if i have already connected to the server. then how i show exception saying it is already connected

how to add in exception showing the input has to be enter, if the user click on connect without entering any input? i would like the have a message box show if the user click on the connect button without entering the name, ip and port. [SOLVED]

using System;
using System.Collections.Generic;
using System.ComponentModel;
using System.Data;
using System.Drawing;
using System.Linq;
using System.Text;
using System.Windows.Forms;
using System.Net.Sockets;
using System.Threading;

namespace SocketClient
{

    public partial class SocketClient : Form
    {
        System.Net.Sockets.TcpClient clientSocket = new System.Net.Sockets.TcpClient();
        NetworkStream serverStream = default(NetworkStream);
        string readData = null;


        public SocketClient()
        {
            InitializeComponent();
        }


        private void getMessage()
        {
            while (true)
            {
                serverStream = clientSocket.GetStream();
                int buffSize = 0;
                byte[] inStream = new byte[10025];
                buffSize = clientSocket.ReceiveBufferSize;
                serverStream.Read(inStream, 0, buffSize);
                string returndata = System.Text.Encoding.ASCII.GetString(inStream);
                readData = "" + returndata;
                msg();
            }
        }


        private void msg()
        {
            if (this.InvokeRequired)
                this.Invoke(new MethodInvoker(msg));
            else
                textDisplay.Text = textDisplay.Text + Environment.NewLine + " >> " + readData;
        }


        private void buttonConnect_Click(object sender, EventArgs e)
        {

            readData = "Conected to NYP Server ...";
            msg();
            //clientSocket.Connect("127.0.0.1", 8888);
            clientSocket.Connect(textIP.Text, Convert.ToInt32(textPort.Text));
            serverStream = clientSocket.GetStream();

            byte[] outStream = System.Text.Encoding.ASCII.GetBytes(textName.Text + "$");
            serverStream.Write(outStream, 0, outStream.Length);
            serverStream.Flush();

            Thread ctThread = new Thread(getMessage);
            ctThread.Start();
        }

        private void buttonSend_Click(object sender, EventArgs e)
        {
            // Show msg box if no server is connected
            if (serverStream == null)
            {
                MessageBox.Show("Please connect to a server first!");
                return;
            }

            // Send text
            byte[开发者_如何转开发] outStream = System.Text.Encoding.ASCII.GetBytes(textSend.Text + "$");
            serverStream.Write(outStream, 0, outStream.Length);
            serverStream.Flush();

            // Clear text
            textSend.Text = "";

        }

        private void textDisplay_TextChanged(object sender, EventArgs e)
        {
            textDisplay.SelectionStart = textDisplay.Text.Length;
            textDisplay.ScrollToCaret();
            textDisplay.Refresh();
        }

        private void textSend_TextChanged(object sender, EventArgs e)
        {
            buttonSend.Enabled = !string.IsNullOrEmpty(textSend.Text);
        }
    }
}

if(string.IsNullOrEmpty(textIP.Text) || string.IsNullOrEmpty(textPort.Text)  || string.IsNullOrEmpty(textName.Text))
{
   MessageBox.Show("Please enter IP address, Port #, and Name");
}
else
{
  //they entered stuff...so, try to connect..
}

You can use TcpClient.Connected property to check whether a connection is already established or not.

if(clientSocket.Connected){
  .... show error message here
}
else{
  ... go ahead  
} 

Your form holds references to both a TcpClient and a NetworkStream instance. Both of these implement IDisposable, so that they must be disposed of explicitly. You must override the Dispose method of the form and dispose of these instances if they are not null.