CODEWITHSUNDEEP
c#httptcpmjpeg
Wojciech Ketrzynski
Wojciech Ketrzynski

Reputation: 2651

How to send Motion JPEG through internet to Windows phone over TCP

I would like to send motion JPEG from my computer to the Windows Phone telephone with global ip 89.232.123.122. How to establish a connection with that mobile phone and push the mjpeg through the connection?

Upvotes: 1

Views: 1687

Answers (2)

moa7amed
moa7amed

Reputation: 16

u must have web service and upload your picture to web service and save it in server or in database but i suggest save it in server ...

in Web Service : 

 [WebMethod]
    public string UploadFile(byte[] f, string fileName)
    {
        try
        {
            MemoryStream ms = new MemoryStream(f);
            FileStream fs = new FileStream
               (System.Web.Hosting.HostingEnvironment.MapPath("~/ArchiveImages/") +
               fileName, FileMode.Create);
            ms.WriteTo(fs);
            ms.Close();
            fs.Close();
            fs.Dispose();

            return "ok";
        }
        catch (Exception ex)
        {
            return ex.Message.ToString();
        }
    }

and in client app : 

private void UploadFile(string filename)
    {
        try
        {
            ArchiveServiceObj.ArchiveServiceSoapClient srv = new ArchiveServiceObj.ArchiveServiceSoapClient();

            MemoryStream stream = new MemoryStream();
            picscannedimage.Image.Save(stream, System.Drawing.Imaging.ImageFormat.Jpeg);
            byte[] pic = stream.ToArray();
            string sTmp = srv.UploadFile(pic, filename + ".jpg");
            MessageBox.Show("File Upload Status: " + sTmp, "File Upload");

        }
        catch (Exception ex)
        {

            MessageBox.Show(ex.Message.ToString(), "Upload Error");
        }
    }

Upvotes: 0

AmirHossein Rezaei
AmirHossein Rezaei

Reputation: 1410

for sending multimedia (like motion JPEG) use UDP instead of TCP.

in the sender side use this code :

  UdpClient sendFrame = new UdpClient();
  // your image is img:
  Bitmap img = new Bitmap("pic.png");


  // always send image
  while (true)
                {
      MemoryStream memory_Stream = new MemoryStream();
      // convert bitmap to jpg
      SaveJPG100(img, memory_Stream);

      byte[] byte_Of_Frame = memory_Stream.ToArray();
     // send data on port 2000 on remote host
     sendFrame.Send(byte_Of_Frame, byte_Of_Frame.Length,"89.232.123.122",2000);

                }

// convert btm to jpg

public void SaveJPG100(Bitmap bmp, System.IO.Stream stream)
{

    EncoderParameters encoderParameters = new EncoderParameters(1);
    encoderParameters.Param[0] = new EncoderParameter(System.Drawing.Imaging.Encoder.Quality, 100L);
    bmp.Save(stream, GetEncoder(ImageFormat.Jpeg), encoderParameters);
}

// generate jpg description
public ImageCodecInfo GetEncoder(ImageFormat format)
{

    ImageCodecInfo[] codecs = ImageCodecInfo.GetImageDecoders();

    foreach (ImageCodecInfo codec in codecs)
    {
        if (codec.FormatID == format.Guid)
        {
            return codec;
        }
    }
    return null;
}

on the receiver side use this code

            UdpClient receiveFrame = new UdpClient(2000);
            // recieve data from any ip address and any port
            IPEndPoint remote = new IPEndPoint(IPAddress.ANY, 0);

            while (true)
            {

                byte[] byte_Of_Frame = receiveFrame.Receive(ref remote);
                MemoryStream ms = new MemoryStream(byte_Of_Frame);

                pictureBox1.Image=(new Bitmap(ms));
            }

Upvotes: 1

Related Questions