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我已經設置了彈出窗口彈出窗口的辦法,但我想居中按鈕(視圖V),需要在下面被點擊打開它:獲取
public void showPopup(Context c, View v){
int[] location = new int[2];
v.getLocationOnScreen(location);
ViewGroup base = (ViewGroup) getView().findViewById(R.id.pup_pattern);
LayoutInflater inflater = (LayoutInflater) c.getSystemService(Context.LAYOUT_INFLATER_SERVICE);
View pupLayout = inflater.inflate(R.layout.linearlayout_popup, base);
final PopupWindow pup = new PopupWindow(pupLayout, android.view.ViewGroup.LayoutParams.WRAP_CONTENT, android.view.ViewGroup.LayoutParams.WRAP_CONTENT);
int x = location[0] - (int) ((pupLayout.getWidth() - v.getWidth())/2); // --> pupLayout.getWidth() gives back -2?
int y = location[1] + v.getHeight() + 10;
pup.setFocusable(true);
pup.showAtLocation(v, Gravity.NO_GRAVITY, x, y);
}
有沒有人一個想法來獲得措施?