Java Smooth 2D矩形碰撞？

Question

大家好，我正在做一個迷宮游戲，我想弄清楚如何使2D碰撞變得流暢。 目前，我有一些非常糟糕的碰撞代碼，因為我還沒有做完這部分，所以它會出現故障，有時會撞到牆內，那么您就無法退出。

我的代碼：

    public void checkCollision() {
    Rectangle player_rectangle = new Rectangle(player.getX(),player.getY(),32,32);

    for(Wall wall : walls) {

        Rectangle wall_rectangle = new Rectangle(wall.getX(), wall.getY(), 32,32);

        if(player_rectangle.intersects(wall_rectangle)) {
            if(player.xspeed == 1) {
                player.xspeed = 0;
                player.x -= 1.2;
            } else {

                if(player.xspeed == -1) {
                    player.xspeed = 0;
                    player.x += 1.2;
                }
                else
                    if(player.yspeed == 1) {
                        player.yspeed = 0;
                        player.y -= 1;
                    } else {
                        if(player.yspeed == -1) {
                            player.yspeed = 0;
                            player.y += 1;
                        }
                    }

            }
        }

    }

}

有什么更好的方法來進行碰撞？ 這是我將要擁有的地圖類型的圖片：

編輯：

新的碰撞代碼：

            if (player_rectangle.intersects(wall_rectangle)) {
            Rectangle intersection = (Rectangle) player_rectangle.createIntersection(wall_rectangle);

            if (player.xspeed > 0) {
                player.x -= intersection.getWidth();
            }

            if (player.yspeed > 0) {
                player.y -= intersection.getHeight();
            }

            if (player.xspeed < 0) {
                player.x += intersection.getWidth();
            }

            if (player.yspeed < 0) {
                player.y += intersection.getHeight(); 
            }

            Print(Integer.toString(intersection.width) + ", " + Integer.toString(intersection.height));

        }

運動代碼：

    public void keyPressed(KeyEvent e) {
    int key = e.getKeyCode();

    if(key == KeyEvent.VK_LEFT) {

        xspeed = -1;

    } else {
        if(key == KeyEvent.VK_RIGHT) {

                xspeed = 1;

        } else {
            if(key == KeyEvent.VK_UP) {
                yspeed = -1;
            } else {
                if(key == KeyEvent.VK_DOWN) {
                    yspeed = 1;
                }
            }
        }
    }
}

感謝你給與我的幫助。

Answer 1

您可以使用Rectangle.createIntersection(Rectangle) 。

如果玩家向正方向移動，只需從其位置減去所得Rectangle的寬度和高度，如果向負方向移動，則將其相加。

不要將速度設置為零，這可能會導致播放器掛起。

編輯

if (playerRectangle.intersects(wallRectangle) {
    Rectangle intersection = (Rectangle) playerRectangle.createIntersection(wallRectangle);

    if (player.xspeed > 0) {
        player.x -= intersection.getWidth();
    }

    if (player.yspeed > 0) {
        player.y -= intersection.getHeight();
    }

    if (player.xspeed < 0) {
        player.x += intersection.getWidth();
    }

    if (player.yspeed < 0) {
        player.y += intersection.getHeight();
    }
}

編輯2

據我所知，keyPressed僅在您按下某個鍵時觸發，而不是在您釋放它時觸發。 像這樣嘗試：

private boolean up, down, left, right;

public void keyPressed(KeyEvent e) {
    if (e.getKeyCode() == KEyEvent.VK_UP) {
        up = true;
    }

    ...
}

public void keyReleased(KeyEvent e) {
    if (e.getKeyCode() == KEyEvent.VK_UP) {
        up = false;
    }

    ...
}


// Call this in the mainthread
public void updateMovement() {
    if (up) {
        player.xspeed = 1;
    }

    ...
}

因此，在每一幀中，您都要更新動作，移動播放器，然后檢查碰撞。