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

Android提供了ConnectivityManager类来了解互联网连接状态。下面的方法将是非常有用的,以了解互联网连接状态。

首先在AndroidManifest.xml中添加INTERNET和ACCESS_NETWORK_STATE权限

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

然后使用下面的方法检查设备是否连接到互联网。如果设备已连接到互联网,此方法将返回true。

public boolean isInternetAvailable(Context context) {
    ConnectivityManager connectivityManager = (ConnectivityManager) 
    context.getSystemService(Context.CONNECTIVITY_SERVICE);
    NetworkInfo activeNetwork = connectivityManager.getActiveNetworkInfo();
    return activeNetwork != null
            && activeNetwork.isConnectedOrConnecting();
}

参考链接:—http://www.androidtutorialshub.com/android-check-internet-connection-status/

其他回答

我做了这个代码,它是最简单的,它只是一个布尔值。 通过询问if(isOnline()){

如果存在连接,并且可以连接到页面,则会得到状态代码200(稳定连接)。

确保添加正确的INTERNET和ACCESS_NETWORK_STATE权限。

public boolean isOnline() {
    ConnectivityManager cm = (ConnectivityManager) getSystemService(Context.CONNECTIVITY_SERVICE);
    NetworkInfo netInfo = cm.getActiveNetworkInfo();
    if (netInfo != null && netInfo.isConnected()) {
        try {
            URL url = new URL("http://www.google.com");
            HttpURLConnection urlc = (HttpURLConnection) url.openConnection();
            urlc.setConnectTimeout(3000);
            urlc.connect();
            if (urlc.getResponseCode() == 200) {
                return new Boolean(true);
            }
        } catch (MalformedURLException e1) {
            // TODO Auto-generated catch block
            e1.printStackTrace();
        } catch (IOException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }
    }
    return false;
}

这个线程中的大多数答案只检查是否有可用的连接,但不检查该连接是否工作,其他答案不是设备范围,我的解决方案应该在每个设备上工作。

你可以在启动应用程序之前在你的主要活动中删除我的代码,它会快速确定是否有实际的互联网连接,如果有对话框将立即删除,应用程序将被启动,如果没有警报会弹出说应用程序需要互联网连接才能工作。

final AlertDialog alertDialog = new AlertDialog.Builder(this).create();
        alertDialog.setTitle("Checking Connection");
        alertDialog.setMessage("Checking...");
        alertDialog.show();
        new CountDownTimer(5000, 1000) {
            @Override
            public void onTick(long millisUntilFinished) {

                new Thread(new Runnable() {
                    public void run() {
                        try {
                            URL url = new URL("http://web.mit.edu/");
                            HttpURLConnection connection = (HttpURLConnection) url.openConnection();
                            connection.setRequestMethod("GET");
                            connection.setConnectTimeout(5000);
                            isConnected = connection.getResponseCode() == HttpURLConnection.HTTP_OK;
                        } catch (Exception e) {
                            e.printStackTrace();
                        }
                    }
                }).start();

                if (isConnected == false){
                    alertDialog.setMessage("Try " +  (5 - millisUntilFinished/1000) + " of 5.");
                } else {
                    alertDialog.dismiss();
                }
            }
            @Override
            public void onFinish() {
                if (isConnected == false) {
                    alertDialog.dismiss();
                    new AlertDialog.Builder(activity)
                            .setTitle("No Internet")
                            .setMessage("Please connect to Internet first.")
                            .setPositiveButton(android.R.string.yes, new DialogInterface.OnClickListener() {
                                public void onClick(DialogInterface dialog, int which) {
                                    // kill the app?
                                }
                            })
                            .setIcon(android.R.drawable.ic_dialog_alert)
                            .show();
                } else {
                    // Launch the app
                }
            }
        }.start();
public boolean isNetworkAvailable(Context context) {
    ConnectivityManager connectivityManager
            = (ConnectivityManager) context.getSystemService(Context.CONNECTIVITY_SERVICE);
    NetworkInfo activeNetworkInfo = connectivityManager.getActiveNetworkInfo();
    return activeNetworkInfo != null && activeNetworkInfo.isConnected();
}

以下是您需要的许可:

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

我已经应用了@Levit提供的解决方案,并创建了不会调用额外Http请求的函数。

它将解决无法解析主机的错误

public static boolean isInternetAvailable(Context context) {
    ConnectivityManager cm = (ConnectivityManager) context.getSystemService(Context.CONNECTIVITY_SERVICE);
    NetworkInfo activeNetwork = cm.getActiveNetworkInfo();
    if (activeNetwork == null) return false;

    switch (activeNetwork.getType()) {
        case ConnectivityManager.TYPE_WIFI:
            if ((activeNetwork.getState() == NetworkInfo.State.CONNECTED ||
                    activeNetwork.getState() == NetworkInfo.State.CONNECTING) &&
                    isInternet())
                return true;
            break;
        case ConnectivityManager.TYPE_MOBILE:
            if ((activeNetwork.getState() == NetworkInfo.State.CONNECTED ||
                    activeNetwork.getState() == NetworkInfo.State.CONNECTING) &&
                    isInternet())
                return true;
            break;
        default:
            return false;
    }
    return false;
}

private static boolean isInternet() {

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

    return false;
}

现在叫它,

if (!isInternetAvailable(getActivity())) {
     //Show message
} else {
     //Perfoem the api request
}

Kotlin和协程

我将函数放置在一个ViewModel中,该ViewModel具有viewModelScope。使用一个可观察的LiveData,我通知一个活动有关连接。

ViewModel

 fun checkInternetConnection(timeoutMs: Int) {
        viewModelScope.launch(Dispatchers.IO) {
            try {
                val socket = Socket()
                val socketAddress = InetSocketAddress("8.8.8.8", 53)

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

                _connection.postValue(true)
            }
            catch(ex: IOException) {
                _connection.postValue(false)
            }
        }
    }
 private val _connection = MutableLiveData<Boolean>()
 val connection: LiveData<Boolean> = _connection

活动

 private fun checkInternetConnection() {
     viewModel.connection.observe(this) { hasInternet ->
         if(!hasInternet) {
             //hasn't connection
         }
         else {
            //has connection
         }
     }
  }