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

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
}

其他回答

最好的方法:

public static boolean isOnline() {
    try {
    InetAddress.getByName("google.com").isReachable(3);

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

这是最简单和简单的方法来检查互联网连接的wifi和移动数据。

public static boolean isConnected(Activity _context) {
        if (_context != null) {
            ConnectivityManager connMgr = (ConnectivityManager) _context.getSystemService(Context.CONNECTIVITY_SERVICE);
            NetworkInfo activeInfo = connMgr.getActiveNetworkInfo();
            if (activeInfo != null && activeInfo.isConnected()) {
                boolean wifiConnected = activeInfo.getType() == ConnectivityManager.TYPE_WIFI;
                boolean mobileConnected = activeInfo.getType() == ConnectivityManager.TYPE_MOBILE;

                if (wifiConnected || mobileConnected) {
                    Log.d(TAG, "Wifi Connected ");
                    return true;
                } else {
                    showAlert(_context,_context.getString(R.string.err_no_internet));
                    return false;
                }
            } else {
                showAlert(_context,_context.getString(R.string.err_no_internet));
                return false;
            }
        } else {
            Log.e(TAG, "networkConnectivity: Context NULL");

        }
        return false;
    }

我已经尝试了近5+不同的android方法,发现这是谷歌提供的最佳解决方案,特别是android:

  try {
  HttpURLConnection urlConnection = (HttpURLConnection)
  (new URL("http://clients3.google.com/generate_204")
  .openConnection());
  urlConnection.setRequestProperty("User-Agent", "Android");
  urlConnection.setRequestProperty("Connection", "close");
  urlConnection.setConnectTimeout(1500);
  urlConnection.connect();
  if (urlConnection.getResponseCode() == 204 &&
  urlConnection.getContentLength() == 0) {
  Log.d("Network Checker", "Successfully connected to internet");
  return true;
  }
  } catch (IOException e) {
  Log.e("Network Checker", "Error checking internet connection", e);
  }

它比任何其他可用的解决方案都更快、高效和准确。

检查这段代码…这对我很有用:)

public static void isNetworkAvailable(final Handler handler, final int timeout) {
    // ask fo message '0' (not connected) or '1' (connected) on 'handler'
    // the answer must be send before before within the 'timeout' (in milliseconds)

    new Thread() {
        private boolean responded = false;   
        @Override
        public void run() { 
            // set 'responded' to TRUE if is able to connect with google mobile (responds fast) 
            new Thread() {      
                @Override
                public void run() {
                    HttpGet requestForTest = new HttpGet("http://m.google.com");
                    try {
                        new DefaultHttpClient().execute(requestForTest); // can last...
                        responded = true;
                    } 
                    catch (Exception e) {
                    }
                } 
            }.start();

            try {
                int waited = 0;
                while(!responded && (waited < timeout)) {
                    sleep(100);
                    if(!responded ) { 
                        waited += 100;
                    }
                }
            } 
            catch(InterruptedException e) {} // do nothing 
            finally { 
                if (!responded) { handler.sendEmptyMessage(0); } 
                else { handler.sendEmptyMessage(1); }
            }
        }
    }.start();
}

然后,我定义处理程序:

Handler h = new Handler() {
    @Override
    public void handleMessage(Message msg) {

        if (msg.what != 1) { // code if not connected

        } else { // code if connected

        }   
    }
};

...并启动测试:

isNetworkAvailable(h,2000); // get the answser within 2000 ms

使用以下方法检查互联网连接,同时ConnectivityManager.getActiveNetworkInfo()在API 28中已弃用

@Suppress("DEPRECATION")
fun isNetworkConnected(context: Context): Boolean {
    val cm = context.getSystemService(Context.CONNECTIVITY_SERVICE) as ConnectivityManager?
    return cm?.run {
        if (Build.VERSION.SDK_INT >= Build.VERSION_CODES.M) {
            cm.getNetworkCapabilities(cm.activeNetwork)?.run {
                when {
                    hasTransport(NetworkCapabilities.TRANSPORT_WIFI) -> true
                    hasTransport(NetworkCapabilities.TRANSPORT_CELLULAR) -> true
                    hasTransport(NetworkCapabilities.TRANSPORT_ETHERNET) -> true
                    else -> false
                }
            }
        } else {
            cm.activeNetworkInfo?.run {
                when (type) {
                    ConnectivityManager.TYPE_WIFI -> true
                    ConnectivityManager.TYPE_MOBILE -> true
                    else -> false
                }
            }
        }
    } ?: false

}

还要向清单添加以下权限

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

hatransport_ethernet (NetworkCapabilities.TRANSPORT_ETHERNET)用于为Android TV开发的应用程序,其中TV可以直接连接到以太网