java - OnCreate 中的 PopupWindow 和 ShowAtLocation - 如何获取父 View ？

Question

我想在屏幕中间显示我的弹出窗口。我在 MainActivity 的 OnCreate 中调用它。我无法理解如何获得 showAtLocation 所要求的父 View :public void showAtLocation(android.view.View父级，int重力，int x，int y)

我拥有的是:

     List<Fragment> list=new ArrayList<>();
        list.add(new PageFragment1());
        list.add(new PageFragment2());
        list.add(new PageFragment3());


        LayoutInflater inflater = (LayoutInflater)this.getSystemService(Context.LAYOUT_INFLATER_SERVICE);
        View v = inflater.inflate(R.layout.onboarding, null);


        pager=v.findViewById(R.id.pager);
        pagerAdapter=new Onboard_SlidePageAdapter(getSupportFragmentManager(),1,list);
        pager.setAdapter(pagerAdapter);

        onboardPopupWindow = new PopupWindow(v,500, RelativeLayout.LayoutParams.WRAP_CONTENT, true);
        onboardPopupWindow.setElevation(10);

        onboardPopupWindow.showAtLocation(this,Gravity.CENTER,0,0);

我知道我无法传递“this”，但没有 this.getView() 或类似的

Answer 1

替换

onboardPopupWindow.showAtLocation(this,Gravity.CENTER,0,0);

与

v.post(new Runnable() {
    public void run() {
        onboardPopupWindow.showAtLocation(v,Gravity.CENTER, 0, 0);
    }
});