REST API - Using the Content-Type: Multipart Method

You can use the Multipart method with the following requests:

To add data using the Multipart method, you must use boundary parameters for the data. For more information on using MultipartContent, see http://www.w3.org/Protocols/rfc1341/7_2_Multipart.html.

Example in C#

{
  //Call REST API
  var fetchRequest = (HttpWebRequest)WebRequest.Create(string.Concat(BaseURI, "/", APICall));                
  fetchRequest.Headers.Add("Authorization", “BlueprintToken” + “ “ + authenticationToken);
                                
  fetchRequest.Accept = "application/xml";
  fetchRequest.Method = "POST";
  fetchRequest.KeepAlive = true;

  fetchRequest.Credentials = System.Net.CredentialCache.DefaultCredentials;

  //Start creating multipart boundary
  string boundary = "---------------------------" + DateTime.Now.Ticks.ToString("x");
  byte[] boundarybytes = System.Text.Encoding.ASCII.GetBytes("\r\n--" + boundary + "\r\n");

  fetchRequest.ContentType = "multipart/form-data; boundary=" + boundary;
  using (Stream rs = fetchRequest.GetRequestStream())
  {
  //Add starting boundary                    
  rs.Write(boundarybytes, 0, boundarybytes.Length);

  //Add Content headers (content disposition header)
  string headerTemplate = "Content-Disposition: form-data; name=\"{0}\"; filename=\"{1}\"\r\nContent-Type: {2}\r\n\r\n";
  string header = string.Format(headerTemplate, paramName, file, contentType);
  byte[] headerbytes = System.Text.Encoding.UTF8.GetBytes(header);
  rs.Write(headerbytes, 0, headerbytes.Length);
                    
  //Add file
  using (var fileStream = new FileStream(file, FileMode.Open, FileAccess.Read))
  {
  var buffer = new byte[4096];
  int bytesRead = 0;
  while ((bytesRead = fileStream.Read(buffer, 0, buffer.Length)) != 0)
  {
  rs.Write(buffer, 0, bytesRead);
  }
  }

  //Add last boundary
  byte[] trailer = System.Text.Encoding.ASCII.GetBytes("\r\n--" + boundary + "--\r\n");
  rs.Write(trailer, 0, trailer.Length);
  }
  //Send request
  var webResponse = fetchRequest.GetResponse() as HttpWebResponse;
                
}