arduino - 如何使用 arduino 将我的 ESP8266 连接到我的 Flutter 应用程序。在这里完成初学者

Question

我想通过 wifi 将一些数据从我的 Arduino-uno 发送到我的颤振应用程序。我正在使用 ESP8266 wifi 模块。我试图找到一些在线图书馆，我找到了wifi|Flutter和connectivity|Flutter。但是我无法让它运行。

我已经安装了这些包，并将它们添加到 pubspec.yaml 文件中。

dependencies:
  flutter:
    sdk: flutter
  wifi: ^0.1.4
  connectivity: ^0.3.2

我还尝试了包存储库中的示例代码，

import 'dart:async';
import 'package:connectivity/connectivity.dart';
import 'package:flutter/material.dart';
import 'package:wifi/wifi.dart';

void main()=>runApp(MyApp());
class MyApp extends StatelessWidget {
  String ssid = await Wifi.ssid;

//Signal strength， 1-3，The bigger the number, the stronger the signal
  int level = await Wifi.level;

  String ip = await Wifi.ip;

  var result = await Wifi.connection('ssid', 'password');
  List<WifiResult> list = await Wifi.list('key');
  @override
  Widget build(BuildContext context) {
    return Container(
      child: Text('$level $list'),
    );
  }
}

但遇到一些错误，例如（行号在每行的末尾）

error: A value of type 'Future<String>' can't be assigned to a variable of type 'String'. (invalid_assignment at [all_tests_here1] lib\main.dart:7)
error: Unexpected text 'await'. (unexpected_token at [all_tests_here1] lib\main.dart:7)
error: Unexpected text 'await'. (unexpected_token at [all_tests_here1] lib\main.dart:10)
error: A value of type 'Future<int>' can't be assigned to a variable of type 'int'. (invalid_assignment at [all_tests_here1] lib\main.dart:10)
error: A value of type 'Future<String>' can't be assigned to a variable of type 'String'. (invalid_assignment at [all_tests_here1] lib\main.dart:12)
error: Unexpected text 'await'. (unexpected_token at [all_tests_here1] lib\main.dart:12)
error: Unexpected text 'await'. (unexpected_token at [all_tests_here1] lib\main.dart:14)
error: A value of type 'Future<List<WifiResult>>' can't be assigned to a variable of type 'List<WifiResult>'. (invalid_assignment at [all_tests_here1] lib\main.dart:15)
error: Unexpected text 'await'. (unexpected_token at [all_tests_here1] lib\main.dart:15)

我想使用 wifi 将所有传感器数据（2 个传感器）发送到 firebase，然后将其发送到颤振应用程序。我和我的团队完全是初学者，所以请原谅任何愚蠢的错误，

标签： arduinoflutteresp8266

Answer 1

检查错误，它是由在 String 变量上分配预期的 Future 类型引起的。声明的其他变量也会引发类似的错误。您可以将 var 设置为dynamic或设置预期的类型。

Future<String> getWifiSSID() async => await Wifi.ssid;
Future<int> getWifiSignalLevel() async => await Wifi.level;
Future<String> getWifiIp() async => await Wifi.ip;
Future<WifiState> getWifiAccess(String wifiSSID, String wifiPass) async =>
  await Wifi.connection(wifiSSID, wifiPass);

Future<List<WifiResult>> getWifiList() async => await Wifi.list('key');

现在要调用异步函数，您可以等待它们的结果，then(value)并可选择使用catchError(onError).

// Connect to wifi
getWifiAccess(SSID, pass)
  .then((wifiAccess) => debugPrint('wifiAccess: $wifiAccess'))
  .catchError((onError) => debugPrint('wifiAccess error: $onError'));

// fetch network SSID of current wifi connection
getWifiSSID()
  .then((wifiSSID) => debugPrint('wifiSSID: $wifiSSID'))
  .catchError((onError) => debugPrint('wifiSSID error: $onError'));

// fetch network signal level of current wifi connection
getWifiSignalLevel()
  .then((wifiSignalLevel) => debugPrint('wifiSignalLevel: $wifiSignalLevel'))
  .catchError((onError) => debugPrint('wifiSignalLevel error: $onError'));

// fetch network IP address
getWifiIp()
  .then((wifiIpAddress) => debugPrint('wifiIpAddress: $wifiIpAddress'))
  .catchError((onError) => debugPrint('wifiIpAddress error: $onError'));

// fetch list of access points (only works on Android)
getWifiList().then((wifiList) {
  wifiList.forEach((wifi) {
    debugPrint('wifiList: ${wifi.ssid} signal: ${wifi.level}');
  });
}).catchError((onError) => debugPrint('wifiList error: $onError'));

编辑：要添加，您可能还需要考虑使用与插件wifi_info_flutter相比更新的wifi插件。