从PictureBox上传图像到服务器

Question

I have one pictureBox and a button on form1 one. When the button is clicked, it should upload the file to the server. For now I am using the below method. First save the image locally and then upload to the server:

Bitmap bmp = new Bitmap(this.form1.pictureBox1.Width, this.form1.pictureBox1.Height);
Graphics g = Graphics.FromImage(bmp);
Rectangle rect = this.form1.pictureBox1.RectangleToScreen(this.form1.pictureBox1.ClientRectangle);
g.CopyFromScreen(rect.Location, Point.Empty, this.form1.pictureBox1.Size);
g.Dispose();
 bmp.Save("filename", ImageFormat.Jpeg);

And then uploading that file:

using (var f = System.IO.File.OpenRead(@"F:\filename.jpg"))
{
    HttpClient client = new HttpClient();
    var content = new StreamContent(f);
    var mpcontent = new MultipartFormDataContent();
    content.Headers.ContentType = new MediaTypeHeaderValue("image/jpeg");
    mpcontent.Add(content);
    client.PostAsync("http://domain.com/upload.php", mpcontent);
}

I can't use the Bitmap type in StreamContent. How can I stream the image from pictureBox directly instead saving it as file first?

I came up with the below code using MemoryStream, but the uploaded file size is 0 using this method. Why?

byte[] data;

using (MemoryStream m = new MemoryStream())
{
    bmp.Save(m, ImageFormat.Png);
    m.ToArray();
    data = new byte[m.Length];
    m.Write(data, 0, data.Length);

    HttpClient client = new HttpClient();
    var content = new StreamContent(m);
    var mpcontent = new MultipartFormDataContent();
    content.Headers.ContentType = new MediaTypeHeaderValue("image/png");
    mpcontent.Add(content, "file", filename + ".png");
    HttpResponseMessage response = await client.PostAsync("http://domain.com/upload.php", mpcontent);
    //response.EnsureSuccessStatusCode();
    string body = await response.Content.ReadAsStringAsync();
    MessageBox.Show(body);
}

Answer 1

I am not sure if it is the correct way to do it, but I have solved it by creating a new stream and then copying the older one to it:

using (MemoryStream m = new MemoryStream())
{
    m.Position = 0;
    bmp.Save(m, ImageFormat.Png);
    bmp.Dispose();
    data = m.ToArray();
    MemoryStream ms = new MemoryStream(data);
    // Upload ms
}

Answer 2

 Image returnImage = Image.FromStream(....);