检查无状态小部件是否乱扔 [英] Check if Stateless widget is disposed in flutter

问题描述

构建无状态小部件时，我使用以下代码按顺序播放一些声音：

When my stateless widget built I play some sounds in sequence order by using this code:

await _audioPlayer.play(contentPath1, isLocal: true);
await Future.delayed(Duration(seconds: 4));
await _audioPlayer.play(contentPath2, isLocal: true);
await Future.delayed(Duration(seconds: 4));
await _audioPlayer.play(contentPath3, isLocal: true);

当用户在结束播放声音之前关闭当前小部件时，即使关闭了使用此代码的当前路线：

when the user closes the current widget before finish playing the sounds, The sounds still work even after closing the current route by using this code:

Navigator.pop(context);

我的解决方法是使用布尔变量来指示关闭操作是否完成。

my workaround is to use a boolean variable to indicate if the closing action has done.

播放声音代码：

await _audioPlayer.play(contentPath1, isLocal: true);
if (closed) return;
await Future.delayed(Duration(seconds: 4));
if (closed) return;
await _audioPlayer.play(contentPath2, isLocal: true);
if (closed) return;
await Future.delayed(Duration(seconds: 4));
if (closed) return;
await _audioPlayer.play(contentPath3, isLocal: true);

关闭当前小部件：

closed = true;
_audioPlayer.stop();

如果我的小部件关闭，还有更好的方法来停止异步方法吗？

Are there a better way to stop the async methods if my widget closed?

推荐答案

如果将窗口小部件更改为StatefulWidget，则可以具有如下功能：

If you change your widget to a StatefulWidget then you can have a function like the following:

void _playSounds() {
  await _audioPlayer.play(contentPath1, isLocal: true);
  await Future.delayed(Duration(seconds: 4));
  if (!mounted) return;

  await _audioPlayer.play(contentPath2, isLocal: true);
  await Future.delayed(Duration(seconds: 4));
  if (!mounted) return;

  await _audioPlayer.play(contentPath3, isLocal: true);
}

，然后在dispose方法中处置玩家：

and then in the dispose method just dispose of the player:

@override
void dispose() {
  super.dispose();
  _audioPlayer?.dispose();
}

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