Question

我有以下三个流：

  final _light1RoomCtlr = BehaviorSubject<String>();
  final _light2RoomCtlr = BehaviorSubject<String>();
  final _light3RoomCtlr = BehaviorSubject<String>();
  Stream<String> get getLight1Room => _light1RoomCtlr.stream;
  Function(String) get setLight1Room => _light1RoomCtlr.sink.add;
  Stream<String> get getLight2Room => _light2RoomCtlr.stream;
  Function(String) get setLight2Room => _light2RoomCtlr.sink.add;
  Stream<String> get getLight3Room => _light3RoomCtlr.stream;
  Function(String) get setLight3Room => _light3RoomCtlr.sink.add;

我想将它们全部合并，以便我可以在需要时获取所有发出的值。

问题是..我正在使用块模式，并且我没有initState（）函数，所以我无法初始化东西..所以dart不允许我在同一位置做类似的事情文件：

 final List<String> _lightRooms = List<String>();
  Observable.merge([getLight1Room, getLight2Room, getLight3Room]).listen((room) =>
    _lightRooms.add(room);
  ));

我已经尝试了很多事情，并且不断收到错误消息：只有静态成员才能初始化..或类似的东西。如何使用反应式编程和块模式进行此操作？如果我在需要的小部件上进行监听，则可能丢失其他信息。我尝试了rxjs示例，但是在这里它不起作用，因为这只是类定义。

Answer 1

您可以使用CombineLatest2，combinateLatest4，combinateLatest5等

例如：

import 'dart:async';

import 'package:login_bloc/src/blocs/validators.dart';
import 'package:rxdart/rxdart.dart';

class Bloc with Validators {
  final _email = BehaviorSubject<String>();
  final _password = BehaviorSubject<String>();

  Stream<String> get email => _email.stream.transform(validateEmail);
  Stream<String> get password => _password.stream.transform(validatePassword);
  Stream<bool> get submitValid =>
      Observable.combineLatest2(email, password, (e, p) => true);

  Function(String) get changeEmail => _email.sink.add;
  Function(String) get changePassword => _password.sink.add;

  submit() {
    final validEmail = _email.value;
    final validPassword = _password.value;

    print('valid data:');
    print(validEmail);
    print(validPassword);
  }

  dispose() {
    _email.close();
    _password.close();
  }
}

我如何才能使其他流的新流返回所有发出的值的列表（bloc模式）？

1 个答案: