How to check if rectangular node is in the window

Question

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?

Answer 1

If one rectangle is outside of the another one it means two things:

• they don't intersect
• bigger one doesn't fully contain the smaller one

So your method can look next way:

private boolean inTheWindow(Rectangle rect) {
    return rect.getBoundsInParent().intersects(root.getLayoutBounds()) 
        || root.getLayoutBounds().contains(rect.getBoundsInParent());
}

Here is MCVE to demonstrate:

public class FXBounds extends Application {

    private Pane root;

    @Override
    public void start(Stage primaryStage) {

        root = new Pane();

        Scene scene = new Scene(root, 300, 250);

        primaryStage.setTitle("Hello World!");
        primaryStage.setScene(scene);
        primaryStage.show();

        check(1,1,50,50, true);
        check(100,100,50,50, true);
        check(-5,-5,30,30, true);
        check(-50,0,30,30, false);
    }

    private void check(int x, int y, int width, int height, boolean in) {
        Rectangle rect = new Rectangle(x, y, width, height);
        root.getChildren().add(rect);
        System.out.println(in == inTheWindow(rect));
    }

    private boolean inTheWindow(Rectangle rect) {
        return rect.getBoundsInParent().intersects(root.getLayoutBounds()) || root.getLayoutBounds().contains(rect.getBoundsInParent());
    }
}