我得到了一个AsyncTask,应该检查对主机名的网络访问。但是doInBackground()永远不会超时。有人知道吗?

public class HostAvailabilityTask extends AsyncTask<String, Void, Boolean> {

    private Main main;

    public HostAvailabilityTask(Main main) {
        this.main = main;
    }

    protected Boolean doInBackground(String... params) {
        Main.Log("doInBackground() isHostAvailable():"+params[0]);

        try {
            return InetAddress.getByName(params[0]).isReachable(30); 
        } catch (UnknownHostException e) {
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        }
        return false;       
    }

    protected void onPostExecute(Boolean... result) {
        Main.Log("onPostExecute()");

        if(result[0] == false) {
            main.setContentView(R.layout.splash);
            return;
        }

        main.continueAfterHostCheck();
    }   
}

当前回答

下面是我使用的方法:

public boolean isNetworkAvailable(final Context context) {
    return ((ConnectivityManager) context.getSystemService(Context.CONNECTIVITY_SERVICE)).getActiveNetworkInfo() != null;
}

更好的是,检查确保它是“连接”的:

public boolean isNetworkAvailable(final Context context) {
    final ConnectivityManager connectivityManager = ((ConnectivityManager) context.getSystemService(Context.CONNECTIVITY_SERVICE));
    return connectivityManager.getActiveNetworkInfo() != null && connectivityManager.getActiveNetworkInfo().isConnected();
}

下面是如何使用该方法:

if (isNetworkAvailable(context)) {
    // code here
} else {
    // code
}

需要许可:

<uses-permission android:name="android.permission.ACCESS_NETWORK_STATE" />

https://stackoverflow.com/a/16124915/950427

其他回答

有不止一种方法

第一,最短但效率低的方法

只需要网络状态权限

<uses-permission android:name="android.permission.ACCESS_NETWORK_STATE" />

然后这个方法,

 public boolean activeNetwork () {
        ConnectivityManager cm =
                (ConnectivityManager)getSystemService(Context.CONNECTIVITY_SERVICE);

        NetworkInfo activeNetwork = cm.getActiveNetworkInfo();
        boolean isConnected = activeNetwork != null &&
                activeNetwork.isConnected();

        return isConnected;

    }

正如在回答中所看到的ConnectivityManager是一个解决方案,我只是在一个方法中添加了它,这是一个简化的方法 ConnectivityManager返回true,如果有网络访问而不是互联网访问,这意味着如果你的WiFi连接到路由器,但路由器没有互联网,它返回true,它检查连接可用性

二、高效的方式

需要网络状态和Internet权限

<uses-permission android:name="android.permission.ACCESS_NETWORK_STATE" />
<uses-permission android:name="android.permission.INTERNET" />

然后这门课,

 public class CheckInternetAsyncTask extends AsyncTask<Void, Integer, Boolean> {

        private Context context;

        public CheckInternetAsyncTask(Context context) {
            this.context = context;
        }

        @Override
        protected Boolean doInBackground(Void... params) {

            ConnectivityManager cm =
                    (ConnectivityManager)context.getSystemService(Context.CONNECTIVITY_SERVICE);

            assert cm != null;
            NetworkInfo activeNetwork = cm.getActiveNetworkInfo();
            boolean isConnected = activeNetwork != null &&
                    activeNetwork.isConnected();


            if (isConnected) {
                try {
                    HttpURLConnection urlc = (HttpURLConnection)
                            (new URL("http://clients3.google.com/generate_204")
                                    .openConnection());
                    urlc.setRequestProperty("User-Agent", "Android");
                    urlc.setRequestProperty("Connection", "close");
                    urlc.setConnectTimeout(1500);
                    urlc.connect();
                    if (urlc.getResponseCode() == 204 &&
                            urlc.getContentLength() == 0)
                        return true;

                } catch (IOException e) {
                    Log.e("TAG", "Error checking internet connection", e);
                    return false;
                }
            } else {
                Log.d("TAG", "No network available!");
                return false;
            }


            return null;
        }

        @Override
        protected void onPostExecute(Boolean result) {
            super.onPostExecute(result);
            Log.d("TAG", "result" + result);

            if(result){
                // do ur code
            }

        }


    }

叫CheckInternetAsyncTask

new CheckInternetAsyncTask(getApplicationContext()).execute();

部分解释:-

you have to check Internet on AsyncTask, otherwise it can throw android.os.NetworkOnMainThreadException in some cases ConnectivityManager used to check the network access if true sends request (Ping) Request send to http://clients3.google.com/generate_204, This well-known URL is known to return an empty page with an HTTP status 204 this is faster and more efficient than http://www.google.com , read this. if you have website it's preferred to put you website instead of google, only if you use it within the app Timeout can be changed range (20ms -> 2000ms), 1500ms is commonly used

网络连接/互联网接入

isConnectedOrConnecting()(在大多数回答中使用)检查任何网络连接 要了解这些网络是否有internet接入,请使用以下方法之一

A) Ping服务器(简单)

// ICMP 
public boolean isOnline() {
    Runtime runtime = Runtime.getRuntime();
    try {
        Process ipProcess = runtime.exec("/system/bin/ping -c 1 8.8.8.8");
        int     exitValue = ipProcess.waitFor();
        return (exitValue == 0);
    }
    catch (IOException e)          { e.printStackTrace(); }
    catch (InterruptedException e) { e.printStackTrace(); }

    return false;
}

+可以在主线程上运行

在一些旧设备上不能工作(Galays S3等),如果没有网络,它会阻塞一段时间。

B)连接到Internet上的Socket(高级)

// TCP/HTTP/DNS (depending on the port, 53=DNS, 80=HTTP, etc.)
public boolean isOnline() {
    try {
        int timeoutMs = 1500;
        Socket sock = new Socket();
        SocketAddress sockaddr = new InetSocketAddress("8.8.8.8", 53);

        sock.connect(sockaddr, timeoutMs);
        sock.close();

        return true;
    } catch (IOException e) { return false; }
}

+非常快(任何一种方式),适用于所有设备,非常可靠

-不能在UI线程上运行

这工作非常可靠,在每个设备上,非常快。它需要在一个单独的任务中运行(例如ScheduledExecutorService或AsyncTask)。

可能的问题

Is it really fast enough? Yes, very fast ;-) Is there no reliable way to check internet, other than testing something on the internet? Not as far as I know, but let me know, and I will edit my answer. What if the DNS is down? Google DNS (e.g. 8.8.8.8) is the largest public DNS in the world. As of 2018 it handled over a trillion queries a day [1]. Let 's just say, your app would probably not be the talk of the day. Which permissions are required? <uses-permission android:name="android.permission.INTERNET" /> Just internet access - surprise ^^ (Btw have you ever thought about, how some of the methods suggested here could even have a remote glue about internet access, without this permission?)

 

额外:一次性RxJava/RxAndroid示例(Kotlin)

fun hasInternetConnection(): Single<Boolean> {
  return Single.fromCallable {
    try {
      // Connect to Google DNS to check for connection
      val timeoutMs = 1500
      val socket = Socket()
      val socketAddress = InetSocketAddress("8.8.8.8", 53)
    
      socket.connect(socketAddress, timeoutMs)
      socket.close()
  
      true
    } catch (e: IOException) {
      false
    }
  }
  .subscribeOn(Schedulers.io())
  .observeOn(AndroidSchedulers.mainThread())
}

///////////////////////////////////////////////////////////////////////////////////
// Usage

    hasInternetConnection().subscribe { hasInternet -> /* do something */}

额外:一次性RxJava/RxAndroid示例(Java)

public static Single<Boolean> hasInternetConnection() {
    return Single.fromCallable(() -> {
        try {
            // Connect to Google DNS to check for connection
            int timeoutMs = 1500;
            Socket socket = new Socket();
            InetSocketAddress socketAddress = new InetSocketAddress("8.8.8.8", 53);

            socket.connect(socketAddress, timeoutMs);
            socket.close();

            return true;
        } catch (IOException e) {
            return false;
        }
    }).subscribeOn(Schedulers.io()).observeOn(AndroidSchedulers.mainThread());
}

///////////////////////////////////////////////////////////////////////////////////
// Usage

    hasInternetConnection().subscribe((hasInternet) -> {
        if(hasInternet) {

        }else {

        }
    });

额外:一次性AsyncTask示例

注意:这是如何执行请求的另一个示例。然而,由于AsyncTask已弃用,它应该被你的应用程序的线程调度,Kotlin协程,Rx,…

class InternetCheck extends AsyncTask<Void,Void,Boolean> {

    private Consumer mConsumer;
    public  interface Consumer { void accept(Boolean internet); }

    public  InternetCheck(Consumer consumer) { mConsumer = consumer; execute(); }

    @Override protected Boolean doInBackground(Void... voids) { try {
        Socket sock = new Socket();
        sock.connect(new InetSocketAddress("8.8.8.8", 53), 1500);
        sock.close();
        return true;
    } catch (IOException e) { return false; } }

    @Override protected void onPostExecute(Boolean internet) { mConsumer.accept(internet); }
}

///////////////////////////////////////////////////////////////////////////////////
// Usage

    new InternetCheck(internet -> { /* do something with boolean response */ });

这种方法为您提供了一个非常快速的方法(用于实时反馈)或较慢的方法(用于需要可靠性的一次性检查)。

public boolean isNetworkAvailable(bool SlowButMoreReliable) {
    bool Result = false; 
    try {
        if(SlowButMoreReliable){
            ConnectivityManager MyConnectivityManager = null;
            MyConnectivityManager = (ConnectivityManager) getSystemService(Context.CONNECTIVITY_SERVICE);

            NetworkInfo MyNetworkInfo = null;
            MyNetworkInfo = MyConnectivityManager.getActiveNetworkInfo();

            Result = MyNetworkInfo != null && MyNetworkInfo.isConnected();

        } else
        {
            Runtime runtime = Runtime.getRuntime();
            Process ipProcess = runtime.exec("/system/bin/ping -c 1 8.8.8.8");

            int i = ipProcess.waitFor();

            Result = i== 0;

        }

    } catch(Exception ex)
    {
        //Common.Exception(ex); //This method is one you should have that displays exceptions in your log
    }
    return Result;
}

非常重要的是检查我们是否与isAvailable()有连接,以及是否可能与isConnected()建立连接

private static ConnectivityManager manager;

public static boolean isOnline(Context context) {
    ConnectivityManager connectivityManager = (ConnectivityManager) context.getSystemService(Context.CONNECTIVITY_SERVICE);
    NetworkInfo networkInfo = connectivityManager.getActiveNetworkInfo();
    return networkInfo != null && networkInfo.isAvailable() && networkInfo.isConnected();
}

你可以取消网络活动WiFi的类型:

public static boolean isConnectedWifi(Context context) {
    ConnectivityManager connectivityManager = (ConnectivityManager) context.getSystemService(Context.CONNECTIVITY_SERVICE);
    NetworkInfo networkInfo = connectivityManager.getActiveNetworkInfo();
    return networkInfo != null && networkInfo.getType() == ConnectivityManager.TYPE_WIFI;
}

或手机Móvil:

public static boolean isConnectedMobile(Context context) {
    ConnectivityManager connectivityManager = (ConnectivityManager) context.getSystemService(Context.CONNECTIVITY_SERVICE);
    NetworkInfo networkInfo = connectivityManager.getActiveNetworkInfo();
    return networkInfo != null && networkInfo.getType() == ConnectivityManager.TYPE_MOBILE;
}

不要忘记权限:

    <uses-permission android:name="android.permission.ACCESS_NETWORK_STATE" />
   <uses-permission android:name="android.permission.INTERNET" />

使用这个Kotlin扩展:

/**
 * Check whether network is available
 *
 * @param context
 * @return Whether device is connected to Network.
 */
fun Context.isNetworkAvailable(): Boolean {
    with(getSystemService(Context.CONNECTIVITY_SERVICE) as ConnectivityManager) {
        if (Build.VERSION.SDK_INT >= Build.VERSION_CODES.M) {
            //Device is running on Marshmallow or later Android OS.
            with(getNetworkCapabilities(activeNetwork)) {
                return hasTransport(NetworkCapabilities.TRANSPORT_WIFI) || hasTransport(
                    NetworkCapabilities.TRANSPORT_CELLULAR
                )
            }
        } else {
            activeNetworkInfo?.let {
                // connected to the internet
                @Suppress("DEPRECATION")
                return listOf(ConnectivityManager.TYPE_WIFI, ConnectivityManager.TYPE_MOBILE).contains(it.type)
            }
        }
    }
    return false
}