onFling和SimpleOnGestureListener麻烦
||
我正在实现一个onFling检测器,类似于以下问题:
网格布局上的翻转手势检测
但是,我只是无法使其正常工作。
如果我通过某种方式设置了我的应用程序而导致了这个问题(也许是asynctask?),则这是我的应用程序的结构:
(试图使它成为格式正确的列表,但是由于某种原因,它搞砸了下面的代码?)
--MapActivity创建一个MapView。
--MapView创建ASyncTask以从URL提取XML。
ASyncTask的--onPostExecute()解析XML并使用获取的数据添加ItemizedOverlay
ItemizedOverlay的--onTap()函数使用LoaderImageView从Web上获取图像(http://www.anddev.org/novice-tutorials-f8/imageview-with-loading-spinner-t49439.html)
--onTap()函数然后调用下面列出的PopupPanel类的show()函数
从下面的代码中可以看到,注释行有效。但是,与MyGestureDetector类一起使用时,永远不会触发onTouch和onFling事件。
class PopupPanel {
View popup;
boolean isVisible = false;
private static final int SWIPE_MIN_DISTANCE = 120;
private static final int SWIPE_MAX_OFF_PATH = 250;
private static final int SWIPE_THRESHOLD_VELOCITY = 200;
private GestureDetector gestureDetector;
public View.OnTouchListener gestureListener;
PopupPanel(Context context, int layout) {
ViewGroup parent = (ViewGroup) map.getParent();
popup = ((MapActivity) context).getLayoutInflater().inflate(layout, parent, false);
SelectFilterActivity selectFilterActivity = new SelectFilterActivity();
popup.setOnClickListener(selectFilterActivity);
// This works!
// popup.setOnTouchListener(new View.OnTouchListener() {
// public boolean onTouch(View v, MotionEvent event) {
// Toast.makeText(map.getContext(), \"Touched\", Toast.LENGTH_SHORT).show();
// return false;
// }
// });
// Gesture detection
gestureDetector = new GestureDetector(new MyGestureDetector());
gestureListener = new View.OnTouchListener() {
public boolean onTouch(View v, MotionEvent event) {
if (gestureDetector.onTouchEvent(event)) {
return true;
}
return false;
}
};
}
class MyGestureDetector extends SimpleOnGestureListener {
@Override
public boolean onFling(MotionEvent e1, MotionEvent e2, float velocityX, float velocityY) {
try {
if (Math.abs(e1.getY() - e2.getY()) > SWIPE_MAX_OFF_PATH)
return false;
// right to left swipe
if(e1.getX() - e2.getX() > SWIPE_MIN_DISTANCE && Math.abs(velocityX) > SWIPE_THRESHOLD_VELOCITY) {
Toast.makeText(map.getContext(), \"Left Swipe\", Toast.LENGTH_SHORT).show();
} else if (e2.getX() - e1.getX() > SWIPE_MIN_DISTANCE && Math.abs(velocityX) > SWIPE_THRESHOLD_VELOCITY) {
Toast.makeText(map.getContext(), \"Right Swipe\", Toast.LENGTH_SHORT).show();
}
} catch (Exception e) {
// nothing
}
return false;
}
@Override
public boolean onDown(MotionEvent e) {
return true;
}
}
View getView() {
return (popup);
}
void show(boolean alignTop) {
RelativeLayout.LayoutParams lp = new RelativeLayout.LayoutParams(
RelativeLayout.LayoutParams.WRAP_CONTENT,
RelativeLayout.LayoutParams.WRAP_CONTENT);
if (alignTop) {
lp.addRule(RelativeLayout.ALIGN_PARENT_TOP);
lp.setMargins(0, 20, 0, 0);
} else {
lp.addRule(RelativeLayout.ALIGN_PARENT_BOTTOM);
lp.setMargins(0, 0, 0, 60);
}
hide();
((ViewGroup) map.getParent()).addView(popup, lp);
isVisible = true;
}
void hide() {
if (isVisible) {
isVisible = false;
((ViewGroup) popup.getParent()).removeView(popup);
}
}
}
没有找到相关结果
已邀请:
2 个回复
渴翅吮斡撤
创建它之后。
羔磺
中退还
。超类的实现是返回“ 4”,然后抑制将来事件的传递,直到释放触摸为止。