Display ActionMode over Toolbar

Question

I am trying to use the android.view.ActionMode with the new android.support.v7.widget.Toolbar, in addition to the traditional android.app.Act

Answer 1

So, after days going thru this thread, I finally got it working. I'd like to summarize what I did.

Note: This is the solution using a Toolbar to replace the default ActionBar in AppCompatActivity.

1. I added this line to my AppTheme: It tells android that you want your action mode to overlay the toolbar

<item name="windowActionModeOverlay">true</item>

2. Use the right imports:

You have to use these imports in order to work with AppCompatActivity:

import androidx.appcompat.view.ActionMode;
// or
import android.support.v7.view.ActionMode;

3. Start the ActionMode on your Activity like so:

actionMode = startSupportActionMode(callback);

And not like so:

actionMode = startActionMode(callback);

You Activity creates the ActionMode on the toolbar automatically, because it's set as the supportActionToolbar. The style handles the dsiplaying as overlay.

Thanks to @Kuffs and @Lefty.

Answer 2

If you see the view tree,You will can write below code:

 ViewGroup decorView = (ViewGroup) getWindow().getDecorView();
    traverseView(decorView);
 /**
 * traverse find actionmodebar
 * @param root  view
 */
public void traverseView(View root) {
    if (root==null){
        return;
    }
    if (root instanceof ActionBarContextView){
        root.setVisibility(View.GONE);
        return;
    }
    if ((root instanceof ViewGroup)) { // If view is ViewGroup, apply this method on it's child views
        ViewGroup viewGroup = (ViewGroup) root;
        for (int i = 0; i < viewGroup.getChildCount(); ++i) {
            traverseView(viewGroup.getChildAt(i));
        }
    }
}

Answer 3

Do not start it on your activity, but on your toolbar. In you activity:

Toolbar toolbar = (Toolbar) findViewById(R.id.toolbar);
toolbar.startActionMode(mActionModeCallback)

and you have to use

<item name="windowActionModeOverlay">true</item>

in your theme as stated by Andre.