Flutter：检查互联网连接并根据输出导航？

Posted 2023-02-23

【中文标题】Flutter：检查互联网连接并根据输出导航？

【英文标题】：Flutter: Check Internet connectivity and navigate based on output?

【发布时间】：2020-05-19 23:22:52

【问题描述】：

我是flutter的新手，我想检查互联网是否可用，根据情况，屏幕必须改变。我已经编写了下面的代码（屏幕开关工作正常），但我无法获得布尔输出（互联网）。当我在检查互联网类中删除 Future 时，它​​会引发错误。你能解决这个问题吗：

class _ScreenState extends State<ChannelScreen> 
bool isInternet;
bool result;

  @override
  void initState() 
    // TODO: implement initState
    result = check();
    super.initState();
  

  @override

  Widget _buildChild() 


        print ("The output “);
        print (result);
       if (result != Null && result == true) 
// if internet is ON
return Container();
                
//if internet is off
       return Container();
  


  Widget build(BuildContext context) 
    return new Container(child: _buildChild());
  



Future<bool> check()   async
  var connectivityResult =   await Connectivity().checkConnectivity();
  if (connectivityResult == ConnectivityResult.mobile) 
    print ("******* Mobile is ON ******");
    return true;
   else if (connectivityResult == ConnectivityResult.wifi) 
    print ("******* Wifi is ON ******");
    return true;
  
  print ("No connectivity");
  return false;

【问题讨论】：

您必须在 _buildChild 方法中使用 FutureBuilder。 api.flutter.dev/flutter/widgets/FutureBuilder-class.html

【参考方案1】：

您可以使用StreamBuilder

StreamBuilder(
  stream: Connectivity().onConnectivityChanged,
  builder: (context, snapshot) 
    // Use this to avoid null exception
    if (snapshot.connectionState == ConnectionState.none) 
      return CircularProgressIndicator();
     else 
      ConnectivityResult result = snapshot.data;

      // Check Connectivity result here and display your widgets
      if(ConnectivityResult.none) 
        yourWidgetForNoInternet();
       else 
        yourWidgetForInternet();
      
    
  ,
)

【参考方案2】：

bool hasInternet = false, isChecking = true;

  @override
  void initState() 
    super.initState();
    check();
  

  @override
  Widget build(BuildContext context) 
    return Container(
        child: isChecking
            ? ListTile(
                leading: CircularProgressIndicator(), title: Text('Checking...'))
            : hasInternet
                ? ListTile(title: Text('Your widget here...'))
                : ListTile(
                    leading: Icon(Icons.info),
                    title: Text('No Internet Conncetion')));
  

  check() async 
    var connectivityResult = await Connectivity().checkConnectivity();
    if (connectivityResult == ConnectivityResult.mobile) 
      print("******* Mobile is ON ******");
      setState(() 
        isChecking = false;
        hasInternet = true;
       //navigate to another screen.....
      );
     else if (connectivityResult == ConnectivityResult.wifi) 
      print("******* Wifi is ON ******");
      setState(() 
        isChecking = false;
        hasInternet = true;
        //navigate to another screen.....
      );
     else 
      setState(() 
        isChecking = false;
        hasInternet = false;
      );
    
  

【参考方案3】：

有一个插件可以使用连接：

https://pub.dev/packages/connectivity#-readme-tab-

我使用以下代码检查互联网是否已连接

static Future<bool> checkInternetConnectivity() async 
    bool isConnected;
    try 
      final result = await InternetAddress.lookup('google.com');
      if (result.isNotEmpty && result[0].rawAddress.isNotEmpty) 
        isConnected = true;
      
     on SocketException catch (_) 
      isConnected = false;
    
    return isConnected;