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();
}
}
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。
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
}
The other answers that use ConnectivityManager are wrong because having a network connection doesn't mean you have internet access. For example, the user might be connected to a coffee shop's WiFi portal but can't get to the internet. To check that the internet is accessible you have to try to connect to an actual server. Normally when you want to do this you have a specific server in mind that you want to connect to, so go ahead and check if you can connect to that server. Here's a simple method for checking connectivity to a server.
private boolean isOnTheInternet() {
try {
URLConnection urlConnection = new URL("http://yourserver").openConnection();
urlConnection.setConnectTimeout(400);
urlConnection.connect();
return true;
} catch (Exception e) {
return false;
}
}
设置ConnectTimeout的原因是，否则它默认为TCP超时，可以有很多秒长。
还要注意的是，Android不允许你在主线程上运行这个程序。
看一下ConnectivityManager类。您可以使用这个类来获取主机上活动连接的信息。http://developer.android.com/reference/android/net/ConnectivityManager.html
编辑:你可以使用
Context.getSystemService(Context.CONNECTIVITY_SERVICE)
.getNetworkInfo(ConnectivityManager.TYPE_MOBILE)
or
Context.getSystemService(Context.CONNECTIVITY_SERVICE)
.getNetworkInfo(ConnectivityManager.TYPE_WIFI)
并解析返回NetworkInfo对象的DetailedState枚举
EDIT EDIT:查看是否可以访问主机
Context.getSystemService(Context.CONNECTIVITY_SERVICE)
.requestRouteToHost(TYPE_WIFI, int hostAddress)
显然，我使用Context.getSystemService(Context.CONNECTIVITY_SERVICE)作为代理来表示
ConnectivityManager cm = Context.getSystemService(Context.CONNECTIVITY_SERVICE);
cm.yourMethodCallHere();
