Question

我正在尝试在java中创建一个老虎机。在这台老虎机中，我完成了项目的基本开始。我有动画师落在一个随机空间（樱桃，空白，七等），背景，按钮开始和赌注在开口，但我必须弄清楚插槽的结尾;如何在没有用户单击按钮的情况下显示投币机的结果。要做到这一点，我认为最好的方法是深入了解线程领域。然而，当我尝试创建一个简单的线程时，动画师停止工作，当我把线程拿出来时，动画师确实工作了。这是一个难题以下是一些代码，如果这有助于解释：

public class SlotMachineOpeningGraphic extends JPanel implements Runnable
{
    JLayeredPane layeredPane= new JLayeredPane ();
JFrame frame = new JFrame();

public Thread thread;
public volatile boolean running = false;

Player plyr = new Player ();

public static final long startTime = System.currentTimeMillis ();

JLabel numberBet = new JLabel ("" + plyr.getBet()); //things that change on JFrame
JLabel numberAccount = new JLabel (""+ plyr.getAccount ());
JButton betButton = new JButton ("Bet ++");
JButton playAgain = new JButton ("Start");

public SlotMachineOpeningGraphic()   
 {
    layeredPane.setPreferredSize(new Dimension(750, 460));//changes size to image with  border for buttons
                                                          //andlabels

    ImageIcon bG = new ImageIcon ("/Users/Documents/slotmachine.png");//background file
    JLabel backGround = new JLabel (bG);
    backGround.setBounds (70,0, bG.getIconWidth(), bG.getIconHeight());//won't display if you do not set bounds
    layeredPane.add (backGround, new Integer (0));//add to first layer

    /*add buttons and labels to give user options
     * and information, placed in layer 2 */

    playAgain.setBounds (110,420, 100, 25);
    layeredPane.add (playAgain, new Integer (2));
    playAgain.addActionListener (new Start());


    betButton.setBounds (320,420, 100, 25);
    layeredPane.add (betButton, new Integer (2));
    betButton.addActionListener (new SlotBB ());


    JButton mainMenu = new JButton ("Main Menu");
    mainMenu.setBounds (520,420, 100, 25);
    layeredPane.add (mainMenu, new Integer (2));
    long checkTime = System.currentTimeMillis ();
    System.out.println ("Total execution time : " + (checkTime - startTime));

    JLabel bet = new JLabel ("Bet:");
    bet.setBounds (620, 320, 100, 30);

    Font font = new Font ("Corsiva Hebrew",bet.getFont().getStyle(),30);  //desired font    & size

    bet.setFont (font);
    layeredPane.add (bet, new Integer (2));


    numberBet.setBounds (620, 340, 100, 30);
    layeredPane.add (numberBet, new Integer (2));
    numberBet.setFont (font);

    JLabel account = new JLabel ("Account: ");
    account.setBounds (5, 320, 150, 30);
    account.setFont (font);
    layeredPane.add (account, new Integer (2));

    numberAccount.setBounds (5, 340, 150, 30);
    numberAccount.setFont (font);
    layeredPane.add (numberAccount, new Integer (2));


     add(layeredPane);

    JComponent newContentPane = layeredPane;

    frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
    newContentPane.setOpaque(true); //content panes must be opaque
    frame.setContentPane(newContentPane);
    frame.setBackground (Color.white);

    //Display the window.
    frame.pack();
    frame.setVisible(true);

}




public void run ()
{
  try {
    Thread.sleep (20);
    System.out.println ("works");
  }
  catch (Exception e) {
    System.out.println ("doesn't work");
  }

}




class Start extends JPanel implements ActionListener
{
  public void actionPerformed (ActionEvent event)
  { 
    SlotAnimator a0 = new SlotAnimator (40);
    a0.setBounds(155, 85, 100, 90); 
    layeredPane.add (a0, new Integer (1));

    SlotAnimator a1 = new SlotAnimator (85);
    a1.setBounds(320, 85, 100, 90); 
    layeredPane.add (a1, new Integer (1));

    SlotAnimator a2 = new SlotAnimator (135);
    a2.setBounds(470, 85, 100, 90); 
    layeredPane.add (a2, new Integer (1));
    playAgain.setText ("play again?");
    hearSound();

    thread = new Thread (new SlotMachineOpeningGraphic ());
    thread.start ();

  }
}



 public static void main (String [] args)
 {
   new SlotMachineOpeningGraphic();
 }
}

有关方法的任何建议吗？非常感谢！

Answer 1

这至少是你的一个错误：

thread = new Thread(new SlotMachineOpeningGraphic());

您正在创建一个新的SlotMachineOpeningGraphic对象，该对象与原始对象完全不同。不要这样做。如果你必须在这里使用SlotMachineOpeningGraphic对象，那么你应该使用对已经存在的原始类的引用并保存Start private内部类，例如：

thread = new Thread(SlotMachineOpeningGraphic.this);

所以这不是一个“线程”问题本身，而是一个引用问题 - 你使用错误的引用作为你的线程的Runnable。

但是我强烈建议你不要在ActionListeners，OtherListeners或Runnables中使用你的GUI类，因为你要求类做太多。最好使用匿名内部类或单独的独立类。

Answer 2

@HFOE正确使用javax.swing.Timer，它管理自己的线程以安全的方式生成定期的GUI事件。您可能还希望考虑采用Model-View-Controller模式，例如here。

Answer 3

您可以尝试使用Future或Runnable类进行处理，并在完成后立即检索结果。只需确保将这些线程放在线程池中即可！

Java线程问题

3 个答案: