java - 我应该如何让我的开关回到主开关？

Question

美好的一天......我正在创建一个地址簿程序......我在我的主菜单中使用了一个开关......并计划为我的编辑菜单创建另一个开关......我的问题是......我不知道如何回到我的主开关……这是我的主程序:

import javax.swing.JOptionPane;
import javax.swing.JTextArea;

public class AddressBook {

    private AddressBookEntry entry[];
    private int counter;

    public static void main(String[] args) {
        AddressBook a = new AddressBook();
        a.entry = new AddressBookEntry[100];
        int option = 0;
        while (option != 6) {
            String content = "Choose an Option\n\n"
                    + "[1] Add an Entry\n"
                    + "[2] Delete an Entry\n"
                    + "[3] Update an Entry\n"
                    + "[4] View all Entries\n"
                    + "[5] View Specific Entry\n"
                    + "[6] Exit";
            option = Integer.parseInt(JOptionPane.showInputDialog(content));
            switch (option) {
                case 1:
                    a.addEntry();
                    break;
                case 2:

                    break;
                case 3:
                    a.editMenu();
                    break;
                case 4:
                    a.viewAll();
                    break;
                case 5:
                    break;
                case 6:
                    System.exit(1);
                    break;
                default:
                    JOptionPane.showMessageDialog(null, "Invalid Choice!");
            }
        }
    }

    public void addEntry() {
        entry[counter] = new AddressBookEntry();
        entry[counter].setName(JOptionPane.showInputDialog("Enter name: "));
        entry[counter].setAdd(JOptionPane.showInputDialog("Enter add: "));
        entry[counter].setPhoneNo(JOptionPane.showInputDialog("Enter Phone No.: "));
        entry[counter].setEmail(JOptionPane.showInputDialog("Enter E-mail: "));
        counter++;
    }

    public void viewAll() {
        int i = 0;
        for (; i < counter; i++) {
            JOptionPane.showMessageDialog(null, new JTextArea(entry[i].getInfo()));
        }
    }

    public void editMenu() {
        int option = 0;
        while (option != 6) {
            String content = "Choose an Option\n\n"
                    + "[1] Edit Name\n"
                    + "[2] Edit Address\n"
                    + "[3] Edit Phone No.\n"
                    + "[4] Edit E-mail address\n"
                    + "[5] Go Back to Main Menu\n";
            option = Integer.parseInt(JOptionPane.showInputDialog(content));
            switch (option) {
                case 1:
                    editEntry();
                    break;
                case 2:
                    break;
                case 3:
                    break;
                case 4:
                    break;
                case 5:
                    break;
                default:
                    JOptionPane.showMessageDialog(null, "Invalid Choice!");
            }
        }
    }

    public void editEntry() {
        String EName;
        EName = JOptionPane.showInputDialog("Enter name to edit: ");
        for (int i = 0; i < counter; i++) {
            if (entry[i].getName().equals(EName)) {
                entry[i].setName(JOptionPane.showInputDialog("Enter new name: "));
            }
        }
    }
}

请帮助...提前致谢:)

Answer 1

要返回给调用者，您可以使用 return;而不是 break;
在编辑菜单中

            case 5:
                return;

但是，我怀疑你的问题是

while (option != 6) {

应该

while (option != 5) {

您还可以使用标签来跳出 while 循环，这在这里会做同样的事情。