如何使球弹出Javafx中的对象？ [英] How to make ball bounce off an object in Javafx?

问题描述

我现在有一个球从画布的墙上弹起来。我在屏幕中间添加了一个矩形。每当球与矩形碰撞时，我也想让它反弹，但我不知道该怎么做。我有一个叫做r的矩形。

我该如何让球体将矩形当作墙来处理，并在它碰到它时改变方向？代码示例将不胜感激。

<$ p
$ b

这是我的代码， c> public void handle（ActionEvent t）{

//根据X和Y的值来移动球体
circle.setLayoutX（circle.getLayoutX（）+ X）;
circle.setLayoutY（circle.getLayoutY（）+ Y）;

final Bounds bounds = canvas.getBoundsInLocal（）;
//检查墙是否被命中的布尔值
boolean leftWall = circle.getLayoutX（）< =（bounds.getMinX（）+ circle.getRadius（））;
boolean topWall = circle.getLayoutY（）< =（bounds.getMinY（）+ circle.getRadius（））;
boolean rightWall = circle.getLayoutX（）> =（bounds.getMaxX（） - circle.getRadius（））;
boolean bottomWall = circle.getLayoutY（）> =（bounds.getMaxY（） - circle.getRadius（））;



//如果触摸了底部或顶部墙，球会反转方向。
if（bottomWall || topWall）{

Y = Y * -1;
}
//如果触摸了左侧或右侧的墙，球会反转方向。
if（leftWall || rightWall）{
X = X * -1;
}
}

}））;

loop.setCycleCount（Timeline.INDEFINITE）;
loop.play（）;
}

解决方案

我不知道JavaFx但是这里是这样的想法：

$ p $ while（1）{
bool btop = pos.y> = top
bool bbottom = pos.y <= bottom
bool bleft = pos.x <= left
bool bright = pos.x> = right
bool rect_btop = pos。 y< = rect_top&& pos.x> = rect_left&& pos.x <= rect_right
bool rect_bbottom = pos.y <= rect_bottom&& pos.x> = rect_left&& pos.x <= rect_right
bool rect_bright = pos.x <= rect_right&& pos.y> = rect_bottom&& pos.y< = rect_top
bool rect_bleft = pos.x> = rect_left&& pos.y> = rect_bottom&& pos.y< = rect_top

if（btop || bottom || rect_btop || rect_bbottom）
vy - = vy

if（bleft || bright || rect_bleft || rect_bright）
vx - = vx
}


是一个更好的和可扩展的解决方案（用于编写突破砖块）。

I currently have a ball bouncing off the walls of the canvas. I added a rectangle in the middle of the screen. Whenever the ball collides with the rectangle, I want it to bounce off it too, but I don't know how to do that. I have a rectangle called "r".

How do I make the ball treat the rectangle as a wall and change direction whenever it hits it? Code examples will be greatly appreciated. Thanks.

Here's my code for the ball bouncing off the walls:

public void handle(ActionEvent t) {

                    // Moves the ball depending on the values of X and Y
                    circle.setLayoutX(circle.getLayoutX() + X);
                    circle.setLayoutY(circle.getLayoutY() + Y);                        

                    final Bounds bounds = canvas.getBoundsInLocal();
           // Boolean values to check if a wall has been hit
                    boolean leftWall = circle.getLayoutX() <= (bounds.getMinX() + circle.getRadius()); 
                    boolean topWall = circle.getLayoutY() <= (bounds.getMinY() + circle.getRadius());
                    boolean rightWall = circle.getLayoutX() >= (bounds.getMaxX() - circle.getRadius());
                    boolean bottomWall = circle.getLayoutY() >= (bounds.getMaxY() - circle.getRadius());



                    // If the bottom or top wall has been touched, the ball reverses direction.
                    if (bottomWall || topWall) {

                        Y = Y * -1;
                    }
                    // If the left or right wall has been touched, the ball reverses direction.
                    if (leftWall || rightWall) {
                        X = X * -1;
                    }
                }              

        }));

        loop.setCycleCount(Timeline.INDEFINITE);
        loop.play();
    }

解决方案

I don't know JavaFx but here is the idea :

while (1) {
    bool btop = pos.y >= top
    bool bbottom = pos.y <= bottom
    bool bleft = pos.x <= left
    bool bright = pos.x >= right
    bool rect_btop = pos.y <= rect_top && pos.x >= rect_left && pos.x <= rect_right
    bool rect_bbottom = pos.y <= rect_bottom && pos.x >= rect_left && pos.x <= rect_right
    bool rect_bright = pos.x <= rect_right && pos.y >= rect_bottom && pos.y <= rect_top
    bool rect_bleft = pos.x >= rect_left && pos.y >= rect_bottom && pos.y <= rect_top

    if (btop || bottom || rect_btop || rect_bbottom)
        vy -= vy

    if (bleft || bright || rect_bleft || rect_bright)
        vx -= vx
}

There are however far better and scalable solution (to code a breakout bricks).

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