如何打开 PopupMenuButton？

Question

如何从第二个小部件打开弹出菜单？

final button = new PopupMenuButton(
    itemBuilder: (_) => <PopupMenuItem<String>>[
          new PopupMenuItem<String>(
              child: const Text('Doge'), value: 'Doge'),
          new PopupMenuItem<String>(
              child: const Text('Lion'), value: 'Lion'),
        ],
    onSelected: _doSomething);

final tile = new ListTile(title: new Text('Doge or lion?'), trailing: button);

我想通过点击 tile 打开 button 的菜单。

Answer 1

我认为没有办法实现这种行为。尽管您可以将 onTap 属性附加到图块，但您无法从 'outside'

访问 MenuButton

您可以采用的一种方法是使用 ExpansionPanels，因为它们看起来像 ListTiles，旨在允许轻松修改和编辑。

Answer 2

我找到了您问题的解决方案。您可以为 PopupMenuButton 提供一个子项，它可以是任何包含 ListTile 的 Widget（请参见下面的代码）。唯一的问题是 PopupMenu 在 ListTile 的左侧打开。

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,
)

Answer 3

这可行，但不够优雅（并且与上面 Rainer 的解决方案有相同的显示问题：

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,
      ),
    );
  }
}

我怀疑您实际要求的是 https://github.com/flutter/flutter/issues/254 or https://github.com/flutter/flutter/issues/8277 跟踪的内容——将标签与控件相关联并使标签可点击的能力——这是一个缺失的功能来自 Flutter 框架。

Answer 4

我认为这样会更好，而不是显示 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,
  );
}

有时您会想要在您按下按钮的位置显示_showPopupMenu 为此使用 GestureDetector

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

然后 _showPopupMenu 会像

_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, ); }

Answer 5

使用 popup_menu 包作为库

在您的 pubspec.yaml 依赖项中添加 popup_menu: ^1.0.5。并导入它：

import 'package:popup_menu/popup_menu.dart';

首先，您应该在代码中的某处设置上下文。如下所示：

PopupMenu.context = context;

然后创建一个 showPopup 小部件并传递所需的参数：

 void showPopup(Offset offset) {
    PopupMenu menu = PopupMenu(
        // backgroundColor: Colors.teal,
        // lineColor: Colors.tealAccent,
        maxColumn: 3,
        items: [
          MenuItem(title: 'Copy', image: Image.asset('assets/copy.png')),
          MenuItem(title: 'Mail', image: Icon(Icons.mail, color: Colors.white)),
          MenuItem(title: 'Power',image: Icon(Icons.power, color: Colors.white,)),
        ],
        onClickMenu: onClickMenu,
        stateChanged: stateChanged,
        onDismiss: onDismiss);
    menu.show(rect: Rect.fromPoints(offset, offset));
  }

  void stateChanged(bool isShow) {
    print('menu is ${isShow ? 'showing' : 'closed'}');
  }

  void onClickMenu(MenuItemProvider item) {
    print('Click menu -> ${item.menuTitle}');
  }

  void onDismiss() {
    print('Menu is dismiss');
  }

这是打开的弹窗

  @override
  Widget build(BuildContext context) {

  PopupMenu.context = context;  // This is the set context

    return Scaffold(
      appBar: AppBar(
        title: Text('Show Popup')),
      body: Stack(
        children: <Widget>[
          ListView.builder(
            itemCount: list.length,
            itemBuilder: (context, index) {
             return MaterialButton(
                child: GestureDetector(
                  onTapUp: (TapUpDetails details) {
                    showPopup(details.globalPosition);
                  },
                  child: ListTile(
                    leading: IconButton(
                      icon: Icon(Icons.restaurant_menu),
                    ),
                    title: Text("Select your categories"),
                    subtitle: Text("Sub Title"),
                    // onTap: showPopup,
                  ),
                ),

              );
            },
          ),
        ],
      ),
    );
  }

Answer 6

截图：

完整代码：

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'),
      ),
    );
  }
}

如何打开 PopupMenuButton？

How to open a PopupMenuButton?

dart

flutter