我正在开发一款双人Pong游戏,并希望用一根手指控制蝙蝠。但是,我的问题是它一次只响应一个手指在屏幕上。整个游戏都在一个视图中运行。首先,我检查指针是否处于“活动状态”,然后检查它是否在屏幕的顶部或底部出现,然后再移动相应的蝙蝠。它只能正常工作,只有一个触摸寄存器。我的方法代码如下:
// variables
int _touchY = (int) event.getY();
boolean isTopSide = true;
// test if motionEvent was on the top or bottom of the screen
if (SCREEN_HEIGHT / 2 - _touchY < 0) {
isTopSide = false;
}
switch (event.getActionMasked()) {
case MotionEvent.ACTION_DOWN:
_activePointer = event.getPointerId(0);
break;
case MotionEvent.ACTION_MOVE:
if(_activePointer != INVALID_POINTER_ID) {
if (isTopSide) { // top bat
_topBat.moveBat(event);
} else { // bottom bat
_botBat.moveBat(event);
}
}
if(_newPointer != INVALID_POINTER_ID){
if (isTopSide) { // top bat
_topBat.moveBat(event);
} else { // bottom bat
_botBat.moveBat(event);
}
}
break;
case MotionEvent.ACTION_UP:
if(_activePointer != INVALID_POINTER_ID){
_activePointer = INVALID_POINTER_ID;
}
break;
case MotionEvent.ACTION_CANCEL:
_activePointer = INVALID_POINTER_ID;
break;
case MotionEvent.ACTION_POINTER_UP:
int pointerIndex = event.getActionIndex();
int pointerId = event.getPointerId(pointerIndex);
if(pointerId == _activePointer){
_activePointer = INVALID_POINTER_ID;
}
case MotionEvent.ACTION_POINTER_DOWN:
int newPointerIndex = event.getActionIndex();
_newPointer = event.getPointerId(newPointerIndex);
break;
}
相关的全球代码
// touched
private static final int INVALID_POINTER_ID = -1;
private int _activePointer = INVALID_POINTER_ID;
private int _newPointer = INVALID_POINTER_ID;
答案 0 :(得分:3)
为什么你只看到单个指针的值的主要问题是因为event.getY()
没有任何参数会给你第一个指针索引的位置,即使实际移动的指针是第二个指针。 / p>
MotionEvent类包含所有活动指针的坐标。
如果您想获得第二个指针的位置,您需要拨打event.getY(1)
。
您可能希望遍历所有活动点,通过event.getPointerCount()
获取总计数,然后相应地发送事件。