dart – 如何打开PopupMenuButton？

概述如何从第二个小部件打开d出菜单？ final button = new PopupMenuButton( itemBuilder: (_) => <PopupMenuItem<String>>[ new PopupMenuItem<String>( child: const Text('Doge'), value: 'Doge'), 如何从第二个小部件打开d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);

我想通过点击图块来打开按钮的菜单.

解决方法 这可行,但不够优雅(并且与Rainer的解决方案具有相同的显示问题：

class _MyHomePageState extends State<MyHomePage> {  final GlobalKey _menuKey = new GlobalKey();  @overrIDe  @R_404_5537@ build(BuildContext context) {    final button = new PopupMenubutton(        key: _menuKey,itemBuilder: (_) => <PopupMenuItem<String>>[              new PopupMenuItem<String>(                  child: const Text('Doge'),new PopupMenuItem<String>(                  child: const Text('lion'),onSelected: (_) {});    final tile =        new ListTile(Title: new Text('Doge or lion?'),trailing: button,onTap: () {          // This is a Hack because _PopupMenubuttonState is private.          dynamic state = _menuKey.currentState;          state.showbuttonMenu();        });    return new Scaffold(      body: new Center(        child: tile,),);  }}

我怀疑你实际要求的是类似于https://github.com/flutter/flutter/issues/254或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 – 并且是Flutter框架中缺少的功能.

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