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

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。

其他回答

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;
    }

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

移动设备上的一个重要用例是确保存在实际连接。当移动用户使用“专属门户”进入Wifi网络时,这是一个常见的问题,他们需要在其中登录。我在后台使用这个阻塞功能来确保连接存在。

/*
 * Not Thread safe. Blocking thread. Returns true if it
 * can connect to URL, false and exception is logged.
 */
public boolean checkConnectionHttps(String url){
    boolean responded = false;
    HttpGet requestTest = new HttpGet(url);
    HttpParams params = new BasicHttpParams();
    HttpConnectionParams.setConnectionTimeout(params, 3000);
    HttpConnectionParams.setSoTimeout(params, 5000);
    DefaultHttpClient client = new DefaultHttpClient(params);
    try {
        client.execute(requestTest);
        responded = true;
    } catch (ClientProtocolException e) {
        Log.w(MainActivity.TAG,"Unable to connect to " + url + " " + e.toString());
    } catch (IOException e) {
        Log.w(MainActivity.TAG,"Unable to connect to " + url + " " + e.toString());
        e.printStackTrace();
    }
    return responded;
}

在我目前所见过的所有方法中,最短、最干净的方法应该是:

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

PS:这不会ping任何主机,它只是检查连接状态,所以如果你的路由器没有互联网连接,而你的设备连接到它,这个方法将返回true,尽管你没有互联网。 对于实际的测试,我建议执行一个HttpHead请求(例如到www.google.com)并检查状态,如果它是200 OK,一切正常,并且您的设备有互联网连接。

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