如何在运行时枚举颤振小部件树？

Question

有没有办法从顶级小部件递归枚举所有子小部件？

Answer 1

所以我找到了答案:)

班级

构建上下文

方法

无效visitChildElements(ElementVisitorvisitor);

示例代码

`

// Returns a map of all of the heroes in context, indexed by hero tag.
  static Map<Object, _HeroState> _allHeroesFor(BuildContext context) {
    assert(context != null);
    final Map<Object, _HeroState> result = <Object, _HeroState>{};
    void visitor(Element element) {
      if (element.widget is Hero) {
        final StatefulElement hero = element;
        final Hero heroWidget = element.widget;
        final Object tag = heroWidget.tag;
        assert(tag != null);
        assert(() {
          if (result.containsKey(tag)) {
            throw new FlutterError(
              'There are multiple heroes that share the same tag within a subtree.\n'
              'Within each subtree for which heroes are to be animated (typically a PageRoute subtree), '
              'each Hero must have a unique non-null tag.\n'
              'In this case, multiple heroes had the following tag: $tag\n'
              'Here is the subtree for one of the offending heroes:\n'
              '${element.toStringDeep(prefixLineOne: "# ")}'
            );
          }
          return true;
        }());
        final _HeroState heroState = hero.state;
        result[tag] = heroState;
      }
      element.visitChildren(visitor);
    }
    context.visitChildElements(visitor);
    return result;
  }

`

链接 - https://medium.com/flutter-community/flutter-heroes-and-villains-bringing-balance-to-the-flutterverse-2e900222de41