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();
}
}
Kotlin和协程
我将函数放置在一个ViewModel中，该ViewModel具有viewModelScope。使用一个可观察的LiveData，我通知一个活动有关连接。
ViewModel
fun checkInternetConnection(timeoutMs: Int) {
viewModelScope.launch(Dispatchers.IO) {
try {
val socket = Socket()
val socketAddress = InetSocketAddress("8.8.8.8", 53)
socket.connect(socketAddress, timeoutMs)
socket.close()
_connection.postValue(true)
}
catch(ex: IOException) {
_connection.postValue(false)
}
}
}
private val _connection = MutableLiveData<Boolean>()
val connection: LiveData<Boolean> = _connection
活动
private fun checkInternetConnection() {
viewModel.connection.observe(this) { hasInternet ->
if(!hasInternet) {
//hasn't connection
}
else {
//has connection
}
}
}
这个线程中的大多数答案只检查是否有可用的连接，但不检查该连接是否工作，其他答案不是设备范围，我的解决方案应该在每个设备上工作。
你可以在启动应用程序之前在你的主要活动中删除我的代码，它会快速确定是否有实际的互联网连接，如果有对话框将立即删除，应用程序将被启动，如果没有警报会弹出说应用程序需要互联网连接才能工作。
final AlertDialog alertDialog = new AlertDialog.Builder(this).create();
alertDialog.setTitle("Checking Connection");
alertDialog.setMessage("Checking...");
alertDialog.show();
new CountDownTimer(5000, 1000) {
@Override
public void onTick(long millisUntilFinished) {
new Thread(new Runnable() {
public void run() {
try {
URL url = new URL("http://web.mit.edu/");
HttpURLConnection connection = (HttpURLConnection) url.openConnection();
connection.setRequestMethod("GET");
connection.setConnectTimeout(5000);
isConnected = connection.getResponseCode() == HttpURLConnection.HTTP_OK;
} catch (Exception e) {
e.printStackTrace();
}
}
}).start();
if (isConnected == false){
alertDialog.setMessage("Try " + (5 - millisUntilFinished/1000) + " of 5.");
} else {
alertDialog.dismiss();
}
}
@Override
public void onFinish() {
if (isConnected == false) {
alertDialog.dismiss();
new AlertDialog.Builder(activity)
.setTitle("No Internet")
.setMessage("Please connect to Internet first.")
.setPositiveButton(android.R.string.yes, new DialogInterface.OnClickListener() {
public void onClick(DialogInterface dialog, int which) {
// kill the app?
}
})
.setIcon(android.R.drawable.ic_dialog_alert)
.show();
} else {
// Launch the app
}
}
}.start();
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
}
这里有一个简单的解决方案，以确保你的应用程序可以访问互联网:
static final String CHECK_INTERNET_ACCESS_URL = "https://www.google.com";
public static void isInternetAccessWorking(Context context) {
StringRequest stringRequest = new StringRequest(Request.Method.GET, CHECK_INTERNET_ACCESS_URL,
new Response.Listener<String>() {
@Override
public void onResponse(String response) {
// Internet access is OK
}
}, new Response.ErrorListener() {
@Override
public void onErrorResponse(VolleyError error) {
// NO internet access
}
});
Volley.newRequestQueue(context).add(stringRequest);
}
这个解决方案使用Android的Volley库，必须在build.gradle中声明:
implementation 'com.android.volley:volley:1.1.1'
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
