How to open a PopupMenuButton?

Screenshot:

---

Full code:

class MyPage extends StatelessWidget {
  final GlobalKey<PopupMenuButtonState<int>> _key = GlobalKey();
  @override
  Widget build(BuildContext context) {
    return Scaffold(
      appBar: AppBar(
        actions: [
          PopupMenuButton<int>(
            key: _key,
            itemBuilder: (context) {
              return <PopupMenuEntry<int>>[
                PopupMenuItem(child: Text('0'), value: 0),
                PopupMenuItem(child: Text('1'), value: 1),
              ];
            },
          ),
        ],
      ),
      body: RaisedButton(
        onPressed: () => _key.currentState.showButtonMenu(),
        child: Text('Open/Close menu'),
      ),
    );
  }
}

I found a solution to your question. You can provide a child to PopupMenuButton which can be any Widget including a ListTile (see code below). Only problem is that the PopupMenu opens on the left side of the ListTile.

final popupMenu = new PopupMenuButton(
  child: new ListTile(
    title: new Text('Doge or lion?'),
    trailing: const Icon(Icons.more_vert),
  ),
  itemBuilder: (_) => <PopupMenuItem<String>>[
            new PopupMenuItem<String>(
                child: new Text('Doge'), value: 'Doge'),
            new PopupMenuItem<String>(
                child: new Text('Lion'), value: 'Lion'),
          ],
  onSelected: _doSomething,
)

I think it would be better do it in this way, rather than showing a PopupMenuButton

void _showPopupMenu() async {
  await showMenu(
    context: context,
    position: RelativeRect.fromLTRB(100, 100, 100, 100),
    items: [
      PopupMenuItem<String>(
          child: const Text('Doge'), value: 'Doge'),
      PopupMenuItem<String>(
          child: const Text('Lion'), value: 'Lion'),
    ],
    elevation: 8.0,
  );
}

---

There will be times when you would want to display _showPopupMenu at the location where you pressed on the button Use GestureDetector for that

final tile = new ListTile(
  title: new Text('Doge or lion?'),
  trailing: GestureDetector(
    onTapDown: (TapDownDetails details) {
      _showPopupMenu(details.globalPosition);
    },
    child: Container(child: Text("Press Me")),
  ),
);

and then _showPopupMenu will be like

_showPopupMenu(Offset offset) async {
    double left = offset.dx;
    double top = offset.dy;
    await showMenu(
    context: context,
    position: RelativeRect.fromLTRB(left, top, 0, 0),
    items: [
      ...,
    elevation: 8.0,
  );
}

This works, but is inelegant (and has the same display problem as Rainer's solution above:

class _MyHomePageState extends State<MyHomePage> {
  final GlobalKey _menuKey = GlobalKey();

  @override
  Widget build(BuildContext context) {
    final button = PopupMenuButton(
        key: _menuKey,
        itemBuilder: (_) => const<PopupMenuItem<String>>[
              PopupMenuItem<String>(
                  child: Text('Doge'), value: 'Doge'),
              PopupMenuItem<String>(
                  child: Text('Lion'), value: 'Lion'),
            ],
        onSelected: (_) {});

    final tile =
        ListTile(title: Text('Doge or lion?'), trailing: button, onTap: () {
          // This is a hack because _PopupMenuButtonState is private.
          dynamic state = _menuKey.currentState;
          state.showButtonMenu();
        });
    return Scaffold(
      body: Center(
        child: tile,
      ),
    );
  }
}

I suspect what you're actually asking for is something like what is tracked by https://github.com/flutter/flutter/issues/254 or https://github.com/flutter/flutter/issues/8277 -- the ability to associated a label with a control and have the label be clickable -- and is a missing feature from the Flutter framework.