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Geekpedia » C# Tutorials » File transfers through networks and the Internet

File transfers through networks and the Internet

This tutorial will teach you how to send and receive files from your local system to a server, either through a network or through the Internet, using C# and streaming connections. This tutorial is also helpful for learning the basics of networking using the .NET Framework.

On Wednesday, May 3rd 2006 at 08:28 PM

By Andrew Pociu (View Profile)

(Rated 4.6 with 22 votes)

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Download this project (Visual Studio 2005)

In this tutorial we're going to build a very interesting .NET application, and based on the code here you will hopefully be able to build various applications or to help you with extending your current work.

Before digging into the code, you should know that the code you see in this tutorial and in the attached project resumes to only the strict functionality of exchanging files between two computers using the .NET Framework. If you want to integrate this into a real-life application, you will need to put more effort into this code since currently it lacks crash prevention and recovery, testing on multiple systems, against firewalls, etc.
For using this code I recommend that you use two different computers, either on a network or on the Internet. The firewall on both computers (but especially on the server computer) should be shut down to prevent the port we're going to use being locked. However, the beauty of this application is that you can use a single computer to test it; you can run the server and the client application on the same computer and transfer files from one path to another.
Speaking of server and client, in this tutorial we're going to build two applications. One is the server that listens for connections on a certain port and receives files in streams and then saves them to the local C:\ drive, and one is the client that connects to the server application on the same port the server is listening to.

The application is developed using .NET Framework 2.0 in Visual Studio 2005, however you should be able to convert it to .NET Framework 1.1 rather easily.

How the file transfer works

First we need to create the server application. The server application pictured below under the arrow will be the one receiving a file from a remote computer and storing it on the hard drive. Prior to receiving the actual file, it will receive information about the file: the file name and size. This information will be passed by the client, and it will be retrievied using the FileInfo class from the .NET Framework.

The first thing you need to do is to enter a port (815 is normally not taken on most computers) and click on "Start Server". You will see a message "Starting the server..." shortly followed by "The server has started. Please connect the client to 192.168.0.1". Of course, the IP will probably differ in your case, so you should write it down because you will need to enter it in the client application later. At this time you should leave the server application running and switch to the client application (Network File Sender).

The client application will first need to connect to the server, so you need to press the "Connect" button. Before pressing the button however, please make sure that you entered the same port as in the server window, and the IP that you were prompted to enter.
Hopefully you will not experience any problems and you will see the "Successfully connected to server" message. After that you will be able to press the "Send New File" button which will let you browse to a file. Right after you pick the file, you will see that the progress bar on the server (Network File Receiver) starts making progress and the file is being transfered. So where is the file? By default the code saves it to C:\ so look for it there. After the file is sent the client and server-side streams and connections are closed to release resources. To prepare for an upcoming connection, the server then starts again (of course, you can change this to your liking).

Client to Server transfer

Receiving the file - creating the server

We will start with creating the server, since we first need an application to listen for connections before we actually send a file.
Begin by creating a new Windows Application project in Visual Studio 2005. Add to it two textboxes: txtPort stores the port that we are listening to, and txtLog will be the MultiLine textbox to which we will append text containing the status of the server. Add two buttons: btnStart and btnStop which are self explanatory, and finally a progress bar entitled prgDownload where we will display the progress of the file being transfered.

File Receiver Server

Switch to the code view and make sure you add these 4 additional using statements:

using System.Net;
using System.Net.Sockets;
using System.IO;
using System.Threading;

Next we need to create some objects we'll be setting and accessing throughout the application:

// The thread in which the file will be received
private Thread thrDownload;
// The stream for writing the file to the hard-drive
private Stream strLocal;
// The network stream that receives the file
private NetworkStream strRemote;
// The TCP listener that will listen for connections
private TcpListener tlsServer;
// Delegate for updating the logging textbox
private delegate void UpdateStatusCallback(string StatusMessage);
// Delegate for updating the progressbar
private delegate void UpdateProgressCallback(Int64 BytesRead, Int64 TotalBytes);
// For storing the progress in percentages
private static int PercentProgress;

Now we need to create the most important part of the application - the StartReceiving() method which listens and establishes connections with the client, and also receives the files. This method will be called from a thread, the thrDownload thread we defined above.

private void StartReceiving()
{
    // There are many lines in here that can cause an exception
    try
    {
        // Get the hostname of the current computer (the server)
        string hstServer = Dns.GetHostName();
        // Get the IP of the first network device, however this can prove unreliable on certain configurations
        IPAddress ipaLocal = Dns.GetHostEntry(hstServer).AddressList[0];
        // If the TCP listener object was not created before, create it
        if (tlsServer == null)
        {
            // Create the TCP listener object using the IP of the server and the specified port
            tlsServer = new TcpListener(ipaLocal, Convert.ToInt32(txtPort.Text));
        }
        // Write the status to the log textbox on the form (txtLog)
        this.Invoke(new UpdateStatusCallback(this.UpdateStatus), new object[] { "Starting the server...\r\n" });
        // Start the TCP listener and listen for connections
        tlsServer.Start();
        // Write the status to the log textbox on the form (txtLog)
        this.Invoke(new UpdateStatusCallback(this.UpdateStatus), new object[] { "The server has started. Please connect the client to " + ipaLocal.ToString() + "\r\n" });
        // Accept a pending connection
        TcpClient tclServer = tlsServer.AcceptTcpClient();
        // Write the status to the log textbox on the form (txtLog)
        this.Invoke(new UpdateStatusCallback(this.UpdateStatus), new object[] { "The server has accepted the client\r\n" });
        // Receive the stream and store it in a NetworkStream object
        strRemote = tclServer.GetStream();
        // Write the status to the log textbox on the form (txtLog)
        this.Invoke(new UpdateStatusCallback(this.UpdateStatus), new object[] { "The server has received the stream\r\n" });
 
        // For holding the number of bytes we are reading at one time from the stream
        int bytesSize = 0;
 
        // The buffer that holds the data received from the client
        byte[] downBuffer = new byte[2048];
        // Read the first buffer (2048 bytes) from the stream - which represents the file name
        bytesSize = strRemote.Read(downBuffer, 0, 2048);
        // Convert the stream to string and store the file name
        string FileName = System.Text.Encoding.ASCII.GetString(downBuffer, 0, bytesSize);
        // Set the file stream to the path C:\ plus the name of the file that was on the sender's computer
        strLocal = new FileStream(@"C:\" + FileName, FileMode.Create, FileAccess.Write, FileShare.ReadWrite);
 
        // The buffer that holds the data received from the client
        downBuffer = new byte[2048];
        // Read the next buffer (2048 bytes) from the stream - which represents the file size
        bytesSize = strRemote.Read(downBuffer, 0, 2048);
        // Convert the file size from bytes to string and then to long (Int64)
        long FileSize = Convert.ToInt64(System.Text.Encoding.ASCII.GetString(downBuffer, 0, bytesSize));
 
        // Write the status to the log textbox on the form (txtLog)
        this.Invoke(new UpdateStatusCallback(this.UpdateStatus), new object[] { "Receiving file " + FileName + " (" + FileSize + " bytes)\r\n" });
 
        // The buffer size for receiving the file
        downBuffer = new byte[2048];
 
        // From now on we read everything that's in the stream's buffer because the file content has started
        while ((bytesSize = strRemote.Read(downBuffer, 0, downBuffer.Length)) > 0)
        {
            // Write the data to the local file stream
            strLocal.Write(downBuffer, 0, bytesSize);
            // Update the progressbar by passing the file size and how much we downloaded so far to UpdateProgress()
            this.Invoke(new UpdateProgressCallback(this.UpdateProgress), new object[] { strLocal.Length, FileSize });
        }
        // When this point is reached, the file has been received and stored successfuly
    }
    finally
    {
        // This part of the method will fire no matter wether an error occured in the above code or not
 
        // Write the status to the log textbox on the form (txtLog)
        this.Invoke(new UpdateStatusCallback(this.UpdateStatus), new object[] { "The file was received. Closing streams.\r\n" });
 
        // Close the streams
        strLocal.Close();
        strRemote.Close();
 
        // Write the status to the log textbox on the form (txtLog)
        this.Invoke(new UpdateStatusCallback(this.UpdateStatus), new object[] { "Streams are now closed.\r\n" });
 
        // Start the server (TCP listener) all over again
        StartReceiving();
    }
}

If you are not used to networking with .NET applications, or if you have only started C# recently you may have found the above code a little overwhelming. If so, you may find the Creating a download manage in C# tutorial I created a short while back, to be more helpful for a beginner, since it has explanations on why we are using delegates to update the form from a thread, and how the streams and buffers work.

In case you are wondering why we are reading the data stream two times, 2048 bytes at a time and then we are suddenly looping through all the data using a while loop in the code above, here is why: in the client application that we are going to build, we will first send the file name within the first 2048 bytes to the server, using the same stream we will be using for transfering the actual file. Thus, first we only read the first 2048 bytes and store then in the FileName variable because the client application will send the file name within the first 2048 bytes. Then we repeat the same thing for transfering the size of the file - we read the second set of 2048 bytes which contain the file size and store this into the FileSize variable. We need to know the file name so that on the server we can save the file using the same file name as it was in the client (this includes the extension); as for the file size, we want to know that so that we can calculate the current progress of the download.
After we pass this, there is no other data coming in the stream other than the file itself, so we loop through the stream until we reach the end, and we continue writing 2048 bytes of the file at a time on the server computer.

The next method we need to implement is the one we created a delegate for - UpdateStatus() which updates the txtLog textbox with the current status of the application:

private void UpdateStatus(string StatusMessage)
{

    // Append the status to the log textbox text 
    txtLog.Text += StatusMessage;
}

There is one more method we created a delegate for, which is also accessed from inside the thread. This method updates the progress bar with the current progress of the download:

private void UpdateProgress(Int64 BytesRead, Int64 TotalBytes)
{
    if (TotalBytes > 0)
    {
        // Calculate the download progress in percentages
        PercentProgress = Convert.ToInt32((BytesRead * 100) / TotalBytes);
        // Make progress on the progress bar
        prgDownload.Value = PercentProgress;
    }
}

Now there's one more important thing left to do - start the thread with the StartReceiving() method, all inside the btnStart click event. To create the Click event of this button you can double click it in form designer view. Inside it, add the following code:

thrDownload = new Thread(StartReceiving);
thrDownload.Start();

You will probably also want to close the streams using the btnStop button, so click this button and add the following code:

strLocal.Close();
strRemote.Close();
txtLog.Text += "Streams are now closed.\r\n";

That is all, you now have a C# server ready to listen for connections and receive files. Below is the entire code for this application in case you prefer an overall look:

using System;
using System.Collections.Generic;
using System.ComponentModel;
using System.Data;
using System.Drawing;
using System.Text;
using System.Windows.Forms;
using System.Net;
using System.Net.Sockets;
using System.IO;
using System.Threading;
 
namespace NetworkFileReceiver
{
    public partial class Form1 : Form
    {
        // The thread in which the file will be received
        private Thread thrDownload;
        // The stream for writing the file to the hard-drive
        private Stream strLocal;
        // The network stream that receives the file
        private NetworkStream strRemote;
        // The TCP listener that will listen for connections
        private TcpListener tlsServer;
        // Delegate for updating the logging textbox
        private delegate void UpdateStatusCallback(string StatusMessage);
        // Delegate for updating the progressbar
        private delegate void UpdateProgressCallback(Int64 BytesRead, Int64 TotalBytes);
        // For storing the progress in percentages
        private static int PercentProgress;
 
        public Form1()
        {
            InitializeComponent();
        }
 
        private void btnStart_Click(object sender, EventArgs e)
        {
            thrDownload = new Thread(StartReceiving);
            thrDownload.Start();
        }
 
        private void StartReceiving()
        {
            // There are many lines in here that can cause an exception
            try
            {
                // Get the hostname of the current computer (the server)
                string hstServer = Dns.GetHostName();
                // Get the IP of the first network device, however this can prove unreliable on certain configurations
                IPAddress ipaLocal = Dns.GetHostEntry(hstServer).AddressList[0];
                // If the TCP listener object was not created before, create it
                if (tlsServer == null)
                {
                    // Create the TCP listener object using the IP of the server and the specified port
                    tlsServer = new TcpListener(ipaLocal, Convert.ToInt32(txtPort.Text));
                }
                // Write the status to the log textbox on the form (txtLog)
                this.Invoke(new UpdateStatusCallback(this.UpdateStatus), new object[] { "Starting the server...\r\n" });
                // Start the TCP listener and listen for connections
                tlsServer.Start();
                // Write the status to the log textbox on the form (txtLog)
                this.Invoke(new UpdateStatusCallback(this.UpdateStatus), new object[] { "The server has started. Please connect the client to " + ipaLocal.ToString() + "\r\n" });
                // Accept a pending connection
                TcpClient tclServer = tlsServer.AcceptTcpClient();
                // Write the status to the log textbox on the form (txtLog)
                this.Invoke(new UpdateStatusCallback(this.UpdateStatus), new object[] { "The server has accepted the client\r\n" });
                // Receive the stream and store it in a NetworkStream object
                strRemote = tclServer.GetStream();
                // Write the status to the log textbox on the form (txtLog)
                this.Invoke(new UpdateStatusCallback(this.UpdateStatus), new object[] { "The server has received the stream\r\n" });
 
                // For holding the number of bytes we are reading at one time from the stream
                int bytesSize = 0;
 
                // The buffer that holds the data received from the client
                byte[] downBuffer = new byte[2048];
                // Read the first buffer (2048 bytes) from the stream - which represents the file name
                bytesSize = strRemote.Read(downBuffer, 0, 2048);
                // Convert the stream to string and store the file name
                string FileName = System.Text.Encoding.ASCII.GetString(downBuffer, 0, bytesSize);
                // Set the file stream to the path C:\ plus the name of the file that was on the sender's computer
                strLocal = new FileStream(@"C:\" + FileName, FileMode.Create, FileAccess.Write, FileShare.ReadWrite);
 
                // The buffer that holds the data received from the client
                downBuffer = new byte[2048];
                // Read the next buffer (2048 bytes) from the stream - which represents the file size
                bytesSize = strRemote.Read(downBuffer, 0, 2048);
                // Convert the file size from bytes to string and then to long (Int64)
                long FileSize = Convert.ToInt64(System.Text.Encoding.ASCII.GetString(downBuffer, 0, bytesSize));
 
                // Write the status to the log textbox on the form (txtLog)
                this.Invoke(new UpdateStatusCallback(this.UpdateStatus), new object[] { "Receiving file " + FileName + " (" + FileSize + " bytes)\r\n" });
 
                // The buffer size for receiving the file
                downBuffer = new byte[2048];
 
                // From now on we read everything that's in the stream's buffer because the file content has started
                while ((bytesSize = strRemote.Read(downBuffer, 0, downBuffer.Length)) > 0)
                {
                    // Write the data to the local file stream
                    strLocal.Write(downBuffer, 0, bytesSize);
                    // Update the progressbar by passing the file size and how much we downloaded so far to UpdateProgress()
                    this.Invoke(new UpdateProgressCallback(this.UpdateProgress), new object[] { strLocal.Length, FileSize });
                }
                // When this point is reached, the file has been received and stored successfuly
            }
            finally
            {
                // This part of the method will fire no matter wether an error occured in the above code or not
 
                // Write the status to the log textbox on the form (txtLog)
                this.Invoke(new UpdateStatusCallback(this.UpdateStatus), new object[] { "The file was received. Closing streams.\r\n" });
 
                // Close the streams
                strLocal.Close();
                strRemote.Close();
 
                // Write the status to the log textbox on the form (txtLog)
                this.Invoke(new UpdateStatusCallback(this.UpdateStatus), new object[] { "Streams are now closed.\r\n" });
 
                // Start the server (TCP listener) all over again
                StartReceiving();
            }
        }
 
        private void UpdateStatus(string StatusMessage)
        {
            // Append the status to the log textbox text 
            txtLog.Text += StatusMessage;
        }
 
        private void UpdateProgress(Int64 BytesRead, Int64 TotalBytes)
        {
            if (TotalBytes > 0)
            {
                // Calculate the download progress in percentages
                PercentProgress = Convert.ToInt32((BytesRead * 100) / TotalBytes);
                // Make progress on the progress bar
                prgDownload.Value = PercentProgress;
            }
        }
 
 
        private void btnStop_Click(object sender, EventArgs e)
        {
            strLocal.Close();
            strRemote.Close();
            txtLog.Text += "Streams are now closed.\r\n";
        }
    }
}

You should now compile and run this application. If everything appears to work fine, your next move is creating the file sender application:

Sending the file - creating the client

Start another Windows Application project in Visual Studio 2005 and add to it 3 textboxes and 3 buttons. The textboxes are: txtPort, txtServer and txtLog. The buttons are btnConnect, btnDisconnect and btnSend. Similar to what we did in the server application, you will need to switch to code view and add the following using statements:

using System.Net;
using System.Net.Sockets;
using System.IO;

Follow by defining the following objects inside the class:

// The TCP client will connect to the server using an IP and a port
TcpClient tcpClient;
// The file stream will read bytes from the local file you are sending
FileStream fstFile;
// The network stream will send bytes to the server application
NetworkStream strRemote;

Before sending a file we need to establish a successful connection to the server. Inside the application's class we're going to build a small method that does just that:

private void ConnectToServer(string ServerIP, int ServerPort)
{
    // Create a new instance of a TCP client
    tcpClient = new TcpClient();
    try
    {
        // Connect the TCP client to the specified IP and port
        tcpClient.Connect(ServerIP, ServerPort);
        txtLog.Text += "Successfully connected to server\r\n";
    }
    catch (Exception exMessage)
    {
        // Display any possible error
        txtLog.Text += exMessage.Message;
    }
}

Now we're going to call this method from the click event of btnConnect, so double click the button in design view so that you get the Click event automatically created, then inside it place the following code:

// Call the ConnectToServer method and pass the parameters entered by the user
ConnectToServer(txtServer.Text, Convert.ToInt32(txtPort.Text));

At this point if you open the server application we created and set it to monitor on port 815, then run the client application and press the btnConnect button, you should see in the server application window how the connection is accepted.

Moving on, we haven't yet seen the code that actually sends the file to the server. We want to have this code inside the Click event of btnSend, thus double click that button and inside the event use the following chunk of code:

// If tclClient is not connected, try a connection
if (tcpClient.Connected == false)
{
    // Call the ConnectToServer method and pass the parameters entered by the user
    ConnectToServer(txtServer.Text, Convert.ToInt32(txtPort.Text));
}
 
// Prompt the user for opening a file
if (openFile.ShowDialog() == DialogResult.OK)
{
    txtLog.Text += "Sending file information\r\n";
    // Get a stream connected to the server
    strRemote = tcpClient.GetStream();
    byte[] byteSend = new byte[tcpClient.ReceiveBufferSize];
    // The file stream will read bytes from the file that the user has chosen
    fstFile = new FileStream(openFile.FileName, FileMode.Open, FileAccess.Read);
    // Read the file as binary
    BinaryReader binFile = new BinaryReader(fstFile);
 
    // Get information about the opened file
    FileInfo fInfo = new FileInfo(openFile.FileName);
 
    // Get and store the file name
    string FileName = fInfo.Name;
    // Store the file name as a sequence of bytes
    byte[] ByteFileName = new byte[2048];
    ByteFileName = System.Text.Encoding.ASCII.GetBytes(FileName.ToCharArray());
    // Write the sequence of bytes (the file name) to the network stream
    strRemote.Write(ByteFileName, 0, ByteFileName.Length);
 
    // Get and store the file size
    long FileSize = fInfo.Length;
    // Store the file size as a sequence of bytes
    byte[] ByteFileSize = new byte[2048];
    ByteFileSize = System.Text.Encoding.ASCII.GetBytes(FileSize.ToString().ToCharArray());
    // Write the sequence of bytes (the file size) to the network stream
    strRemote.Write(ByteFileSize, 0, ByteFileSize.Length);
 
    txtLog.Text += "Sending the file " + FileName + " (" + FileSize + " bytes)\r\n";
 
    // Reset the number of read bytes
    int bytesSize = 0;
    // Define the buffer size
    byte[] downBuffer = new byte[2048];
 
    // Loop through the file stream of the local file
    while ((bytesSize = fstFile.Read(downBuffer, 0, downBuffer.Length)) > 0)
    {
        // Write the data that composes the file to the network stream
        strRemote.Write(downBuffer, 0, bytesSize);
    }
 
    // Update the log textbox and close the connections and streams
    txtLog.Text += "File sent. Closing streams and connections.\r\n";
    tcpClient.Close();
    strRemote.Close();
    fstFile.Close();
    txtLog.Text += "Streams and connections are now closed.\r\n";
}

As you can see here, using the stream we are first sending to the server the file name; the server on its side will read this piece of stream and store the file name in a variable for later use. Same happens for the file size, which is retrieved just like the file name using the FileInfo class and written to the stream. After we send these two pices of information, we start sending the actual content of the file (in binary). The content is sent by looping through yet another stream - the file stream, a local stream that reads the file from the hard drive. While we read pieces of this local stream we write pieces to the network stream so that the server can receive them and save them, until the very last byte of the file when the while loop ends.
After that the streams and connections are closed, and the client has received the file.

There's one more piece of this application that we haven't implemented yet, the disconnection from the server. This happens when btnDisconnect is clicked:

// Close connections and streams and update the log textbox
tcpClient.Close();
strRemote.Close();
fstFile.Close();
txtLog.Text += "Disconnected from server.\r\n";

We're pretty much finished with this tutorial. I hope you enjoyed it and didn't experience too many problems with using this code on your system.
Below is the complete code of the network file sender application (the client):

using System;
using System.Collections.Generic;
using System.ComponentModel;
using System.Data;
using System.Drawing;
using System.Text;
using System.Windows.Forms;
using System.Net;
using System.Net.Sockets;
using System.IO;
 
namespace NetworkFileSender
{
    public partial class Form1 : Form
    {
        // The TCP client will connect to the server using an IP and a port
        TcpClient tcpClient;
        // The file stream will read bytes from the local file you are sending
        FileStream fstFile;
        // The network stream will send bytes to the server application
        NetworkStream strRemote;
 
        public Form1()
        {
            InitializeComponent();
        }
 
        private void ConnectToServer(string ServerIP, int ServerPort)
        {
            // Create a new instance of a TCP client
            tcpClient = new TcpClient();
            try
            {
                // Connect the TCP client to the specified IP and port
                tcpClient.Connect(ServerIP, ServerPort);
                txtLog.Text += "Successfully connected to server\r\n";
            }
            catch (Exception exMessage)
            {
                // Display any possible error
                txtLog.Text += exMessage.Message;
            }
        }
 
        private void btnConnect_Click(object sender, EventArgs e)
        {
            // Call the ConnectToServer method and pass the parameters entered by the user
            ConnectToServer(txtServer.Text, Convert.ToInt32(txtPort.Text));
        }
 
        private void btnSend_Click(object sender, EventArgs e)
        {
            // If tclClient is not connected, try a connection
            if (tcpClient.Connected == false)
            {
                // Call the ConnectToServer method and pass the parameters entered by the user
                ConnectToServer(txtServer.Text, Convert.ToInt32(txtPort.Text));
            }
 
            // Prompt the user for opening a file
            if (openFile.ShowDialog() == DialogResult.OK)
            {
                txtLog.Text += "Sending file information\r\n";
                // Get a stream connected to the server
                strRemote = tcpClient.GetStream();
                byte[] byteSend = new byte[tcpClient.ReceiveBufferSize];
                // The file stream will read bytes from the file that the user has chosen
                fstFile = new FileStream(openFile.FileName, FileMode.Open, FileAccess.Read);
                // Read the file as binary
                BinaryReader binFile = new BinaryReader(fstFile);
 
                // Get information about the opened file
                FileInfo fInfo = new FileInfo(openFile.FileName);
 
                // Get and store the file name
                string FileName = fInfo.Name;
                // Store the file name as a sequence of bytes
                byte[] ByteFileName = new byte[2048];
                ByteFileName = System.Text.Encoding.ASCII.GetBytes(FileName.ToCharArray());
                // Write the sequence of bytes (the file name) to the network stream
                strRemote.Write(ByteFileName, 0, ByteFileName.Length);
 
                // Get and store the file size
                long FileSize = fInfo.Length;
                // Store the file size as a sequence of bytes
                byte[] ByteFileSize = new byte[2048];
                ByteFileSize = System.Text.Encoding.ASCII.GetBytes(FileSize.ToString().ToCharArray());
                // Write the sequence of bytes (the file size) to the network stream
                strRemote.Write(ByteFileSize, 0, ByteFileSize.Length);
 
                txtLog.Text += "Sending the file " + FileName + " (" + FileSize + " bytes)\r\n";
 
                // Reset the number of read bytes
                int bytesSize = 0;
                // Define the buffer size
                byte[] downBuffer = new byte[2048];
 
                // Loop through the file stream of the local file
                while ((bytesSize = fstFile.Read(downBuffer, 0, downBuffer.Length)) > 0)
                {
                    // Write the data that composes the file to the network stream
                    strRemote.Write(downBuffer, 0, bytesSize);
                }
 
                // Update the log textbox and close the connections and streams
                txtLog.Text += "File sent. Closing streams and connections.\r\n";
                tcpClient.Close();
                strRemote.Close();
                fstFile.Close();
                txtLog.Text += "Streams and connections are now closed.\r\n";
            }
        }
 
        private void btnDisconnect_Click(object sender, EventArgs e)
        {
            // Close connections and streams and update the log textbox
            tcpClient.Close();
            strRemote.Close();
            fstFile.Close();
            txtLog.Text += "Disconnected from server.\r\n";
        }
    }
}

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Current Comments

by MyrO on Sunday, May 21st 2006 at 08:22 AM

man.. u are great!

thx!

by Mavarok on Friday, June 2nd 2006 at 01:47 PM

im sorry to rain on you but i have been programing C# for 2days now[WOW] and i have a problem with your code when it comes to creating the thread:
you need to use a slightly longer but safer method of

{
thrDownload = new Thread( new ThreadStart(StartReceiving));
thrDownload.Start();
}

sorry bout that!
also you may want to terminate the thread when you stop the server!

by piterfromwawa on Sunday, June 18th 2006 at 06:27 PM

Its great.
But you avoid situations like here:
Server is able to accept new connection only when it\'s not receiving data.
You should rather accept connection, pass Client to separate thread, and in while loop start accepting new connection, than start accepting new connection after job is done.

by rpG on Saturday, May 26th 2007 at 05:04 AM

Hi! This tutorial is great!
But if I connect more clients to the server (Client1, Client2 and Client3), and send data with the Client1, how can the server send this data to Client2 and Client3? E.g. in a messenger program, where the client1 writes something, send it to the server, and the server sends it to all of the connected clients. How can I do this? Please help!

by prabhat on Tuesday, June 26th 2007 at 07:00 AM

Good One..Thank You

by mahmoud faress on Saturday, July 7th 2007 at 12:06 PM

so smart and so great ...
but you had missed some thing that is ,
if you try to open the task manager an look at the processes you will notice that the client application is still running(this case happens if you close the client and still keep server running ) ,
so i suggest to add this to your application in[ from closing event]
Environment.Exit(0);
this will make your application work efficiently
good luck

by cshrpd00d on Wednesday, September 5th 2007 at 08:48 PM

Nice.

piterfromwawa/others:
it is only an example to show how to transfer files from 1 server to 1 client. You would surely implement more threads if programming an an actual file transfer server... Using the example here, you could easily throw something together that would allow multiple client connections, etc.

rpG:
As piter said: after a client connects, hand off the connection to another thread (make sure the client stays connected) and listen for other connections, when another client connects, hand that off to another thread, etc. Keep track of the connected clients (in a list/array, if the client disconnects remove it from array) then send the file to all connected clients when the send button is pressed.

by Newbie on Friday, September 7th 2007 at 04:46 AM

Sorry to ask you these elimentary questions but I am but a novice in C#.

Do I run the server and client code on seperate instances of VS?

Then I get an error when I try to send any file type other than txt. Something with the int64 part of the code. It says "string not in correct form". Does this code send all types of files, i.e. images etc.?

Thanks!

by skhan on Tuesday, December 11th 2007 at 05:37 AM

C it iz working fine on the loacal host ,but the same code whenever i run on 2 different machines ,i get an eror ,that input is not in correct format .it gives the error for file size

by Trapper on Monday, January 28th 2008 at 06:22 PM

First of all I would like to thank Andrei Pociu for this great program and explanation... I am trying to utilize it for my needs...

As Mahmoud Faress said, the Environment.Exit(0); is needed...

Morever, I don\'t know about the others but the file transfer did not work correctly... I am not an advanced C# user, but after a lot of time spent debugging the program, I came to better understand it\'s logic, and thus understand the error of FormatException...

Same occurs to me, and what the reason seems to be that first a byte array is created to store 2048 bytes from which the filename will be taken to inform the Server...

After which, the client should send another 2048 bytes containing the file size. The problem (at least in my case) is that apart from the bytes used to store the file size, the rest are used to store the other file data.. Ex. if I want to send a txt file of 2048 bytes starting as \"Trapper\" the 2048 bytes would contain (in their decimal format):

[0] = 2
[1] = 0
[2] = 4
[3] = 8
[4] = T
[5] = r
[6] = a
[7] = p
[8] = p
[9] = e
[10] = r

The first four elements correctly contain the file size, but the rest should be zeros, and not contain the file data... That is why the FormatException occurs, since text cannot be converted to integer... I still haven\'t quite solved the problem, but either the user has to enter the number of bytes that he want to be read (thus the number of bytes used to store the file size) or else correct the buffer data...

Apart from this, even though I haven\'t started working on it, at first sight, the progress bar uploading seems to be incorrect, since the amount of bytes transfered seems to be linked to the string containing the file location...

I don\'t know if this is just me, but I think that is how everyone should be having the program... I\'ll send later (if I\'ll have the problem solved), in the mean time I\'d love to hear if anyone has had the program working fine ... after being tested, or if anyone have solved any of the problems (as mine).

by Trapper on Monday, January 28th 2008 at 06:23 PM

by sundar on Tuesday, April 15th 2008 at 11:27 PM

hi Trapper

I am also experiencing the same problem and found ur explanation to be interesting. Thank u for finding out the error in the coding. If u have solved the prob pl send me the corresponding code to me, as i am in great need of it

thanks, bye

by sundar on Wednesday, April 16th 2008 at 10:02 AM

hi guys,

I found a way to solve the problem of Format Exception on RECEIVER side. It was probably due to the contents of FILE SIZE mixing with the CONTENT of the FILE. It is due to not allowing sufficient delay in transporting the FILESIZE & THE CONTENT OF FILE.
So insert a MessageBox("are you sure you want to send the file"), on the SENDER side after the line notifying "SENDING FILE ....."

by sceva on Monday, June 9th 2008 at 03:35 AM

cool

by Mihai on Monday, July 7th 2008 at 09:23 AM

Hey all,
Without a few glitches here and there the program/tutorial is fine.Now, the name of the article is "File transfers through networks and the INTERNET". Has any1 tried to send files over the internet(not just LAN). I did and got an error(Exception).
If any of you know any links where i can get/see some code that does the file transfer correctly over the internet please let me know.
Thanks.

by disgruntled on Monday, February 23rd 2009 at 09:14 PM

has anyone solved trappers problem as I have the same issue with the tutorial

by disgruntled on Monday, February 23rd 2009 at 09:18 PM

has anyone solved trappers problem as I have the same issue with the tutorial and the "delay" did not fix it

by Ankur on Wednesday, April 22nd 2009 at 06:16 AM

Sir this has helped me a lot but , i am getting confuse at the point where i have to send the file to the seperate clients ..and also deciding the folder to store the file at run time (as here c:\).. so please help me out for this ..

thanks ..

by Nameless on Monday, November 2nd 2009 at 05:31 PM

Hi all.
I'm new to the sharp so my solution is... stupid, but for now its the only one :D
Don't use file size... at the client - comment (//):
strRemote.Write(ByteFileSize, 0, ByteFileSize.Length);

Then at the server side, comment:
bytesSize = strRemote.Read(downBuffer, 0, 2048);

And:
long FileSize = Convert.ToInt64(System.Text.Encoding.ASCII.GetString(downBuffer, 0, bytesSize));

Then remove all FileSize and its done.
It works fine to me :)

by Hrishank Jhildiyal on Tuesday, November 24th 2009 at 11:21 PM

Here's a better way to fix trappers problem:

We use two helper functions, one on the server side and the other on the client side {client side uses byteEncode(), while the server side uses returnValue()}

private static byte[] byteEncode(string s, int size)
{
byte[] b = new byte[size];
byte[] c = System.Text.Encoding.ASCII.GetBytes(s.ToCharArray());
for (int k = 0; k < c.Length; k )
b[k] = c[k];
//Console.WriteLine("" (char)b[0] (char)b[1]);
try
{
for (int i = s.Length; i < size; i )
b[i] = (byte)'*';
}
catch (Exception) { }

return b;
}

private static string returnValue(byte[] arr)
{
int idx = -1;
for (int i = 0; i < arr.Length; i )
{
if (arr[i] == '*')
{
idx = i;
break;
}
}

string strBuilder = "";
for (int j = 0; j < idx; j )
strBuilder = arr[j];

return strBuilder;
}

What this does is it fills up the remaining bytes, which is 2048-something with '*'s. The other side rejects those '*'s and takes the other stuff out. Then you can do with it as you like.

by Hrishank Jhildiyal on Wednesday, November 25th 2009 at 12:18 AM

Also note that to decode it on the server side, you would need to use an instance of the MemoryStream class. For example (on the server side):

MemoryStream memstream = new MemoryStream();

byte[] byteData = new byte[2048];
int bytesRead = strRemote.Read(byteData, 0, byteData.Length);
memstream.Write(byteData, 0, bytesRead);

Then by using "memstream.GetBuffer()", we can get the desired byte array to be used for the returnValue() method.

Lastly, we shall update the bytesRead variable:

bytesRead = strRemote.Read(byteData, 0, byteData.Length);

by Hrishank Jhildiyal on Wednesday, November 25th 2009 at 03:06 AM

For some reason the returnValue() function copied incorrectly:

Here's a slight tweak (the correct version):

private static string returnValue(byte[] arr)
{
int idx = -1;
for (int i = 0; i < arr.Length; i )
{
if (arr[i] == '*')
{
idx = i;
break;
}
}

string strBuilder = "";
for (int j = 0; j < idx; j )
strBuilder = (char)arr[j];

return strBuilder;
}

by Ryan on Wednesday, April 14th 2010 at 03:20 AM

How exactly may the above code be implemented? ie how do I use the value returned by memstream.GetBuffer? Also, where should I put

MemoryStream memstream = new MemoryStream();

byte[] byteData = new byte[2048];
int bytesRead = strRemote.Read(byteData, 0, byteData.Length);
memstream.Write(byteData, 0, bytesRead);

? I assume within the StartListening() method.

by Valerijs on Friday, June 11th 2010 at 02:26 AM

I have traded to keep to you but still there are lots of plases thet I dont undestend pless could you upload the sample project.

by Hrishank Jhildiyal on Thursday, June 24th 2010 at 01:35 PM

Hey Valerijs

I have this from quite some time back. Not sure if this works perfectly, but this should be at least 95% working. It'll help for sure if you debug it and go through it.

Download the file at:
http://hrishank.weebly.com/attachments-miscellaneous-work.html

There's only one file on that page called: "ftpuploadmodule.zip". It's a Visual Studio 2008 project.

by Dejan on Tuesday, June 29th 2010 at 07:12 AM

Andrew,

Thank you for the wonderful code.
I do have one question, I'm trying to send two files one after another to same connection like
send.SendFile("sample-1.m2v");
send.SendFile("sample-2.m2v");

And the first file is received ok, while the name of the second file gets mixed with content so it looks like
FileName: sample-2.m2v\0\0[]?\f\x....MPEG-1, 25 frames/sec\0\0...

Do you know what this might be?
Thank you

by Sachet on Thursday, July 1st 2010 at 06:28 AM

As a part of the application which m making..i want to calculate the total data transfer between particular interval of time..ie the total amont of download n uploaded data..by the user..cn any1 provide me with d code to calculate the total data transfer in asp.net for web application