private byte[] MakeFileInputContentOpen(string boundary, string fileFormName, string fileName)
{
string format = string.Format("--{0}\r\nContent-Disposition: form-data; name=\"{1}\"; filename=\"{2}\"\r\nContent-Type: {3}\r\n\r\n",
boundary, fileFormName, fileName, ZAppHelper.GetMimeType(fileName));
return(Encoding.UTF8.GetBytes(format));
}