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How do I make a client-server Java application to send messages on one port but receive them on another?

I am currently trying to make an application that will send messages to a server using one port, but will receive messages on another port. However, based on tutorials I have followed, it looks like the act of connecting to the server is where ports come into play and my client is receiving and sending messages on the same port. How do I make it so it sends on one port but receives on the other?
Here is the code that I think is relevant from the client side (I put some stuff that seems unrelated because I think they are things that would be altered by receiving on one port but sending on another, and ignore the comment about replacing inetaddress, that is just me working on implementing this in a gui):
public void startRunning(){
try{
connectToServer();
setupStreams();
whileChatting();
}catch(EOFException eofException){
showMessage("\n Client terminated connection");
}catch(IOException ioException){
ioException.printStackTrace();
}finally{
closeStuff();
}
}
//connect to server
private void connectToServer() throws IOException{
showMessage("Attempting connection... \n");
connection = new Socket(InetAddress.getByName(serverIP), 480);//replace serverIP with ipTextField.getText or set serverIP to equal ipTextField.getText? Same with port number.
showMessage("Connected to: " + connection.getInetAddress().getHostName() );
}
//set up streams to send and receive messages
private void setupStreams() throws IOException{
output = new ObjectOutputStream(connection.getOutputStream());
output.flush();
input = new ObjectInputStream(connection.getInputStream());
showMessage("\n Streams are good! \n");
}
//while talking with server
private void whileChatting() throws IOException{
ableToType(true);
do{
try{
message = (String) input.readObject();
showMessage("\n" + message);
}catch(ClassNotFoundException classNotfoundException){
showMessage("\n Don't know that object type");
}
}while(!message.equals("SERVER - END"));
}
//send messages to server
private void sendMessage(String message){
try{
output.writeObject("CLIENT - " + message);
output.flush();
showMessage("\nCLIENT - " + message);
}catch(IOException ioException){
messageWindow.append("\n something messed up ");
}
}
//change/update message window
private void showMessage(final String m){
SwingUtilities.invokeLater(
new Runnable(){
public void run(){
messageWindow.append(m);
}
}
);
}
EDIT/UPDATE: To help clarify some things, here is some more information. The device that sends the first message is connected to a sensor, and it sends information when that sensor detects something to the other device. The receiving device sends a message back on a different port telling the original sending device how to respond. Lets name these two devices the "reporter-action taker" and the "decision maker-commander".

If you want to use TCP/IP sockets you can't use a a socket to send and another to read. That's not what they are for.
If you use a centralized distributed algorithm (server/client communication) you have to set the server to listen on a single socket port with the ServerSocket class: then the server tries to accept clients through that socket.
Example:
ServerSocket listener = new ServerSocket(Port)
While (true) {
new Clienthandler(listener.accept());
}
The server will listen on that port, and when a client tries to connect to that port if it is accepted the server launches its handler. On this handler constructor the Socket object used on the client is received on an argument and can then be used to get the writers and the readers. The reader on this handler class will be the writer on the client class and vice-versa, maybe that's what you were looking for.

Your question about using two ports in this manner is a bit strange. You state that you have a client and a server and that they should communicate on different ports.
Just to clarify picture the server as a hanging rack for jackets with several hooks in a row. Each port the server listened on represents a hook. When it comes to the client server relationship the client or jacket knows where to find its hook, however the hook is blind and have no idea where to find jackets.
Now, the client selects a port or a hook and connects to it. The connection is like a pipeline with two pipes. One for the client to deliver data to the server with and the other to send data from the server back to the client. When the connection is established data can be transferred both ways. This means that we only need one port open on the server to send data both from the client to the server and in the opposite direction.
The reason for only having one open port open on the server for the clients to connect to is that holding an open port for connections is hard to do on a regular client computer. The normal desktop user will be behind several firewalls blocking incoming connections. If that wasn't the case the client would probably be hacked senseless from malicious viruses.
Moving on with the two port solution we could not call this a client server connection per say. It would be more like a peer to peer connection or something like that. But if this is what you want to do, the application connecting first would have to start by telling the other application what ip and port to use for connecting back, it should probably also want to give some kind of token that are to be used to pair the new incoming connection when connecting back.
You should take note that making such an implementation is not a good idea most of the time as it complicates things a whole lot for simple data transfer between a client and server application.

how to run a server socket on amazon EC2

I'm trying to write an android app that interacts with a server. I'm trying now to write the server side with sockets. I created an instance of ec2 and ran it. I connected to it with putty and ran a simple "hello world" java program. Now I'm trying to run a server which use a socket, but I get this as the server socket: ServerSocket[addr=0.0.0.0/0.0.0.0,localport=5667].
This is my code, very basic so far:
public class EchoServer {
public static void main(String[] args) throws IOException {
int portNumber = 5667;
System.out.println("server socket main");
try (ServerSocket serverSocket = new ServerSocket(portNumber);
) {
System.out.println(serverSocket.toString());
} catch (IOException e) {
System.out
.println("Exception caught when trying to listen on port "
+ portNumber + " or listening for a connection");
System.out.println(e.getMessage());
}
}
}
What should I do to run the server so I could connect to it via my andriod device?
Thanks!

You are on the right track. 0.0.0.0 simply means:
A way to specify "any IPv4-interface at all". It is used in this way when configuring servers (i.e. when binding listening sockets).
Next, your server will need to listen for incoming connections by using SocketServer.accept().
And then of course you'll want to receive and send data on the socket. This tutorial should help.
Finally, if you plan on serving multiple clients simultaneously with your server, you will want to consider concurrency and scalability, and perhaps consider using a framework like Netty.

Java.net.SocketException: Permission denied: connect

I have two PCs in one network that I want to connect. One of them should send a notification to the other via TCP. One the one PC I have a "server" (Python script) socket which waits for the "client"(Jar file) to send a specific String and then it gives me a notification. This works perfectly fine when I'm trying it out one one PC. But when I want to do the intended action the "client" PC's .jar gives me an error that the connection is refused. Do I have to open a specific port on the other PC or what else could cause trouble? One PC runs Fedora the other Windows 8
"Server Code"
s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
s.bind(("", 5005))
s.listen(1)
try:
while True:
komm, addr = s.accept()
while True:
data = komm.recv(1024)
if data == "$":
noty()
if not data:
komm.close()
break
finally:
s.close()
"Client" Code
public static void main(String[] args) throws Exception {
Socket socket = new Socket("192.168.178.25", 5005);
OutputStream out = socket.getOutputStream();
String dat = "$";
out.write(dat.getBytes());
socket.close();
}

Your server is probably binding to the wrong interface,
calling
s.bind(("", 5005))
Without setting an interface will allow the program to pick what ip address / interface it will connect to.
Since your client is trying to connect to ("192.168.178.25", 5005); you may want to put an IP address into the bind call to prevent the server picking the wrong ip interface.
Example:
s.bind(("192.168.178.25", 5005))

if its permission denied then something is blocking your connection with the computer. i would try to open a port and see if that works. if you want an example of java sockets you can take a look at my SUPER Tic-Tac-Toe Multiplayer it uses java sockets to send strings to the clients as a way to represent what actions the clients should take.

Java Threadpool TCP Server (Port keeps changing!!)

Good Day,
I was taking a look at this tutorial to do a TCP Threadpool server.
http://tutorials.jenkov.com/java-multithreaded-servers/thread-pooled-server.html
It works great for listening/RECEIVING to clients and processing, and returning a response. There is a class inside that I pass in WorkerRunnable into, and that basically prints out the remote socket address (who it was sent from)
public void run(){
synchronized(this){
this.runningThread = Thread.currentThread();
}
openServerSocket();
while(! isStopped()){
Socket clientSocket = null;
try {
clientSocket = this.serverSocket.accept();
} catch (IOException e) {
if(isStopped()) {
System.out.println("Server Stopped.") ;
return;
}
throw new RuntimeException(
"Error accepting client connection", e);
}
this.threadPool.execute(
new WorkerRunnable(clientSocket,
"Thread Pooled Server"));
}
this.threadPool.shutdown();
System.out.println("Server Stopped.") ;
}
The problem is. The remote address is supposed to stay fixed (I am working within my own home wifi router). However, the IP address of the sender stays the same, but the port keeps changing!!
This is a big problem for me..as I need to be able to return a response to the user for future tasks and I actually save this address to use again to send data. When I ran this in a single TCP thread..it stayed fixed (the port).
Why does the threadpool cause the TCP remote address port to keep changing?

With TCP, the client socket port is most of the time (almost 99%, except for specific protocols) randomly chosen. But to you don't have to know it, the only thing you have to do is to keep the clientSocket reference to write back data to the client. If you want to send data to the other host after that the connection is closed, you have to start a ServerSocket on both sides with a fixed port.

Even if you test from same machine the client port will be random by default. I am not sure if there is any way to set the client source port. However, if you use netstat or capture the packet you can be sure the source port is different for every connection.

Java Socket - How to send and receive information between 3 points?

In my application, multiple PCs will connect to the server. The server will receive the message then send to a system. The system may send a message back to the server, and this message should be sent to a PC by the server. Basically it's like this:
PC 0 --------------------> Server ----> Send PC msg to system.
System ------> Send msg to PC 0 Server ----> PC 0
The same with PC 1, 2, 3 etc...
But how can I do this? I can already get PC msg to Server and Server to System, like this:
This is running from the ServerThread:
try {
Socket socket = serverSocket.accept();
if (socket != null) {
clientList.add(new ClientThread(socket));
}
} catch (IOException ex) {
System.out.println(DEBUG
+ ": Failed creating client socket / "
+ ex.getMessage());
}
But how can I know if the incoming socket is meant to send a message to the Server or the socket should send a message to the PC?

You are basically implementing instant messaging, where your system is simply another endpoint (along with PC0 ...). Why not use something like XMPP if your server is simply being used as a message router.
You can use Smack for a client solution and any off the shelf XMPP server will easily suit your needs. If you need to customize the server you can choose one of the Java based ones like Openfire.

This approach is called a proxy. In order to distinguish between clients (PCs in your question) and servers (systems) you need to keep connections that were accepted from clients, and connections initiated by you to servers separate, but cross-referenced (the easiest is one-to-one mapping). Get a message on the client connection, optionally transform/filter, send on corresponding server connection, and the other way around.