弹出窗口的位置 activity

Question

我有一个 activity 想要显示为弹出窗口 window。

我有以下代码，但 activity 显示在中间。我想在不同的位置展示它。可能在我的工具栏下面。

     protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_popup_message);

    DisplayMetrics dm = new DisplayMetrics();
    getWindowManager().getDefaultDisplay().getMetrics(dm);

    int width = dm.widthPixels;
    int height = 120;

    getWindow().setLayout((int) (width * .98), height);

Answer 1

您需要获取当前视图 window 并设置该视图的 x 和 y 参数 window。所以它将设置在所需的位置。

以下是我如何在所需位置设置对话框的示例。

    Window window = getWindow();
    // set "origin" to top left corner, so to speak
    window.setGravity(Gravity.TOP | Gravity.START);
    // after that, setting values for x and y works "naturally"
    WindowManager.LayoutParams params = window.getAttributes();
    params.x =selectedViewWidth;
    params.y = selectedViewHeight/2;

    window.setAttributes(params);

I personally think that if you want to show popup window then instead of using activity you can also use PopupWindow or DialogFragment. You can manage by this controls more effectively. There are so many tutorials available for same.

弹出窗口的位置 activity

Location for pop up activity

android-studio

android-6.0-marshmallow