I'm saying i'm not a programmer but a guy who has been learning to program with java for a while. I hope to find the solution to my problem here. I'm trying to program my home automation system and remote control and to do this, I chose to use Kryonet. My problem is that every time I send the data to the server, the client opens a new connection. It's been 3 weeks since googlo and I try to figure out how to do it but with no results.
Every help is seriously appreciated. This is my code. Thank you.
This code work in my home network.
Sorry for my english...
public class MainActivity extends AppCompatActivity {
Button button;
String IP = "";
EditText editText;
TextView textView;
EditText editText3;
public static String msg_response;
public static String msg_request;
#Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
final Handler handler = new MyHandler();
button = (Button) findViewById(R.id.button);
editText = (EditText) findViewById(R.id.editText);
editText3 = (EditText) findViewById(R.id.editText3);
textView = (TextView) findViewById(R.id.textView);
int MY_PERMISSIONS_REQUEST_INTERNET = 1;
int MY_PERMISSIONS_REQUEST_ACCESS_NETWORK_STATE = 1;
ActivityCompat.requestPermissions(this,
new String[]{Manifest.permission.INTERNET},
MY_PERMISSIONS_REQUEST_INTERNET);
ActivityCompat.requestPermissions(this,
new String[]{Manifest.permission.ACCESS_NETWORK_STATE},
MY_PERMISSIONS_REQUEST_ACCESS_NETWORK_STATE);
int MY_PERMISSIONS_REQUEST_ACCESS_WIFY_STATE = 1;
ActivityCompat.requestPermissions(this,
new String[]{Manifest.permission.ACCESS_WIFI_STATE},
MY_PERMISSIONS_REQUEST_ACCESS_WIFY_STATE);
textView.setText(msg_response);
button.setOnClickListener(new View.OnClickListener() {
#Override
public void onClick(View v) {
try {
msg_request = valueOf(editText3.getText().toString());
} catch (Exception e) {
e.printStackTrace();
}
MyThread myThread = new MyThread(handler);
myThread.start();
}
});
}
private class MyHandler extends Handler {
#Override
public void handleMessage(Message msg) {
Bundle bundle = msg.getData();
if (bundle.containsKey("msg da server")) {
String msgin = bundle.getString("msg da server");
textView.setText(msgin);
}
}
}
class MyThread extends Thread {
private Handler handler;
public MyThread(Handler handler) {
this.handler = handler;
}
public void run() {
System.out.println("MyThread running");
Client client = new Client();
client.start();
Kryo kryoClient = client.getKryo();
kryoClient.register(SampleRequest.class);
kryoClient.register(SampleResponse.class);
try {
client.connect(5000, "192.168.0.101", 54555, 54666);
} catch (IOException e) {
e.printStackTrace();
}
client.addListener(new Listener() {
public void received(Connection connection, Object object) {
if (object instanceof SampleResponse) {
SampleResponse response = (SampleResponse) object;
System.out.println(response.text);
msg_response = response.text.toString();
invia_activity(msg_response);
}
}
});
SampleRequest request = new SampleRequest();
request.text = msg_request;
client.sendTCP(request);
}
private void invia_activity(String invia) {
Message msg = handler.obtainMessage();
Bundle b = new Bundle();
b.putString("msg da server", "" + invia);
msg.setData(b);
handler.sendMessage(msg);
}
}
}
I dont have an direct solution, but i have an tutorial for it. I used the same one. So there the connections keeps open, and you can send as many packets as you need. Its without audio, but the code works well. After that you can experiment with the code. It works fine for me. This is the tutorial
I hope i can help you with this.
EDIT:
Maybe you can make an
public static Connection conn;
and you could use that object again and again as your connection to the server.
Related
I am developing extending WebSocketBehavior in order to send logging data to a client.. have generated the logging handler and it fires as and when needed.
I am having trouble understanding how exactly to push the log entries to the clients and update the console panel. I already know the onMessage method is what I need to override with the console taking the WeSocketRequestHandler as an argument along with the message I want to send. How exactly do I get the onMessage to fire properly?? Here is the code I am using:
public class LogWebSocketBehavior extends WebSocketBehavior {
private static final long serialVersionUID = 1L;
Console console;
private Handler logHandler;
private Model model;
public LogWebSocketBehavior(Console console) {
super();
configureLogger();
this.console = console;
}
private void configureLogger() {
Logger l = Logger.getLogger(AppUtils.loggerName);
logHandler = getLoggerHandler();
l.addHandler(logHandler);
}
#Override
protected void onMessage(WebSocketRequestHandler handler, TextMessage message) {
console.info(handler, model.getObject());
}
private Handler getLoggerHandler() {
return new Handler() {
#Override
public void publish(LogRecord record) {
model.setObject(record);
}
#Override
public void flush() {
throw new UnsupportedOperationException("Not supported yet."); //To change body of generated methods, choose Tools | Templates.
}
#Override
public void close() throws SecurityException {
throw new UnsupportedOperationException("Not supported yet."); //To change body of generated methods, choose Tools | Templates.
}
};
}
private Collection<IWebSocketConnection> getConnectedClients() {
IWebSocketConnectionRegistry registry = new SimpleWebSocketConnectionRegistry();
return registry.getConnections(getApplication());
}
private void sendToAllConnectedClients(String message) {
Collection<IWebSocketConnection> wsConnections = getConnectedClients();
for (IWebSocketConnection wsConnection : wsConnections) {
if (wsConnection != null && wsConnection.isOpen()) {
try {
wsConnection.sendMessage("test");
} catch (IOException e) {
}
}
}
}
}
The logger works as I want it to, providing messages as needed, but I cannot find how to actually fire the onMessage method to update my console. Any help is appreciated...
#onMessage() is called by Wicket whenever the browser pushes a message via Wicket.WebSocket.send("some message").
It is not very clear but I guess you need to push messages from the server to the clients (the browsers). If this is the case then you need to get a handle to IWebSocketRequestHandler and use its #push(String) method. You can do this with WebSocketSettings.Holder.get(Application.get()).getConnectionRegistry().getConnection(...).push("message").
Here is the class working as I need. Thank you Martin!!
public class LogWebSocketBehavior extends WebSocketBehavior {
private static final long serialVersionUID = 1L;
Console console;
private Handler logHandler;
private IModel model;
public LogWebSocketBehavior(Console console, IModel model) {
super();
configureLogger();
this.console = console;
this.model = model;
}
private void configureLogger() {
Logger l = Logger.getLogger(AppUtils.loggerName);
logHandler = getLoggerHandler();
l.addHandler(logHandler);
}
#Override
protected void onPush(WebSocketRequestHandler handler, IWebSocketPushMessage message) {
super.onPush(handler, message);
console.info(handler, model);
}
private Handler getLoggerHandler() {
return new Handler() {
#Override
public void publish(LogRecord record) {
model.setObject(record);
sendToAllConnectedClients(record.toString());
}
#Override
public void flush() {
throw new UnsupportedOperationException("Not supported yet."); //To change body of generated methods, choose Tools | Templates.
}
#Override
public void close() throws SecurityException {
throw new UnsupportedOperationException("Not supported yet."); //To change body of generated methods, choose Tools | Templates.
}
};
}
private Collection<IWebSocketConnection> getConnectedClients() {
IWebSocketConnectionRegistry registry = new SimpleWebSocketConnectionRegistry();
return registry.getConnections(getApplication());
}
private void sendToAllConnectedClients(String message) {
IWebSocketConnectionRegistry registry = new SimpleWebSocketConnectionRegistry();
WebSocketPushBroadcaster b = new WebSocketPushBroadcaster(registry);
IWebSocketPushMessage msg = new Message();
b.broadcastAll(getApplication(), msg);
}
class Message implements IWebSocketPushMessage {
public Message(){
}
}
}
I am trying to share image and text via whatsapp using implicit intent but I'm not able to share. I have searched the net but could not find any proper explanation. I have attached the code below.
... the code works without errors.but it dose not share content with whatsapp. i have searched all the places in google and on stackoverflow .but could not encounter with any proper explanation
public class MainActivity extends Activity implements View.OnClickListener {
TextView title_text,text_description;
ImageButton main_image;
Button button;
private static String url = "random url ";
private static final String TAG_ID = "title";
private static final String TAG_IMAGE = "image_url";
private static final String TAG_DESC = "product_desc";
String id;
String name;
String image;
#Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
text_description=(TextView)findViewById(R.id.text_description);
title_text=(TextView)findViewById(R.id.text_title);
main_image=(ImageButton)findViewById(R.id.image_main);
button= (Button) findViewById(R.id.button);
button.setOnClickListener(this);
new GetContacts().execute();
}
#Override
public void onClick(View v) {
Toast.makeText(getApplicationContext(),"whatsappsharing",Toast.LENGTH_SHORT).show();
Intent whatsappintent=new Intent();
whatsappintent.setAction(Intent.ACTION_SEND);
Uri uri= Uri.parse(TAG_IMAGE);
whatsappintent.setType("text/plain");
whatsappintent.putExtra(Intent.EXTRA_TEXT, "hello nathar");
whatsappintent.setType("image/jpeg");
whatsappintent.putExtra(Intent.EXTRA_STREAM,uri);
whatsappintent.setPackage("com.whatsapp");
startActivity(whatsappintent);
}
private class GetContacts extends AsyncTask<Void, Void, Void>
{
#Override
protected void onPreExecute()
{
super.onPreExecute();
}
#Override
protected Void doInBackground(Void... arg0) {
ServiceHandler sh = new ServiceHandler();
String jsonStr = sh.makeServiceCall(url, ServiceHandler.POST);
Log.d("Response: ", "> " + jsonStr);
if (jsonStr != null) {
try {
JSONObject jsonObj = new JSONObject(jsonStr);
id = jsonObj.getString(TAG_ID);
Log.d(TAG_ID,"title");
name = jsonObj.getString(TAG_DESC);
Log.d(TAG_DESC,"name");
image=jsonObj.getString(TAG_IMAGE);
Log.d(TAG_IMAGE,"image");
} catch (JSONException e) {
e.printStackTrace();
}
} else {
Log.e("ServiceHandler", "Couldn't get any data from the url");
}
return null;
}
#Override
protected void onPostExecute(Void result) {
super.onPostExecute(result);
if (image!=null && !image.equalsIgnoreCase(""))
Picasso.with(MainActivity.this).load(image).fit().into(main_image);
title_text.setText(id);
text_description.setText(name);
}
}
}
Try this piece of code:
Uri imageUri = Uri.parse(TAG_IMAGE);
Intent shareIntent = new Intent();
shareIntent.setAction(Intent.ACTION_SEND);
// Target whatsapp:
shareIntent.setPackage("com.whatsapp");
// Add text and then Image URI
shareIntent.putExtra(Intent.EXTRA_TEXT, <Message_text>);
shareIntent.putExtra(Intent.EXTRA_STREAM, imageUri);
shareIntent.addFlags(Intent.FLAG_GRANT_READ_URI_PERMISSION);
try {
startActivity(shareIntent);
} catch (android.content.ActivityNotFoundException ex) {
ex.printStackTrace();
}
where Message_text will be sent as a image caption
I am trying to finish this program and i am stuck. This is my first program and now it wont work. I keep getting this error when i add query.fromLocalDatastore(); The code runs fine until i try to get it from the local storage. This is telling me there is nothing there for it to retrieve and i don't know why. When i added my test data it worked fine but when i try to pull data from another table i get the error above. Apparently when i added the test data the server synced with the local datastore. Now it is not. Can someone tell me what I did wrong?
public class DataHolder extends Application {
int age;
#Override
public void onCreate() {
super.onCreate();
Parse.enableLocalDatastore(getApplicationContext());
Parse.initialize(this,key, key);
ParseUser.enableAutomaticUser();
ParseACL defaultACL = new ParseACL();
ParseACL.setDefaultACL(defaultACL, true);
}
public class MainActivity extends ActionBarActivity implements Disclaimer.DisclaimerListener {
protected void continueToRun() {
spinner1.setOnItemSelectedListener(new AdapterView.OnItemSelectedListener() {
public void onItemSelected(AdapterView<?> adapter, View v, int x, long lng) {
final ParseQuery<ParseObject> query = ParseQuery.getQuery("Phone_Numbers");
query.fromLocalDatastore();
if (x == 1) {
final Intent intent = new Intent(getBaseContext(), Protocol_Template.class);
query.fromLocalDatastore();
query.whereEqualTo("objectId", "uGANULyrdL");
startActivity(intent);
}
}
public class Protocol_Template extends Activity {
DataHolder global;
#Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_protocol__template);
final TextView protocol = (TextView) findViewById(R.id.txt02);
findViewById(R.id.btn2timesUpperLeft);
final ParseQuery<ParseObject> query = ParseQuery.getQuery("Phone_Numbers");
query.fromLocalDatastore();
query.getFirstInBackground(new GetCallback<ParseObject>() {
public void done(ParseObject object, ParseException e) {
if (e == null) {
final String protocols = object.get("PhoneNumber").toString();
protocol.setText(protocols);
} else {
protocol.setText(e.getMessage());
}
}
});
}
I'm a Android & Java newbie and here is my situation:
I'm trying to create an app which connects to a Beaglebone Black using a ssh connection and then controls some peripherals connected to the BBB by issuing commands coming from an Android device.
I'm opening (successfully) an ssh session in an AsyncTask while the user sees an splash screen, if the connection was successful the user will get a confirmation and then will be able to send predefined commands by clicking some available buttons.
What I want to do next is left the session opened and then create a new channel (exec or shell ) each time I wish to issue a command and wait for the response from the BBB, but I donĀ“t know how to reuse such ssh session outside the AsynkTask.
is that even possible?
I'm using Android Studio 0.8.2 and Jsch 0.1.51, my code is as follows:
public class SplashScreen extends ActionBarActivity {
public static final int segundos =10;
public static final int milisegundos =segundos*1000;
public static final int delay=2;
private ProgressBar pbprogreso;
#Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_splash_screen);
pbprogreso= (ProgressBar)findViewById(R.id.pbprogreso);
pbprogreso.setMax(maximo_progreso());
empezaranimacion();
}
public void empezaranimacion()
{
sshConnect task = new sshConnect();
task.execute(new String[] {"http:"});
new CountDownTimer(milisegundos,1000)
{
#Override
public void onTick(long milisUntilFinished){
pbprogreso.setProgress(establecer_progreso(milisUntilFinished));
}
#Override
public void onFinish(){
finish();
}
}.start();
}
public int establecer_progreso (long miliseconds)
{
return (int)((milisegundos-miliseconds)/1000);
}
public int maximo_progreso () {
return segundos-delay;
}
public class sshConnect extends AsyncTask <String, Void, String>{
ByteArrayOutputStream Baos=new ByteArrayOutputStream();
ByteArrayInputStream Bais = new ByteArrayInputStream(new byte[1000]);
#Override
protected String doInBackground(String... data) {
String host = "xxxxxxx";
String user = "root";
String pwd = "";
int port = 22;
JSch jsch = new JSch();
try {
Session session = jsch.getSession(user, host, port);
session.setPassword(pwd);
session.setConfig("StrictHostKeyChecking", "no");
session.connect();
ChannelExec channel = (ChannelExec)session.openChannel("exec");
channel.setOutputStream(Baos);
channel.setInputStream(Bais);
//Run Command
channel.setCommand("python ~/BBB_test/testconnect.py");
channel.connect();
try{Thread.sleep(3500);}catch (Exception ee){}
channel.disconnect();
//session.disconnect();
}catch (Exception e){
System.out.println(e.getMessage());
}
return Baos.toString();
}
#Override
protected void onPostExecute(String result) {
if (result.equals("Connected to BBB Baby!!\n")) {
Intent nuevofrom = new Intent(SplashScreen.this, Principal.class);
startActivity(nuevofrom);
finish();
} else {
Intent newfrom = new Intent(SplashScreen.this, ConnecError.class);
startActivity(newfrom);
finish();
}
}
}
//Here is where I want to reuse the opened session and create a new channel
public class sendCommand extends AsyncTask <String, Void, String >{
ByteArrayOutputStream Baosc=new ByteArrayOutputStream();
ByteArrayInputStream Baisc = new ByteArrayInputStream(new byte[1000])
protected String doInBackground (String... command){
try {
ChannelExec channel2 = (ChannelExec)session.openChannel("exec");
channel2.setOutputStream(Baosc);
channel2.setInputStream(Baisc);
//Run Command
channel2.setCommand("python ~/BBB_test/testgpio.py");
channel2.connect();
try{Thread.sleep(3500);}catch (Exception ee){}
channel2.disconnect();
}catch (Exception e) {
System.out.println(e.getMessage());
}
return Baosc.toString();
}
protected void onPostExecute(Long result) {
TextView txt = (TextView) findViewById(R.id.infotext);
txt.setText(result);
}
}
If something else is needed let me know! (it is the first time I ask something in a forum)
Thanks a lot for your time and support!
I managed to get what I wanted by using the recommendation from DamienKnight of creating the session outside the Asynktask class. I create a public classwith three methods to create, return and disconnect the session:
public static class cSession {
String host = "xxx.xxx.xxx.xxx";
String user = "root";
String pwd = "";
int port = 22;
JSch jsch = new JSch();
public Session Met1 (){
try {
session = jsch.getSession(user, host, port);
session.setPassword(pwd);
session.setConfig("StrictHostKeyChecking", "no");
} catch (Exception e2){
System.out.println(e2.getMessage());
}return session;
}
public Session damesession (){
return session;
}
public void close_ses(){
session.disconnect();
}
}
By doing this so, the creation of channels is flexible and I can use the methods from Jsch too.
public class sshConnect extends AsyncTask <String, Void, String>{
ByteArrayOutputStream Baos=new ByteArrayOutputStream();
ByteArrayInputStream Bais = new ByteArrayInputStream(new byte[1000]);
#Override
protected String doInBackground(String... data) {
cSession jschses = new cSession();
Session ses =null;
ses = jschses.Met1();
try {
ses.connect();
ChannelExec channel = (ChannelExec)ses.openChannel("exec");
channel.setOutputStream(Baos);
channel.setInputStream(Bais);
//Run Command
channel.setCommand("python ~/BBB_test/testconnect.py");
channel.connect();
try{Thread.sleep(3500);}catch (Exception ee){}
channel.disconnect();
//session.disconnect();
}catch (Exception e){
System.out.println(e.getMessage());
}
return Baos.toString();
}
Thanks #Damienknight!
Regards
If you are wanting to reuse the session, you dont need to reconect the channel each time. Connect it once as a shell, plugging an input and output stream into it. Use the streams to pass commands and capture output.
See the JSCH example on the JCraft website.
Channel channel=session.openChannel("shell");
channel.setInputStream(System.in);
channel.setOutputStream(System.out);
channel.connect();
I want to create httpconnection using asyntask.three parameters are posted to the server
username,password and a search item.the search is provided by the user in an EditText such so that when the user clicks a button,the search item is sent to the server.I want to execute the doInbackground() method in the OnclickListener and display the response from the server on listviews.This is the AsyncTask Class
public class PostToServer extends AsyncTask<String, Void, String> {
#Override
protected void onPostExecute(String result) {
}
#Override
protected String doInBackground(String... arg0) {
try {
HttpClient client = new DefaultHttpClient();
String postURL = "url";
String username ="username";
String password = "password";
HttpPost post = new HttpPost(postURL);
List<NameValuePair> params = new ArrayList<NameValuePair>();
params.add(new BasicNameValuePair("user", username));
params.add(new BasicNameValuePair("pass", password));
UrlEncodedFormEntity ent = new UrlEncodedFormEntity(params,HTTP.UTF_8);
post.setEntity(ent);
HttpResponse responsePOST = client.execute(post);
HttpEntity resEntity = responsePOST.getEntity();
if (resEntity != null) {
Log.i("RESPONSE",EntityUtils.toString(resEntity));
}
} catch (Exception e) {
e.printStackTrace();
}
return null;
}
This the class where the click event is called
public class StartPost extends Activity {
#Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.result_page);
}
Button Submit = (Button) findViewById(R.id.btn_search);
EditText textvalue = (EditText)findViewById(R.id.searcheditText);
String value = textvalue.getText().toString();
PostToServer post = new PostToServer();
CheckInternetConnection check = new CheckInternetConnection(null);
private OnClickListener click = new OnClickListener() {
#Override
public void onClick(final View v) {
switch(v.getId()){
case R.id.btn_search:
post.execute();
break;
}
}
};
}
Questions
1.What am I doing wrong because it seems the post is not working and How can I display the server results from the onPostExecute()?
Thank You.
Create a onPostExecute method to display the results on your textview.
protected onPostExecute (String result){
TextView tv = (TextView) findViewById(R.id.textview1);
if(result != null){
tv.setText(result);
}
}