我如何连接我的ESP8266使用arduino到我的Flutter应用程序。在这里完成初学者

Question

我想通过wifi从我的Arduino-uno发送一些数据到我的flutter应用程序。我使用ESP8266 wifi模块相同。我试图找到一些在线图书馆，我找到了wifi|Flutter和连接|Flutter。然而，我无法让它运行。

我已经安装了这些软件包，并将它们添加到pubspec中。yaml文件。

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

我还尝试了包存储库中发现的示例代码，

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'),
    );
  }
}

但是会遇到一些错误，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Flatter应用程序。我和我的团队都是初学者，所以请原谅我犯的愚蠢错误，

Answer 1

检查错误时，它是由字符串变量上指定的预期未来类型引起的。声明的其他变量也会出现类似的错误。您可以将var设置为dynamic，也可以设置预期的类型。

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，您可以在上等待它们的结果，然后（value）并可以选择使用catchError（onError）捕获错误。

// 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<代码> WiFixIfFoFiffs插件，它比 WiFi插件更为新近。