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

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

其他回答

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
}

这对我很有用。试试吧。

public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.main);
    try {
        URL url = new URL("http://stackoverflow.com/posts/11642475/edit" );
        //URL url = new URL("http://www.nofoundwebsite.com/" );
        executeReq(url);
        Toast.makeText(getApplicationContext(), "Webpage is available!", Toast.LENGTH_SHORT).show();
    }
    catch(Exception e) {
        Toast.makeText(getApplicationContext(), "oops! webpage is not available!", Toast.LENGTH_SHORT).show();
    }
}

private void executeReq(URL urlObject) throws IOException
{
    HttpURLConnection conn = null;
    conn = (HttpURLConnection) urlObject.openConnection();
    conn.setReadTimeout(30000);//milliseconds
    conn.setConnectTimeout(3500);//milliseconds
    conn.setRequestMethod("GET");
    conn.setDoInput(true);

    // Start connect
    conn.connect();
    InputStream response =conn.getInputStream();
    Log.d("Response:", response.toString());
}}

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。

以下是来自我的Utils类的代码:

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

你可以遍历所有的网络连接,检查是否至少有一个可用的连接:

public boolean isConnected() {
    boolean connected = false;

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

    if (cm != null) {
        NetworkInfo[] netInfo = cm.getAllNetworkInfo();

        for (NetworkInfo ni : netInfo) {
            if ((ni.getTypeName().equalsIgnoreCase("WIFI")
                    || ni.getTypeName().equalsIgnoreCase("MOBILE"))
                    && ni.isConnected() && ni.isAvailable()) {
                connected = true;
            }

        }
    }

    return connected;
}