在 JavaFX 中检测单键按下

Question

我在检测 JavaFX 中的单键按下时遇到问题。我必须检测箭头键，但每次我按下这些键中的任何一个时，我的代码部分都会被多次调用。我意识到这是因为 AnimationTimer() 是一个循环，所以这就是原因，但我不知道如何检测单键点击。无论如何，这是代码：

import javafx.animation.AnimationTimer;
import javafx.application.Application;
import javafx.event.EventHandler;
import javafx.scene.Group;
import javafx.scene.Scene;
import javafx.scene.canvas.Canvas;
import javafx.scene.canvas.GraphicsContext;
import javafx.scene.input.KeyEvent;
import javafx.scene.paint.Color;
import javafx.stage.Stage;

import java.util.HashSet;

public class Main extends Application {

    Stage window;
    static Scene scene;

    private static char[][] mapa = {
            {'X','X','X','X','X'},
            {'X','.','.','.','X'},
            {'X','.','M','.','X'},
            {'X','.','.','.','X'},
            {'X','X','X','X','X'}
    };

    private final int sizeX = 16;
    private final int sizeY = 16;

    static HashSet<String> currentlyActiveKeys;

    @Override
    public void start(Stage primaryStage) throws Exception {

        window = primaryStage;
        window.setTitle("Hello World");

        Group root = new Group();
        scene = new Scene(root);

        window.setScene(scene);

        Canvas canvas = new Canvas(512 - 64, 256);
        root.getChildren().add(canvas);

        prepareActionHandlers();

        GraphicsContext gc = canvas.getGraphicsContext2D();

        new AnimationTimer() {
            @Override
            public void handle(long now) {
                gc.clearRect(0,0,512,512);

                for(int i = 0; i < mapa.length; i++) {
                    for(int j = 0; j < mapa[i].length; j++) {
                        if(mapa[i][j] == 'X') {
                            gc.setLineWidth(5);
                            gc.setFill(Color.RED);
                            gc.fillRect(sizeX + i*sizeX, sizeY + j*sizeY, sizeX, sizeY);
                        } else if(mapa[i][j] == '.') {
                            gc.setLineWidth(5);
                            gc.setFill(Color.GREEN);
                            gc.fillRect(sizeX + i*sizeX, sizeY + j*sizeY, sizeX, sizeY);
                        } else if(mapa[i][j] == 'M') {
                            gc.setLineWidth(5);
                            gc.setFill(Color.BLACK);
                            gc.fillRect(sizeX + i*sizeX, sizeY + j*sizeY, sizeX, sizeY);
                        }
                    }
                }

                if(currentlyActiveKeys.contains("LEFT")) {
                    System.out.println("left");
                }

                if(currentlyActiveKeys.contains("RIGHT")) {
                    System.out.println("right");
                }

                if(currentlyActiveKeys.contains("UP")) {
                    System.out.println("up");
                }

                if(currentlyActiveKeys.contains("DOWN")) {
                    System.out.println("down");
                }
            }
        }.start();

        window.show();
    }

    private static void prepareActionHandlers()
    {
        currentlyActiveKeys = new HashSet<String>();
        scene.setOnKeyPressed(new EventHandler<KeyEvent>() {
            @Override
            public void handle(KeyEvent event)
            {
                currentlyActiveKeys.add(event.getCode().toString());

            }
        });
        scene.setOnKeyReleased(new EventHandler<KeyEvent>() {
            @Override
            public void handle(KeyEvent event)
            {
                currentlyActiveKeys.remove(event.getCode().toString());

            }
        });
    }

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

当我按下箭头按钮时（当然其他键也是如此），在范围内我得到如下结果：

down
down
down
down
down
down

显然只要我按下按钮就会发生这种情况。一旦我停止按下它，打印就结束了。我想要实现的是，一旦我按下一个按钮（无论我按住它多久），我都会得到 down 一次。我需要这个，因为我想更新 canvas.

中矩形的颜色

Answer 1

您可以声明一个全局布尔值并将其切换为 press = !press。然后每 2 秒将其重置为 true 或类似的东西。

if(currentlyActiveKeys.contains("DOWN") && press) {
       press = !press;//sets it false
       System.out.println("down");
 }

Answer 2

发生的事情是 OS 具有一种自动键入功能，这样当您按住某个键时，它会不断生成按键事件，即使您并没有真正按下该键不止一次。

通过添加按键和按键释放处理程序以及每个键的布尔状态，您可以跟踪自按下键后是否处理过该键。然后，只要释放键，您就可以重置已处理的状态，以便下次真正按下时可以处理它。

示例应用程序

import javafx.animation.AnimationTimer;
import javafx.application.Application;
import javafx.scene.Group;
import javafx.scene.Scene;
import javafx.stage.Stage;

import java.util.HashMap;

public class Main extends Application {
    private HashMap<String, Boolean> currentlyActiveKeys = new HashMap<>();

    @Override
    public void start(Stage stage) throws Exception {
        final Scene scene = new Scene(new Group(), 100, 100);
        stage.setScene(scene);

        scene.setOnKeyPressed(event -> {
            String codeString = event.getCode().toString();
            if (!currentlyActiveKeys.containsKey(codeString)) {
                currentlyActiveKeys.put(codeString, true);
            }
        });
        scene.setOnKeyReleased(event -> 
            currentlyActiveKeys.remove(event.getCode().toString())
        );

        new AnimationTimer() {
            @Override
            public void handle(long now) {
                if (removeActiveKey("LEFT")) {
                    System.out.println("left");
                }

                if (removeActiveKey("RIGHT")) {
                    System.out.println("right");
                }

                if (removeActiveKey("UP")) {
                    System.out.println("up");
                }

                if (removeActiveKey("DOWN")) {
                    System.out.println("down");
                }
            }
        }.start();

        stage.show();
    }

    private boolean removeActiveKey(String codeString) {
        Boolean isActive = currentlyActiveKeys.get(codeString);

        if (isActive != null && isActive) {
            currentlyActiveKeys.put(codeString, false);
            return true;
        } else {
            return false;
        }
    }

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

Answer 3

我知道这已经是一个老问题了。经过数小时的努力，我只想分享我的发现。我终于找到了一个简单的解决方案，即添加 clear 函数。这就是我的意思

if(currentlyActiveKeys.contains("LEFT")) {
    System.out.println("left");
}

if(currentlyActiveKeys.contains("RIGHT")) {
    System.out.println("right");
}

if(currentlyActiveKeys.contains("UP")) {
    System.out.println("up");
}

if(currentlyActiveKeys.contains("DOWN")) {
    System.out.println("down");
}

currentlyActiveKeys.clear();

这最后一行将清除当前活动的密钥，从而防止重复post

在 JavaFX 中检测单键按下

Detect single key press in JavaFX

java

javafx

keyboard-events