So far I coded a JavaFX application in which some rectangles move around. Now I want to create a method to check whether a rectangle is still visible in the window or already moved out of it. My code looks like that:
import javafx.animation.AnimationTimer;
import javafx.application.Application;
import javafx.geometry.Point2D;
import javafx.scene.Node;
import javafx.scene.Scene;
import javafx.scene.layout.Pane;
import javafx.scene.shape.Rectangle;
import javafx.stage.Stage;
public class Test extends Application {
private Pane root = new Pane();
private Rectangle rect = new Rectangle(150,150,15,15);
private Point2D velocity = new Point2D(2,1);
private Pane createContent(){
root.setPrefSize(500,500);
root.getChildren().add(rect);
AnimationTimer timer = new AnimationTimer() {
@Override
public void handle(long now) {
update();
}
};
timer.start();
return root;
}
private void update(){
if (outOfWindow(rect)) {
System.out.println("out of window...\n");
}else {
System.out.println("in window...\n");
}
rect.setTranslateX(rect.getTranslateX() + velocity.getX());
rect.setTranslateY(rect.getTranslateY() + velocity.getY());
}
private boolean outOfWindow(Node node) {
if (node.getBoundsInParent().intersects(node.getBoundsInParent().getWidth(), node.getBoundsInParent().getHeight(),
root.getPrefWidth() - node.getBoundsInParent().getWidth() * 2,
root.getPrefHeight() - node.getBoundsInParent().getHeight() * 2)){
return false;
}
return true;
}
@Override
public void start(Stage primaryStage) throws Exception {
primaryStage.setScene(new Scene(createContent()));
primaryStage.show();
}
public static void main(String[] args) {
launch(args);
}
}
The outOfWindow() method is my attempt to check if the rectangle's position is still in the window. It works. But is there a better way or a way to detect which window border the rectangle crossed?