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();
}
}
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)
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。
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