切换已经膨胀的 stubview 布局的正确方法是什么？

Question

I have a simple relative layout included two stubview looks like below:

<RelativeLayout
    android:layout_width="match_parent"
    android:layout_height="wrap_content">

    <ViewStub
        android:layout_width="match_parent"
        android:layout_height="wrap_content"
        android:id="@+id/layout_login"
        android:layout_alignParentTop="true"
        android:layout_centerHorizontal="true"
        android:layout="@layout/layout_main_login"/>

    <ViewStub
        android:layout_width="match_parent"
        android:layout_height="wrap_content"
        android:id="@+id/layout_member"
        android:layout_alignParentTop="true"
        android:layout_centerHorizontal="true"
        android:layout="@layout/layout_main_member"/>

</RelativeLayout>

after two stubviews has been inflated, what is the proper way to switch inflated layout visible or gone agagin? I am a little confused, since the stubview removed from parent view hierarchy, the stubview will be null at that point.

I have took a look at https://developer.android.com/training/improving-layouts/loading-ondemand.html and http://android-developers.blogspot.com/2009/03/android-layout-tricks-3-optimize-with.html

am I misunderstand some points, that stubview once inflated will not be able to set visible or gone again?

my walkaround is wrap the stubview by famelayout or other layout, and change the visiblity of that layout.

Answer 1

In the similar situation I use the following code:

View layoutInflated;

void setStubViewVisibility(final boolean visible) {
    if (null == layoutInflated) {
        if (visible) {
            layoutInflated= ((ViewStub) view.findViewById(R.id.layout_login)).inflate();
            layoutInflated.setVisibility(View.VISIBLE);
        }
    } else {
        layoutInflated.setVisibility(visible ? View.VISIBLE : View.GONE);
    }
}

So you don't have to place ViewStub into another layout. Just check if layout has been inflated.