java - 实现MVC

Question

我正在为我的类使用 MVC 实现代码，但是我遇到了一个我似乎无法弄清楚的错误。我应该更改的作业的唯一部分是模型、 View 和 Controller ，因此其余的类已经为我们完成了。

当我运行代码时， Controller 中出现错误:

“错误:不兼容的类型:学生无法转换为字符串 view.DisplayOnButton(model.getInfoForStudent(0), 0);"

如何解决这个问题？

这是我所拥有的:

Controller -

public class Controller
    {
    Model model;
    View view;

    public Controller(Model model, View view)
    {
        this.model = model;
        this.view = view;
        SetViewFromModel();
    }

    public void SetViewFromModel()
    {
        view.DisplayOnButton(model.getInfoForStudent(0), 0);
        view.DisplayOnButton(model.getInfoForStudent(1), 1);
        view.DisplayOnButton(model.getInfoForStudent(2), 2);
    }
    }

型号 -

    public class Model
    {
    ArrayList<Student> sts = new ArrayList<>();

    public Model()
    {
        //creates 3 students
        MailAddress addr1 = new MailAddress("107 W College Avenue", "State College", "PA", 16801);
        Student st1 = new Student("Emily", "Smith", 20, addr1);
        MailAddress addr2 = new MailAddress("200 W College Avenue", "State College", "PA", 16801);
        Student st2 = new Student("Mary", "Doe", 20, addr2);
        MailAddress addr3 = new MailAddress("300 W College Avenue", "State College", "PA", 16801);
        Student st3 = new Student("John", "Doe", 20, addr3);
        //add them to the array of students
        sts.add(st1);
        sts.add(st2);
        sts.add(st3);        
    }

    public Student getInfoForStudent (int studentIndex)
        {
            return sts.get(studentIndex);
        }  
}

查看 -

public void DisplayOnButton(String infoToDisplay, int onButton)
    {
        mf.getIp().DisplayOnButton(infoToDisplay, onButton);
    }

Answer 1

您的方法 DisplayOnButton 需要一个字符串，而您正在传递一个 Student，您必须转换您的 getInfoForStudent 以返回字符串或将 Student 转换为字符串。

我建议您为您的邮件和学生覆盖 toString() ，以便它创建一个包含您期望的信息的字符串。

例如:

public class Student {
private final String name;
private final String surname;
private final int age;
private final MailAddress mailAddress;

public Student(String name, String surname, int age, MailAddress mailAddress) {
    this.name = name;
    this.surname = surname;
    this.age = age;
    this.mailAddress=mailAddress;
}

 @Override
public String toString() {

    return String.format("name: %s surname:%s age:%d mail address:%s", name, surname, age, mailAddress);
}
}

您必须对 mailAddress 执行类似的操作，因此您的 getInfoForStudent 如下所示:

 public String getInfoForStudent (int studentIndex)
    {
        return sts.get(studentIndex).toString();
    }