Solved

problem using applet (and socket)

Posted on 2001-08-29
6
403 Views
Last Modified: 2012-06-21
hi, experts. i have a problem with applet. i make a program that consist of client (applet) and server (application).
if the client using application too, than my program has no problem. the problem in applet is when i want it to loop forever ( using while(true) ) than it's hang (keep download). i know this because if i blocked the 'while' looping (that's will execute only once) than there is no problem occured.


here are the 2 files :
_______________________________________________________
server2.java
-------------------------------------------------
mport java.io.IOException;
import java.net.*;
import java.io.*;

class server2
{
    private static String str = "start receiving data";
    private Socket sock;
    private int counter;

    public static void main(String[] args)
            throws IOException
    {
        ServerSocket servSock = new ServerSocket(8189);
        System.out.println("Server started");
        Socket sock = servSock.accept();

        try
        {
            BufferedReader in = new BufferedReader(
                                new InputStreamReader(sock.getInputStream()));
            PrintStream out = new PrintStream(sock.getOutputStream());
            System.out.println(in.readLine());

            while(true)
                out.println(str);

        } catch(Exception e) {
            System.out.println(e);
        }
    }
}

______________________________________________________
the applet file
-------------------------------------------------
import java.applet.Applet;
import java.awt.*;
import java.awt.event.*;
import java.util.*;
import java.net.Socket;
import java.io.*;

public class clientApplet extends Applet {
    TextArea display;
    private static String host;
    private static Socket sock = null;
    private static PrintStream out = null;
    private static BufferedReader in = null;
     
    public void init() {
         // set the layout
        GridBagLayout gridBag = new GridBagLayout();
        GridBagConstraints c = new GridBagConstraints();
        setLayout(gridBag);

     // add textarea
        display = new TextArea(10, 20);
        display.setEditable(false);
        c.weightx = 1.0;
        c.weighty = 1.0;
        c.fill = GridBagConstraints.BOTH;
        gridBag.setConstraints(display, c);
        add(display);
       
        getData();
    }
   
    public void getData() {
        try {
            host = getCodeBase().getHost();
            sock = new Socket(host,8189);
            out  = new PrintStream(sock.getOutputStream());
            in   = new BufferedReader(new InputStreamReader(sock.getInputStream()));

            display.append("Waiting Data from server...");
            out.println("client");

            while (true)                 //the problem
                display.append("\n--> Data from server : " + in.readLine());

        } catch (java.net.UnknownHostException e) {
            display.append("Can't find host");
        } catch(java.io.IOException e) {
            display.append(e.toString());
        } catch(java.lang.Exception e) {
            display.append(e.toString());
        }
    }
}
_______________________________________________________


if u have test it, u will agree with me that the problem is in the 'while' loop, inside the loop, i put 'in.readLine()' command that will waiting for server input, so the applet should be waiting , not running.

hope u guys can help me, thanx.


regards,
sky.
0
Comment
Question by:skyjamz
[X]
Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people just like you are talking about what matters.

  • Help others & share knowledge
  • Earn cash & points
  • Learn & ask questions
  • 3
  • 3
6 Comments
 
LVL 92

Expert Comment

by:objects
ID: 6435737
Your applet will never get past initialising, and start() won't get called.

Call getData() in start() not init().
0
 

Author Comment

by:skyjamz
ID: 6439288
i have change the code like this :

________________________________________________________
import java.applet.Applet;
import java.awt.*;
import java.awt.event.*;
import java.util.*;
import java.net.Socket;
import java.io.*;

public class clientApplet extends Applet {
    TextArea display;
    private static String host;                                        // setting localhost name
    private static Socket sock = null;
    private static PrintStream out = null;
    private static BufferedReader in = null;
     
    public void init() {
         // set the layout
        GridBagLayout gridBag = new GridBagLayout();
        GridBagConstraints c = new GridBagConstraints();
        setLayout(gridBag);

     // add textarea
        display = new TextArea(10, 20);
        display.setEditable(false);
        c.weightx = 1.0;
        c.weighty = 1.0;
        c.fill = GridBagConstraints.BOTH;
        gridBag.setConstraints(display, c);
        add(display);
    }
   
    public void start() {
        getData();
    }
   
    public void getData() {
        try {
            host = getCodeBase().getHost();                                              //Get the address of the host we came from.
            sock = new Socket(host,8189);                              // connect to the server
            out  = new PrintStream(sock.getOutputStream());                    // get stream
            in   = new BufferedReader(new InputStreamReader(sock.getInputStream()));

            display.append("Waiting Data from server...");
            out.println("client");                                   // send data to server telling that itself is a client
           
            while (true)                                         // looping forever
                display.append("\n--> Data from server : " + in.readLine());          // getting data from server
               
        } catch (java.net.UnknownHostException e) {
            display.append("Can't find host");
        } catch(java.io.IOException e) {
            display.append(e.toString());
        } catch(java.lang.Exception e) {
            display.append(e.toString());
        }
    }
}
________________________________________________________

it's getting better, it's stop initialized, but still there are no respond, and if i blocked or delete the 'while(true)' than it goes normal.

regards,
sky.
0
 
LVL 92

Accepted Solution

by:
objects earned 200 total points
ID: 6439337
Sorry I missed the full extent of the problem originally.

What you need to do is run getData() in a seperate thread.

Something like:

public void start()
{
  new Thread(new Runnable()
    { public void run() { getData(); }}).start();
}


0
What Is Transaction Monitoring and who needs it?

Synthetic Transaction Monitoring that you need for the day to day, which ensures your business website keeps running optimally, and that there is no downtime to impact your customer experience.

 

Author Comment

by:skyjamz
ID: 6439606
wow, it's really work ! i'm really really happy, thanks alot, i think i adore u, OBJECTS, :) .

u have ease my job. i have dealed with that error for three days.

but could u explain more why it hangs ? and why i must run getData() is a separate method ?

i really appreciate your help, thank a lot.

regards,
sky.
0
 
LVL 92

Expert Comment

by:objects
ID: 6439639
> thanks alot, i think i adore u

You're 2 kind.

> but could u explain more why it hangs ?

You were calling getData from the thread that handles updating your gui. So because getData never returns your gui never gets updated.

> and why i must run getData() is a separate method ?

By running getData in a seperate thread, you have two threads of execution, one handling talking to the server, and the other handling your gui.


Happy to be of assistance.
Thanks for the points :)
0
 

Author Comment

by:skyjamz
ID: 6439683
thank u for your explanation, it's really help me, hope u can help me next time :)

regards,
sky.
0

Featured Post

Industry Leaders: We Want Your Opinion!

We value your feedback.

Take our survey and automatically be enter to win anyone of the following:
Yeti Cooler, Amazon eGift Card, and Movie eGift Card!

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Java contains several comparison operators (e.g., <, <=, >, >=, ==, !=) that allow you to compare primitive values. However, these operators cannot be used to compare the contents of objects. Interface Comparable is used to allow objects of a cl…
In this post we will learn different types of Android Layout and some basics of an Android App.
Viewers will learn about if statements in Java and their use The if statement: The condition required to create an if statement: Variations of if statements: An example using if statements:
This theoretical tutorial explains exceptions, reasons for exceptions, different categories of exception and exception hierarchy.

717 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question