如何在Android中设置弹出菜单边距？

Question

我在屏幕的右上角有一个弹出菜单。我想在右边加一些边距。怎么做？我无法找到任何解决方案。每次使用弹出窗口完成。但我想用弹出菜单。请帮忙。建议使用正确的代码，非常有帮助。

我想在此弹出菜单的右侧添加一些边距

Answer 1

你可以试试这个

<style name="PopupMenu" parent="@style/Widget.AppCompat.PopupMenu">
    <item name="android:dropDownHorizontalOffset">-4dp</item>
    <item name="android:dropDownVerticalOffset">4dp</item>
</style>

在下面的代码中应用上面的样式

   PopupMenu popup = null;
        if (android.os.Build.VERSION.SDK_INT >= android.os.Build.VERSION_CODES.LOLLIPOP_MR1) {
            popup = new PopupMenu(context, anchorView, Gravity.END, 0, R.style.MyPopupMenu);
        }else{
            popup = new PopupMenu(context, anchorView);
        }

Answer 2

try {
    Field fListPopup = menuHelper.getClass().getDeclaredField("mPopup");
    fListPopup.setAccessible(true);
    Object listPopup = fListPopup.get(menuHelper);
    argTypes = new Class[] { int.class };
    Class listPopupClass = listPopup.getClass();

    // Get the width of the popup window
    int width = (Integer) listPopupClass.getDeclaredMethod("getWidth").invoke(listPopup);

    // Invoke setHorizontalOffset() with the negative width to move left by that distance
    listPopupClass.getDeclaredMethod("setHorizontalOffset", argTypes).invoke(listPopup, -width);

    // Invoke show() to update the window's position
    listPopupClass.getDeclaredMethod("show").invoke(listPopup);
} catch (Exception e) {
    // Again, an exception here indicates a programming error rather than an exceptional condition
    // at runtime
    Log.w(TAG, "Unable to force offset", e);
}