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。

其他回答

只需创建下面的类来检查internet连接:

public class ConnectionStatus {

    private Context _context;

    public ConnectionStatus(Context context) {
        this._context = context;
    }

    public boolean isConnectionAvailable() {
        ConnectivityManager connectivity = (ConnectivityManager) _context
                .getSystemService(Context.CONNECTIVITY_SERVICE);
        if (connectivity != null) {
            NetworkInfo[] info = connectivity.getAllNetworkInfo();
            if (info != null)
                for (int i = 0; i < info.length; i++)
                    if (info[i].getState() == NetworkInfo.State.CONNECTED) {
                        return true;
                    }
        }
        return false;
    }
}

该类仅包含一个返回连接状态布尔值的方法。因此,简单来说,如果该方法找到一个到Internet的有效连接,则返回值为true,否则为false,如果没有找到有效连接。

MainActivity中的下面的方法调用前面描述的方法的结果,并提示用户进行相应的操作:

public void addListenerOnWifiButton() {
        Button btnWifi = (Button)findViewById(R.id.btnWifi);

        iia = new ConnectionStatus(getApplicationContext());

        isConnected = iia.isConnectionAvailable();
        if (!isConnected) {
            btnWifi.setOnClickListener(new View.OnClickListener() {

                @Override
                public void onClick(View v) {
                    startActivity(new Intent(Settings.ACTION_WIFI_SETTINGS));
                    Toast.makeText(getBaseContext(), "Please connect to a hotspot",
                            Toast.LENGTH_SHORT).show();
                }
            });
        }
        else {
            btnWifi.setVisibility(4);
            warning.setText("This app may use your mobile data to update events and get their details.");
        }
    }

在上面的代码中,如果结果为假,(因此没有互联网连接,用户将被带到Android wi-fi面板,在那里他将被提示连接到wi-fi热点。

对我来说,在Activity类中检查连接状态并不是一个好的实践,因为

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

应该在那里调用,或者您需要下推您的活动实例(上下文)到连接处理程序类,以能够检查那里的连接状态 当没有可用的连接(wifi,网络)时,我捕捉到UnknownHostException异常:

JSONObject jObj = null;
Boolean responded = false;
HttpGet requestForTest = new HttpGet("http://myserver.com");
try {
    new DefaultHttpClient().execute(requestForTest);
    responded = true;
} catch (UnknownHostException e) {
    jObj = new JSONObject();
    try {
        jObj.put("answer_code", 1);
        jObj.put("answer_text", "No available connection");
    } catch (Exception e1) {}
    return jObj;
} catch (IOException e) {
    e.printStackTrace();
}

通过这种方式,我可以处理这种情况连同其他情况在同一类(我的服务器总是响应回json字符串)

public boolean isOnline() {
    boolean var = false;
    ConnectivityManager cm = (ConnectivityManager) getSystemService(Context.CONNECTIVITY_SERVICE);
    if ( cm.getActiveNetworkInfo() != null ) {
        var = true;
    }
    return var;
} 

我是这样做的。 我想会更短更有可读性。

干杯!

Saiyan

Jetpack组成/芬兰湾的科特林

根据Levite的回答,我们可以在Jetpack Compose中使用这个组合:

val DNS_SERVERS = listOf("8.8.8.8", "1.1.1.1", "4.2.2.4")
const val INTERNET_CHECK_DELAY = 3000L
@Composable
fun InternetAwareComposable(
    dnsServers: List<String> = DNS_SERVERS,
    delay: Long = INTERNET_CHECK_DELAY,
    successContent: (@Composable () -> Unit)? = null,
    errorContent: (@Composable () -> Unit)? = null,
    onlineChanged: ((Boolean) -> Unit)? = null
) {
    suspend fun dnsAccessible(
        dnsServer: String
    ) = try {
        withContext(Dispatchers.IO) {
            Runtime.getRuntime().exec("/system/bin/ping -c 1 $dnsServer").waitFor()
        } == 0
    } catch (e: Exception) {
        false
    }

    var isOnline by remember { mutableStateOf(false) }
    LaunchedEffect(Unit) {
        while (true) {
            isOnline = dnsServers.any { dnsAccessible(it) }
            onlineChanged?.invoke(isOnline)
            delay(delay)
        }
    }
    if (isOnline) successContent?.invoke()
    else errorContent?.invoke()
}
 public static boolean isNetworkAvailable(Context context) {
    boolean flag = checkNetworkAvailable(context);

    if (!flag) {
        Log.d("", "No network available!");
    } 
    return flag;
}


private static boolean checkNetworkAvailable(Context context) {
    ConnectivityManager connectivityManager
            = (ConnectivityManager) context.getSystemService(Context.CONNECTIVITY_SERVICE);
    NetworkInfo activeNetworkInfo = connectivityManager.getActiveNetworkInfo();
    return activeNetworkInfo != null;
}