setting the grade
        }
    }

    public static void main(String[] args) {
        Work.doWork();
    }

}

希望这会有所帮助。

谢谢！

Answer 3

Grade 应该在 Student 类中。如果你这样做，你的类应该是这样的：

class Person {
private String name; 
private int age; 

//getters and setters
}

class Student extends Person { // here you have name and age from Person
private String grade;

//getters and setters
}

现在，您想要学生列表中的人员列表吗？你可以这样做：

for (int i = 0; i < listOfStudents.size(); i++){
arrayOfPersons[i] = (Person)listOfStudents.get(i);
}

Answer 4

类似这样的：

    List<Student> studentItems = new ArrayList<Student>();
    Student stud1 = new Student();
    Student stud2 = new Student();

    stud1.name = "ABC";
    stud1.age = "28";
    stud2.name = "XYZ";
    stud2.age = "38";

    studentItems.add(stud1);
    studentItems.add(stud2);

    for (int i = 0; i < studentItems.size(); i++) {
        Student student = studentItems.get(i);
        Person person = new Person();
        person.name = student.name;
        person.age = student.age;
        // person.grade = something - set grade here
        pers[i] = person;
    }

但请注意，您不应该使用公共字段......所以它应该看起来像这样：

for (int i = 0; i < studentItems.size(); i++) {
    Student student = studentItems.get(i);
    Person person = new Person();
    person.setName(student.getName());
    person.setAge(student.getAge());
    // person.setGrade(computeGradeSomehow()); - set grade here
    persons[i] = person;
}

Answer 5

如果您经常从学生对象转换为个人对象，请添加以下构造函数和 setter/getter 方法

class Person {

        String name;

        String age;

        String grade;

        public Person() {

        }

        Person(Student student) {
            this.name = student.getName();
            this.age = student.getAge();
        }

        public String getGrade() {
            return grade;
        }

        public void setGrade(String grade) {
            this.grade = grade;
        }
    }

    class Student {
        private String name;

        private String age;

        public void setName(String name) {
            this.name = name;
        }

        public String getName() {
            return name;
        }

        public String getAge() {
            return age;
        }

        public void setAge(String age) {
            this.age = age;
        }
    }

在 Work 类中，创建 Person 对象，

List<Student> studentItems = new ArrayList<Student>();
 List<Person> personItems = new ArrayList<Person>();

for (int i = 0; i < studentItems.size(); i++) {
        Student student = studentItems.get(i);
        Person person = new Person(student);
        person.setGrade(your_formula_for grade);  
        personItems.add(person); 
}

Answer 6

你可以这样做：

pers[i].name = StudentItems.get(i).name;