Java Swing取消无限循环

时间:2012-11-10 23:28:21

标签: java swing

我遇到了Swing中的无限循环问题。完成一些研究并遇到SwingWorker线程,但不确定如何实现它们。我把一个显示问题的简单程序组合在一起。一个按钮启动无限循环,我希望另一个按钮停止它,但当然由于Swing单线程问题,另一个按钮已冻结。下面的代码,并帮助赞赏: -

public class Model
{
    private int counter;
    private boolean go = true;

    public void go()
    {
        counter = 0;

        while(go)
        {
            counter++;
            System.out.println(counter);
        }
    }

    public int getCounter()
    {
        return counter;
    }

    public void setGo(boolean value)
    {
        this.go = value;
    }
}

public class View extends JFrame
{
    private JPanel                  topPanel, bottomPanel;
    private JTextArea               messageArea;
    private JButton                 startButton, cancelButton;
    private JLabel                  messageLabel;
    private JScrollPane             scrollPane;

    public View()
    {
        setSize(250, 220);
        setTitle("View");
        setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
        topPanel = new JPanel();
        bottomPanel = new JPanel();
        messageArea = new JTextArea(8, 20);
        messageArea.setEditable(false);
        scrollPane = new JScrollPane(messageArea);
        messageLabel = new JLabel("Message Area");
        topPanel.setLayout(new BorderLayout());
        topPanel.add(messageLabel, "North");
        topPanel.add(scrollPane, "South");
        startButton = new JButton("START");
        cancelButton = new JButton("CANCEL");
        bottomPanel.setLayout(new GridLayout(1, 2));
        bottomPanel.add(startButton);
        bottomPanel.add(cancelButton);
        Container cp = getContentPane();
        cp.add(topPanel, BorderLayout.NORTH);
        cp.add(bottomPanel, BorderLayout.SOUTH);
    }

    public JButton getStartButton()
    {
        return startButton;
    }

    public JButton getCancelButton()
    {
        return cancelButton;
    }

    public void setMessageArea(String message)
    {
        messageArea.append(message + "\n");
    }
}


public class Controller implements ActionListener
{
    private Model theModel;
    private View  theView;

    public Controller(Model model, View view)
    {
        this.theModel = model;
        this.theView = view;
        view.getStartButton().addActionListener(this);
        view.getCancelButton().addActionListener(this);
    }

    public void actionPerformed(ActionEvent ae)
    {
        Object buttonClicked = ae.getSource();
        if(buttonClicked.equals(theView.getStartButton()))
        {
            theModel.go();
        }
        else if(buttonClicked.equals(theView.getCancelButton()))
        {
            theModel.setGo(false);
        }
    }
}



public class Main
{
    public static void main(String[] args)
    {
        Model model = new Model();
        View view = new View();
        Controller controller = new Controller(model, view);
        view.setVisible(true);
    }
}

4 个答案:

答案 0 :(得分:3)

您正在屏蔽Event Dispatch Thread(EDT)。该线程负责处理绘画和其他UI相关请求。一旦EDT被阻止,UI将被冻结,因为它无法处理任何事件。有关更多详细信息,请参阅The Event Dispatch Thread教程。

考虑使用计时器(How to Use Swing Timers),SwingWorker或辅助后台线程。后台线程可以使用SwingUtilities.invokeLater()与EDT通信。此机制已在SwingWorker中实现,因此可能更容易使用它。这取决于所需的功能。

答案 1 :(得分:3)

在事件处理中使用javax.swing.Timer和[{1}},使用go()工作start()一次(有一些可选的延迟)。

答案 2 :(得分:3)

您无需实施任何计时器即可轻松完成,只需在 actionPerformed 方法中添加两行:

public void actionPerformed(ActionEvent ae)
{
    Object buttonClicked = ae.getSource();
    if(buttonClicked.equals(theView.getStartButton()))
    {
      theModel.setGo(true); //make it continue if it's just stopped
      Thread t = new Thread(new Runnable() { public void run() {theModel.go();}}); //This separate thread will start the new go...
      t.start(); //...when you start the thread! go!
    }
    else if(buttonClicked.equals(theView.getCancelButton()))
    {
        theModel.setGo(false);
    }
}

Model.go()在一个单独的线程中运行时,事件调度线程可以自由地执行其操作,例如再次绘制按钮,而不是按下按钮。

有一个问题! 但是,因为运行 Model.go()的线程会疯狂地运行 !< / em> ,它几乎被称为系统可以每秒多次。

如果您计划实施某些动画等,则需要:

  • 使用计时器,

如果您使用线程示例:

public void go()
{
    counter = 0;
        while(go)
    {
        counter++;
        System.out.println(counter);
        try {
            Thread.sleep(1500); //Sleep for 1.5 seconds
        } catch (InterruptedException e) {
            e.printStackTrace();
        }
    }
}

正如您所看到的,我将 Thread.sleep(1500)添加为1500毫秒(1.5秒)。由于某些原因,Thread.sleep可能会被中断,因此您必须捕获 InterruptedException

在这种特殊情况下,没有必要更深入地处理 InterruptedException ,但如果您对此感到好奇,可以阅读nice article

答案 3 :(得分:0)

我决定使用SwingWorker线程,下面是更新的Controller类。它做了我需要做的事情,但我的问题是,它是正确的方式,它是干净的代码吗?另外,我已经尝试将model.go()方法的输出按照注释掉的行添加到视图的textarea中,但没有成功,任何人都知道如何?

public class Controller implements ActionListener
{
private Model theModel;
private View  theView;
private SwingWorker<Void, Void> worker;

public Controller(Model model, View view)
{
    this.theModel = model;
    this.theView = view;
    view.getStartButton().addActionListener(this);
    view.getCancelButton().addActionListener(this);
}

public void actionPerformed(ActionEvent ae)
{
    Object buttonClicked = ae.getSource();
    if(buttonClicked.equals(theView.getStartButton()))
    {
        theModel.setGo(true);
        worker = new SwingWorker<Void, Void>()
        {
            @Override
            protected Void doInBackground()
            {
                // theView.setMessageArea(theModel.getCounterToString());
                return theModel.go();
            }
            @Override
            protected void done()
            {
                // theView.setMessageArea(theModel.getCounterToString());
            }
        };
        worker.execute();

    }
    else if(buttonClicked.equals(theView.getCancelButton()))
    {
        theModel.setGo(false);
    }
}
}


public class Model
{
    public Void go()
    {
        counter = 0;

        while(go)
        {
            counter++;
            System.out.println(counter);
        }
        return null;
    }