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

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;
}
public class Network {

Context context;

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

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

    NetworkInfo activeNetwork = cm.getActiveNetworkInfo();
    return activeNetwork != null &&
                          activeNetwork.isConnectedOrConnecting();
}

}

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。

不需要太复杂。最简单的框架方式是使用ACCESS_NETWORK_STATE权限并创建一个连接的方法

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

    return cm.getActiveNetworkInfo() != null && 
       cm.getActiveNetworkInfo().isConnectedOrConnecting();
}

如果您有特定的主机和连接类型(wifi/移动),也可以使用requestRouteToHost。

你还需要:

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

在你的android清单中。

这段代码将帮助您找到互联网是否打开。

public final boolean isInternetOn() {
        ConnectivityManager conMgr = (ConnectivityManager) this.con
                .getSystemService(Context.CONNECTIVITY_SERVICE);
        NetworkInfo info = conMgr.getActiveNetworkInfo();
        return (info != null && info.isConnected());
}

此外,您应该提供以下权限

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