Socket Programming java using datagram multiple packets

Question

I am trying to make a client-server program. Where in client i am getting multiple inputs from user and i have to pass it to server. Server uses that info and does some calculation and sends back the result. important parts from Client code:

private byte[] buf = new byte[256];
private byte[] buf2 = new byte[256];
private byte[] buf3 = new byte[256];
while (true) {

Arrays.fill(buf, (byte) 0); 

buf = jTextField1.getText().trim().getBytes();
sendPacket = new DatagramPacket(buf, buf.length, address, 8000);
socket.send(sendPacket);

buf2 = jTextField2.getText().trim().getBytes();
sendPacket = new DatagramPacket(buf2, buf2.length, address, 8000);
socket.send(sendPacket);

buf3 = jTextField3.getText().trim().getBytes();
sendPacket = new DatagramPacket(buf3, buf3.length, address, 8000);
socket.send(sendPacket);

//Get result from Server
socket.receive(recivePacket);
double result  = Double.parseDouble(new String(buf).trim());
System.out.println("Result" +result);

}

On Server side I am trying to recieve the data(which is not working)

   byte[] buf = new byte[256];
   DatagramPacket  recivePacket = new DatagramPacket(buf, buf.length);
   DatagramPacket dgp = new DatagramPacket(buf, buf.length);
    while (true) {
            Arrays.fill(buf, (byte) 0);
            serverSocket.receive(recivePacket);

            System.out.println("host Name is " + 
   recivePacket.getAddress().getHostName() + '\n');

            while (true) {
                serverSocket.receive(dgp);
                System.out.println("String=====" + new String(buf));
            }

Please help me out here. What is wrong with this code ... I am completely stuck So, its strange that sometimes it works. But it gives me value from last buffer. Other values are lost. Thanks in advance

Just dont use datagrams for that. Take a look into the standard TCP protocol. Makes more sense for you. This should be done via ServerSockets. (very easy to use) — Finn Eggers, Jun 20 '18 at 15:05
It is mandatory to use datagram. Thats the part of assignment. — Jeena, Jun 20 '18 at 15:14
Don't attach the same `buf` to two different `DatagramPacket` instances. — rustyx, Jun 20 '18 at 15:34
on both sides. Always create a new byte[]. Just look at my answer. — Finn Eggers, Jun 20 '18 at 15:59
What you can do is using the same instance but always create a new byte[] with: new byte[...]... — Finn Eggers, Jun 20 '18 at 16:00

Finn Eggers · Accepted Answer · 2018-06-20T15:56:40.533

So I think each time you recieve something on the server, you need to create a new DatagramPacket/byte[] for that. What I do is creating e a new DatagramPacket each time I recieve something like this:

DatagramPacket packet = new DatagramPacket(new byte[max_data_size], max_data_size);
socket.receive(packet);

My max_data_size is normally around 1024 bits.

Then you could read stuff like client, port, length etc:

InetAddress address = packet.getAddress();
int port = packet.getPort();
int len = packet.getLength();
byte[] data = packet.getData();

And usually, you should then be able to simply send a new datagramPacket back to the given InetAddress/port.

You want multiple inputs from the user? Pack everything in only one package.

private YourClass newInstance(byte[] bytes){
    ByteArrayInputStream bis = new ByteArrayInputStream(bytes);
    ObjectInput in = null;
    YourClass o = null;
    try {
        in = new ObjectInputStream(bis);
        o = (YourClass)in.readObject();
    } catch (IOException e) {
        e.printStackTrace();
    } catch (ClassNotFoundException e) {
        e.printStackTrace();
    } catch (ClassCastException e){
        e.printStackTrace();
    }finally {
        try {
            if (in != null) {
                in.close();
            }
        } catch (IOException ex) {
            // ignore close exception
        }
    }
    return o;
}

private byte[] toBytes(YourClass object){
    ByteArrayOutputStream bos = new ByteArrayOutputStream();
    ObjectOutput out = null;
    byte[] bytes = new byte[1];
    try {
        out = new ObjectOutputStream(bos);

        object.setTimestamp(System.currentTimeMillis());

        out.writeObject(object);
        out.flush();
        bytes = bos.toByteArray();
    } catch (IOException e) {
        e.printStackTrace();
    } finally {
        try {
            bos.close();
        } catch (IOException ex) {
            // ignore close exception
        }
    }
    return bytes;
}

So here is what i tried in server: while (true) { byte[] buf = new byte[256]; DatagramPacket dgp = new DatagramPacket(buf, buf.length); serverSocket.receive(dgp); InetAddress address = dgp.getAddress(); int port = dgp.getPort(); int len = dgp.getLength(); byte[] data = dgp.getData(); System.out.println("String=====" + new String(data)); } But no change. I get only one result. — Jeena, Jun 20 '18 at 15:48
I was also thinking about sending some serialized object but I have no clue how to deserialised it on server side — Jeena, Jun 20 '18 at 15:50
there you go. This is what I use. Just make sure that YourClass implements Serializable. — Finn Eggers, Jun 20 '18 at 15:57
Yes atlast it worked with Serialized object. Thank u so much @Finn for your help.. Thanks!! — Jeena, Jun 20 '18 at 16:39

Socket Programming java using datagram multiple packets

1 Answers1