Flutter / Dart异步未等待

Top*_*zoo 5 future async-await dart flutter

我正在构建我的第一个Flutter应用程序,但遇到了一个异步问题。

当我的应用程序执行时,我希望它请求权限并等待它们被授予。我的main()函数如下所示:

import 'permission_manager.dart' as Perm_Manager;

void main() async
{
  //Ensure valid permissions
  Perm_Manager.Permission_Manager pm = Perm_Manager.Permission_Manager();
  var res = await pm.get_permissions();
  print(res);

  return runApp(MyApp());
} 
Run Code Online (Sandbox Code Playgroud)

权限管理器类的get_permissions()函数使用Flutter Simple Permissions包检查并询问权限。

import 'package:simple_permissions/simple_permissions.dart';
import 'dart:io' as IO;
import 'dart:async';

class Permission_Manager {
  /* Get user permissions */

  Future<bool> get_permissions() async
  {
    //Android handler
    if (IO.Platform.isAndroid)
    {
        //Check for read permissions
        SimplePermissions.checkPermission(Permission.ReadExternalStorage).then((result)
        {
          //If granted
          if (result)
            return true;

          //Otherwise request them
          else
          {
            SimplePermissions.requestPermission(Permission.ReadExternalStorage)
            .then((result)
            {
              // Determine if they were granted
              if (result == PermissionStatus.authorized)
                return true;
              else
                IO.exit(0); //TODO - display a message
            });
          }
        });
    }

    else
      return true;
  }
}
Run Code Online (Sandbox Code Playgroud)

当我运行该应用程序时,它不会等待功能按预期完成,并会在更新Future之前打印“ res”的值。

Launching lib\main.dart on Android SDK built for x86 in debug mode...
Built build\app\outputs\apk\debug\app-debug.apk.
I/SimplePermission(15066): Checking permission : android.permission.READ_EXTERNAL_STORAGE
I/flutter (15066): null
I/SimplePermission(15066): Requesting permission : android.permission.READ_EXTERNAL_STORAGE
Run Code Online (Sandbox Code Playgroud)

Future在函数中途返回一个值!有人知道我在做什么错吗?

Rém*_*let 6

为了等待,您必须await在将来调用关键字,而不是.then

final result = await future;
// do something
Run Code Online (Sandbox Code Playgroud)

代替

future.then((result) {
  // do something
});
Run Code Online (Sandbox Code Playgroud)

如果您真的想使用,.then则可以等待生成的将来:

await future.then((result) {
  // do something
});
Run Code Online (Sandbox Code Playgroud)

只要确保在使用嵌套异步调用时在每个异步调用上使用async关键字即可:

await future.then((result) async{
    // do something
    await future.then((result_2) {
       // do something else
    });
});
Run Code Online (Sandbox Code Playgroud)