我正在用 java 制作一个非常简单的乒乓球游戏，我正在使用 KeyListener 来做这个。我想要它，所以当用户按下键盘上的左右键时，乒乓球块会朝那个方向移动。这是一个足够简单的任务，但我发现当用户按住键时，块移动一次，停止一小段时间，然后继续移动直到用户释放键。我注意到当您尝试按住计算机上的字母键时会发生这种情况。如果我尝试按住“a”键，计算机将执行以下操作：
[暂停] aaaaaaaaaaaaaaa
有什么办法可以禁用这种口吃，因为它妨碍了我的小游戏流畅的游戏玩法。快速修复将不胜感激。
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5705 次
3 回答
9
我最初有一个关于 Key Bindings 的答案，但经过一些测试后，我发现他们仍然有同样的口吃问题。
不要依赖操作系统的重复率。每个平台都可能不同，用户也可以自定义它。
而是使用计时器来安排事件。您在 keyPressed 上启动 Timer，在 keyReleased 上停止 Timer。
import java.awt.*;
import java.awt.event.*;
import java.net.*;
import java.util.Map;
import java.util.HashMap;
import javax.imageio.ImageIO;
import javax.swing.*;
public class KeyboardAnimation implements ActionListener
{
private final static String PRESSED = "pressed ";
private final static String RELEASED = "released ";
private final static Point RELEASED_POINT = new Point(0, 0);
private JComponent component;
private Timer timer;
private Map<String, Point> pressedKeys = new HashMap<String, Point>();
public KeyboardAnimation(JComponent component, int delay)
{
this.component = component;
timer = new Timer(delay, this);
timer.setInitialDelay( 0 );
}
public void addAction(String keyStroke, int deltaX, int deltaY)
{
// InputMap inputMap = component.getInputMap(JComponent.WHEN_IN_FOCUSED_WINDOW);
InputMap inputMap = component.getInputMap();
ActionMap actionMap = component.getActionMap();
String pressedKey = PRESSED + keyStroke;
KeyStroke pressedKeyStroke = KeyStroke.getKeyStroke( pressedKey );
Action pressedAction = new AnimationAction(keyStroke, new Point(deltaX, deltaY));
inputMap.put(pressedKeyStroke, pressedKey);
actionMap.put(pressedKey, pressedAction);
String releasedKey = RELEASED + keyStroke;
KeyStroke releasedKeyStroke = KeyStroke.getKeyStroke( releasedKey );
Action releasedAction = new AnimationAction(keyStroke, RELEASED_POINT);
inputMap.put(releasedKeyStroke, releasedKey);
actionMap.put(releasedKey, releasedAction);
}
private void handleKeyEvent(String keyStroke, Point moveDelta)
{
// Keep track of which keys are pressed
if (RELEASED_POINT == moveDelta)
pressedKeys.remove( keyStroke );
else
pressedKeys.put(keyStroke, moveDelta);
// Start the Timer when the first key is pressed
if (pressedKeys.size() == 1)
{
timer.start();
}
// Stop the Timer when all keys have been released
if (pressedKeys.size() == 0)
{
timer.stop();
}
}
// Invoked when the Timer fires
public void actionPerformed(ActionEvent e)
{
moveComponent();
}
// Move the component to its new location
private void moveComponent()
{
int componentWidth = component.getSize().width;
int componentHeight = component.getSize().height;
Dimension parentSize = component.getParent().getSize();
int parentWidth = parentSize.width;
int parentHeight = parentSize.height;
// Calculate new move
int deltaX = 0;
int deltaY = 0;
for (Point delta : pressedKeys.values())
{
deltaX += delta.x;
deltaY += delta.y;
}
// Determine next X position
int nextX = Math.max(component.getLocation().x + deltaX, 0);
if ( nextX + componentWidth > parentWidth)
{
nextX = parentWidth - componentWidth;
}
// Determine next Y position
int nextY = Math.max(component.getLocation().y + deltaY, 0);
if ( nextY + componentHeight > parentHeight)
{
nextY = parentHeight - componentHeight;
}
// Move the component
component.setLocation(nextX, nextY);
}
private class AnimationAction extends AbstractAction implements ActionListener
{
private Point moveDelta;
public AnimationAction(String keyStroke, Point moveDelta)
{
super(PRESSED + keyStroke);
putValue(ACTION_COMMAND_KEY, keyStroke);
this.moveDelta = moveDelta;
}
public void actionPerformed(ActionEvent e)
{
handleKeyEvent((String)getValue(ACTION_COMMAND_KEY), moveDelta);
}
}
public static void main(String[] args)
{
JPanel contentPane = new JPanel();
contentPane.setLayout( null );
Icon dukeIcon = null;
try
{
dukeIcon = new ImageIcon( "dukewavered.gif" );
// dukeIcon = new ImageIcon( ImageIO.read( new URL("http://duke.kenai.com/iconSized/duke4.gif") ) );
}
catch(Exception e)
{
System.out.println(e);
}
JLabel duke = new JLabel( dukeIcon );
duke.setSize( duke.getPreferredSize() );
duke.setLocation(100, 100);
contentPane.add( duke );
KeyboardAnimation navigation = new KeyboardAnimation(duke, 24);
navigation.addAction("LEFT", -3, 0);
navigation.addAction("RIGHT", 3, 0);
navigation.addAction("UP", 0, -3);
navigation.addAction("DOWN", 0, 3);
navigation.addAction("A", -5, 0);
navigation.addAction("S", 5, 0);
navigation.addAction("Z", 0, -5);
navigation.addAction("X", 0, 5);
navigation.addAction("V", 5, 5);
JFrame frame = new JFrame();
frame.setDefaultCloseOperation( JFrame.EXIT_ON_CLOSE );
// frame.getContentPane().add(new JTextField(), BorderLayout.SOUTH);
frame.getContentPane().add(contentPane);
frame.setSize(600, 600);
frame.setLocationRelativeTo( null );
frame.setVisible(true);
}
}
此代码在 Windows 上进行了测试，其中事件的顺序为 keyPressed、keyPressed、keyPressed...keyReleased。
但是，我认为在 Mac（或 Unix）上，事件的顺序是 keyPressed、keyReleased、keyPressed、keyReleased ......所以我不确定这段代码是否会比你当前的代码更好。
于 2013-05-02T02:12:37.580 回答
5
- 您应该使用键绑定，因为它们解决了大多数与焦点相关的问题，并且通常更灵活......
- 您需要定义一个标志来指示何时按下某个键。按下时，您不应执行任何其他任务...
举些例子...
- Java 对象移动
- 使用键盘在java中以角度移动圆圈
- Java 的 Paint 方法的问题，荒谬的刷新速度，它显示了一种在按下键时加速对象的机制；）
用简单的例子更新
在大多数游戏中，您应该对“状态变化”做出反应，而不是对实际的关键事件做出反应。这意味着实际改变状态的事件可以是可变的（想想自定义键）
import java.awt.BorderLayout;
import java.awt.Dimension;
import java.awt.EventQueue;
import java.awt.GridBagLayout;
import java.awt.event.ActionEvent;
import java.awt.event.KeyEvent;
import javax.swing.AbstractAction;
import javax.swing.ActionMap;
import javax.swing.InputMap;
import javax.swing.JFrame;
import javax.swing.JLabel;
import javax.swing.JPanel;
import javax.swing.KeyStroke;
import javax.swing.UIManager;
import javax.swing.UnsupportedLookAndFeelException;
public class SinglePressKeyBinding {
public static void main(String[] args) {
new SinglePressKeyBinding();
}
public SinglePressKeyBinding() {
EventQueue.invokeLater(new Runnable() {
@Override
public void run() {
try {
UIManager.setLookAndFeel(UIManager.getSystemLookAndFeelClassName());
} catch (ClassNotFoundException | InstantiationException | IllegalAccessException | UnsupportedLookAndFeelException ex) {
}
JFrame frame = new JFrame("Testing");
frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
frame.setLayout(new BorderLayout());
frame.add(new TestPane());
frame.pack();
frame.setLocationRelativeTo(null);
frame.setVisible(true);
}
});
}
public class TestPane extends JPanel {
private JLabel message;
private boolean spacedOut = false;
public TestPane() {
message = new JLabel("Waiting");
setLayout(new GridBagLayout());
add(message);
InputMap im = getInputMap(WHEN_IN_FOCUSED_WINDOW);
ActionMap am = getActionMap();
im.put(KeyStroke.getKeyStroke(KeyEvent.VK_SPACE, 0, false), "space-pressed");
im.put(KeyStroke.getKeyStroke(KeyEvent.VK_SPACE, 0, true), "space-released");
am.put("space-pressed", new AbstractAction() {
@Override
public void actionPerformed(ActionEvent e) {
if (spacedOut) {
message.setText("I'm ignoring you");
} else {
spacedOut = true;
message.setText("Spaced out");
}
}
});
am.put("space-released", new AbstractAction() {
@Override
public void actionPerformed(ActionEvent e) {
spacedOut = false;
message.setText("Back to earth");
}
});
}
@Override
public Dimension getPreferredSize() {
return new Dimension(200, 200);
}
}
}
于 2013-05-02T01:34:57.070 回答
1
一个好主意是为要跟踪的键设置布尔值，然后在 keypressed 事件上激活其中一个布尔值，并在释放键时停用它。它将消除按键的滞后并允许多次按键！
于 2013-08-24T17:06:15.863 回答