Android自定义弹出窗口获取视图位置（弹出窗口内）

Question

我试图在弹出窗口中获取View的位置，但View.getX（）方法返回0.0 我试图将removeAdsIm视图的位置设置为textCheck视图

 public void showAdsOfferPopup(View v, Activity activity) {
    if (!getInAppBillingStatus()) {
        int width = display.getWidth(); // deprecated
        int height = display.getHeight();

        final PopupWindow popup = new PopupWindow(context);
        final View layout = activity.getLayoutInflater().inflate(R.layout.popup_ads_offer_new, null);
        .....
        textCheck = (ImageView) layout.findViewById(R.id.popupAdsOffer_checkTestIm);
        removeAdsIm = (ImageView) layout.findViewById(R.id.popupAdsOffer_removeAdsIm);
        playInBackgroundIm = (ImageView) layout.findViewById(R.id.popupAdsOffer_playInBackgroundIm);


        popup.setContentView(layout);
    ........

        popup.showAtLocation(v, Gravity.CENTER, 0, 0);

        int test1[] = new int[2];
        removeAdsIm.getLocationOnScreen(test1);
        textCheck.setVisibility(View.VISIBLE);
        removeAdsIm.getX();
        removeAdsIm.getY();
        playInBackgroundIm.getX();
        playInBackgroundIm.getY();
        textCheck.setX(180);
        textCheck.setY(280);


    }