Question

如果调用theView.continueToGame()，为什么屏幕不会使用新面板（由cont()调用）更新？如果我评论ask()调用cont()的位置，它似乎有效。有人可以解释为什么会这样吗？似乎循环中的东西搞砸了。

Driver.java

public class Driver {

    public static void main(String[] args)
    {
        Controller con = new Controller();
        con.ask();
    }

}

Controller.java

public class Controller {

    private View theView = new View();
    private Model theModel = new Model();

    public void ask()
    {
        theView.displayMenu();

        cont();

        System.out.println("ready");

        theView.continueToGame();


    }

    private void cont()
    {
        Scanner stdin = new Scanner(System.in);

        int input = 0;

        while(!(input == 1))
        {
            System.out.println("Enter 1 to continue");
            input = 0;
            try 
            {
                input = stdin.nextInt();
            } 
            catch (InputMismatchException e) 
            {
                System.out.println("error");
                stdin.next();
            }

        }

        stdin.close();
    }
}

View.java

public class View extends JFrame {

    /**
     * Serial id
     */
    private static final long serialVersionUID = 1L;

    private String String1 = "1";
    private String String2 = "2";

    View()
    {
        setVisible(true);
        setTitle("Tic-Tac-Toe");
        setSize(400,400);
        setDefaultCloseOperation(EXIT_ON_CLOSE);
    }

    public void displayMenu()
    {
        this.add(new startMenu());
    }

    public void continueToGame()
    {
        this.getContentPane().removeAll();
        this.add(new gameScreen());
    }

    class startMenu extends JPanel{

        /**
         * Serial id
         */
        private static final long serialVersionUID = 1L;

        private startMenu()
        {
            setVisible(true);
        }

        public void paintComponent(Graphics g)
        {
            super.paintComponent(g);
            setBackground(Color.blue);

            g.setColor(Color.black);
            g.drawString(String1, this.getWidth()/2, this.getHeight()/2);
        }

    }

    class gameScreen extends JPanel
    {

        /**
         * Serial id
         */
        private static final long serialVersionUID = 1L;

        private gameScreen()
        {
            setVisible(true);
        }

        public void paintComponent(Graphics g)
        {
            super.paintComponent(g);
            setBackground(Color.green);

            g.setColor(Color.black);

            g.drawString(String2, this.getWidth()/2, this.getHeight()/2);
        }

    }



}

编辑：

将cont()更改为

private void cont()
{
    Integer input = -1;

    while(!(input == 0))
    {

        input = JOptionPane.showConfirmDialog(theView, "Continue?", null, JOptionPane.YES_OPTION);
        System.out.println(input);
    }


}

无论是

还是

Answer 1

您遇到了一个线程问题，阻塞方法会阻止您的GUI事件线程冻结您的程序。这是因为您正在尝试将控制台程序与其线性程序逻辑与事件驱动的GUI程序相结合。

解决方案很简单：不要这样做。摆脱你的new Scanner(System.in)，只能通过GUI以事件驱动的方式获取用户输入。您可以使用JOptionPane或JDialog来获取此输入，要么效果很好，要么不是new Scanner(System.in)。我自己，我只是使用JOptionPane.showConfirmDialog(...)。

作为旁注，您正在使用类名称，例如View和Controller，就像您计划进行模型 - 视图 - 控制类型的程序设计一样， GREAT < / strong>想法，如果你问我，但是控件应该处理GUI的用户输入，而不是控制台。

修改
我错了 - 你的问题是你在交换组件后没有在容器上调用revalidate()和repaint()。即，

public void continueToGame() { this.getContentPane().removeAll(); this.add(new gameScreen()); revalidate(); // tells layout managers to layout new components repaint(); // redraw everything }

最好不要担心这些事情并使用CardLayout来交换您的视图JPanel。

编辑2
CardLayout实际上非常易于使用，但如果您将其添加到JFrame，您实际上是将其添加到contentPane，并且在调用CardLayout对象的show方法时必须使用contentPane。例如：

import java.awt.CardLayout; import java.awt.Color; import java.awt.Dimension; import java.awt.Graphics; import javax.swing.*; public class Driver { public static void main(String[] args) { Controller con = new Controller(); con.ask(); } } class Controller { private View theView = new View(); private Model theModel = new Model(); public void ask() { theView.displayMenu(); if (cont(theView)) { System.out.println("ready"); theView.setView(View.GAME); } } private boolean cont(View theView) { int result = JOptionPane.showConfirmDialog(theView, "Go on to game?"); return result == JOptionPane.YES_OPTION; } } class View extends JFrame { private static final long serialVersionUID = 1L; public static final String START = "start"; public static final String GAME = "game"; private String String1 = "1"; private String String2 = "2"; private CardLayout cardLayout = new CardLayout(); View() { // setVisible(true); // don't call this til all added to gui setTitle("Tic-Tac-Toe"); // setSize(400, 400); setDefaultCloseOperation(EXIT_ON_CLOSE); getContentPane().setLayout(cardLayout); add(new StartMenu(), START); add(new GameScreen(), GAME); pack(); setVisible(true); } public void displayMenu() { this.add(new StartMenu()); } public void setView(String constant) { cardLayout.show(getContentPane(), constant); } // class names should begin with an upper case letter class StartMenu extends JPanel { private static final int PREF_W = 400; private static final int PREF_H = PREF_W; private static final long serialVersionUID = 1L; private StartMenu() { setVisible(true); } @Override public Dimension getPreferredSize() { return new Dimension(PREF_W, PREF_H); } public void paintComponent(Graphics g) { super.paintComponent(g); setBackground(Color.blue); g.setColor(Color.black); g.drawString(String1, this.getWidth() / 2, this.getHeight() / 2); } } // class names should begin with an upper case letter class GameScreen extends JPanel { private static final long serialVersionUID = 1L; private GameScreen() { setVisible(true); } public void paintComponent(Graphics g) { super.paintComponent(g); setBackground(Color.green); g.setColor(Color.black); g.drawString(String2, this.getWidth() / 2, this.getHeight() / 2); } } } class Model { }

在调用扫描程序后，方法不会更新GUI

1 个答案: