flutter - 减少Flutter中的dispose()样板？

Question

在Flutter中，我们需要为在dispose()中创建的许多内容写下State，例如

  final _nameController = TextEditingController();

  @override
  void dispose() {
    _nameController.dispose();
    super.dispose();
  }

因此，我想知道是否有一种方法可以消除这种需求，并自动调用该处理程序？

谢谢!

Answer 1

我找到了另一个解决方案:flutter_hooks

优点:几乎没有样板!

缺点:需要从其他基类(HookWidget，而不是StatefulWidget)扩展

一个样本:

之前-

class Example extends StatefulWidget {
  final Duration duration;

  const Example({Key key, @required this.duration})
      : assert(duration != null),
        super(key: key);

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

class _ExampleState extends State<Example> with SingleTickerProviderStateMixin {
  AnimationController _controller;

  @override
  void initState() {
    super.initState();
    _controller = AnimationController(vsync: this, duration: widget.duration);
  }

  @override
  void didUpdateWidget(Example oldWidget) {
    super.didUpdateWidget(oldWidget);
    if (widget.duration != oldWidget.duration) {
      _controller.duration = widget.duration;
    }
  }

  @override
  void dispose() {
    _controller.dispose();
    super.dispose();
  }

  @override
  Widget build(BuildContext context) {
    return Container();
  }
}

之后-

class Example extends HookWidget {
  final Duration duration;

  const Example({Key key, @required this.duration})
      : assert(duration != null),
        super(key: key);

  @override
  Widget build(BuildContext context) {
    final controller = useAnimationController(duration: duration);
    return Container();
  }
}