如何在选择项目后不解散PopUpMenuButton？

Question

我正在使用flutter PopUpMenuButton。所有我想要的是当我选择菜单上的任何项目时，弹出式菜单不应该被驳回，而是让我从弹出式菜单中选择多个值.文档中说你可以覆盖handleTap属性，但我不清楚如何做到这一点？

 ///The [handleTap] method can be overridden to adjust exactly what happens when
/// the item is tapped. By default, it uses [Navigator.pop] to return the
/// [PopupMenuItem.value] from the menu route.

    void handleTap() {
    Navigator.pop<T>(context, widget.value);
  }

Answer 1

class PopupItem extends PopupMenuItem {
  const PopupItem({
    Key key,
    Widget child,
  }) : super(key: key, child: child);

  @override
  _PopupItemState createState() => _PopupItemState();
}

class _PopupItemState extends PopupMenuItemState {
  @override
  void handleTap() {}
}

使用方法

PopupMenuButton(
itemBuilder: (_) {
  return [
    PopupItem(child: ...),
  ];
}

Answer 2

你可以像这样使用CheckedPopupMenuItem... 就像在下面提到的那样。官方 文件资料

PopupMenuButton<Commands>(
      onSelected: (Commands result) {
        switch (result) {
          case Commands.heroAndScholar:
            setState(() { _heroAndScholar = !_heroAndScholar; });
            break;
          case Commands.hurricaneCame:
            // ...handle hurricane option
            break;
          // ...other items handled here
        }
      },
      itemBuilder: (BuildContext context) => <PopupMenuEntry<Commands>>[
        CheckedPopupMenuItem<Commands>(
          checked: _heroAndScholar,
          value: Commands.heroAndScholar,
          child: const Text('Hero and scholar'),
        ),
        const PopupMenuDivider(),
        const PopupMenuItem<Commands>(
          value: Commands.hurricaneCame,
          child: ListTile(leading: Icon(null), title: Text('Bring hurricane')),
        ),
        // ...other items listed here
      ],
    )