了JavaFx Shape相交始終返回false /真

Question

所以，我創建了2矩形，並檢查他們是否有交叉，但每次都得到不同的結果：

Rectangle a = new Rectangle(10.00, 10.00); 
    Rectangle b = new Rectangle(30.00, 20.00); 

    a.setFill(Color.RED); 
    _centerPane.getChildren().add(a); 
    _centerPane.getChildren().add(b); 

    //both at 0.00 
    System.out.println(a.intersects(b.getLayoutX(), b.getLayoutY(), b.getWidth(), b.getHeight())); //true 
    System.out.println(b.intersects(a.getLayoutX(), a.getLayoutY(), a.getWidth(), a.getHeight())); //true 

    a.setLayoutX(100.00); 
    a.setLayoutY(100.00); 

    //only a at different position and not visually intersecting 
    System.out.println(a.intersects(b.getLayoutX(), b.getLayoutY(), b.getWidth(), b.getHeight())); //true 
    System.out.println(b.intersects(a.getLayoutX(), a.getLayoutY(), a.getWidth(), a.getHeight())); //false 

    b.setLayoutX(73.00); 
    b.setLayoutY(100.00); 

    //Now b is set near a and intersects a visually 
    System.out.println(a.intersects(b.getLayoutX(), b.getLayoutY(), b.getWidth(), b.getHeight())); //false 
    System.out.println(b.intersects(a.getLayoutX(), a.getLayoutY(), a.getWidth(), a.getHeight())); //false 

Answer 1

這是因爲你混合layoutX和layoutY與x和y。如果您運行下面的代碼（我修改原密碼，並增加了一些打印語句），你會看到，雖然你改變layoutX和layoutY當你調用a.intersects(b.getLayoutX(), ...)這是測試如果a這是在（0,0），並有大小（10,10）與在（100,100）一個矩形相交，答案是，當然，沒有。

代替使用setLayoutX和getLayoutX（或Y）只使用setX/getX和setY/getY的。

public static void test(Rectangle a, Rectangle b) { 
    System.out.println(a.intersects(b.getLayoutX(), b.getLayoutY(), b.getWidth(), b.getHeight())); 
    System.out.println(b.intersects(a.getLayoutX(), a.getLayoutY(), a.getWidth(), a.getHeight())); 
} 

public static void print(String name, Rectangle r) { 
    System.out.println(name + " : " + r.toString() + " layoutX: " + r.getLayoutX() + " layoutY: " + r.getLayoutY()); 
} 

public static void main(String[] args) { 
    Rectangle a = new Rectangle(10.00, 10.00); 
    Rectangle b = new Rectangle(30.00, 20.00); 

    //both at 0.00 
    print("a", a); 
    print("b", b); 
    test(a, b); // true, true 

    a.setLayoutX(100.00); 
    a.setLayoutY(100.00); 

    //only a at different position and not visually intersecting 
    print("a", a); 
    print("b", b); 
    test(a, b); // true, false 

    b.setLayoutX(73.00); 
    b.setLayoutY(100.00); 

    //Now b is set near a and intersects a visually 
    print("a", a); 
    print("b", b); 
    test(a, b); // false, false 
}