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

public static boolean isNetworkAvailable(Context ctx) {
ConnectivityManager connMgr = (ConnectivityManager)ctx.getSystemService(Context.CONNECTIVITY_SERVICE);
if(connMgr.getNetworkInfo(ConnectivityManager.TYPE_WIFI).isConnected() ||
    connMgr.getNetworkInfo(ConnectivityManager.TYPE_MOBILE).isConnected()){
        return true;
}

return false;
 } 

用户这

其他回答

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)

这在android文档中有涉及 http://developer.android.com/training/monitoring-device-state/connectivity-monitoring.html

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;
}

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" />
public static boolean isNetworkAvailable(Context ctx) {
ConnectivityManager connMgr = (ConnectivityManager)ctx.getSystemService(Context.CONNECTIVITY_SERVICE);
if(connMgr.getNetworkInfo(ConnectivityManager.TYPE_WIFI).isConnected() ||
    connMgr.getNetworkInfo(ConnectivityManager.TYPE_MOBILE).isConnected()){
        return true;
}

return false;
 } 

用户这