Can RestSharp send binary data without using a multipart content type?

Question

I have been using AddParameter to include XML bodies in my HTTP requests:

request.AddParameter(contentType, body, ParameterType.RequestBody);
         


        

Answer 1

I ran into the same issue. I had to upload exactly one file and use a specific content type for communicating with an REST Interface. You could modify Http.RequestBody to byte[] (and all dependencies on that), but i think its easier this way:

I modified RestSharp, so that it only use Multipart Encoding when number of Files > 1 or number of Files = 1 and there is also body or other post data set.

You have to modify Http.cs on Line 288 from

if(HasFiles)

to

if(Files.Count > 1 || (Files.Count == 1 && (HasBody || Parameters.Any())))

For Http.Sync.cs modify PreparePostData from

private void PreparePostData(HttpWebRequest webRequest)
{
    if (HasFiles)
    {
        webRequest.ContentType = GetMultipartFormContentType();
        using (var requestStream = webRequest.GetRequestStream())
        {
            WriteMultipartFormData(requestStream);
        }
    }

    PreparePostBody(webRequest);
}

to

private void PreparePostData(HttpWebRequest webRequest)
{
    // Multiple Files or 1 file and body and / or parameters
    if (Files.Count > 1 || (Files.Count == 1 && (HasBody || Parameters.Any())))
    {
        webRequest.ContentType = GetMultipartFormContentType();
        using (var requestStream = webRequest.GetRequestStream())
        {
            WriteMultipartFormData(requestStream);
        }
    }
    else if (Files.Count == 1)
    {
        using (var requestStream = webRequest.GetRequestStream())
        {
            Files.Single().Writer(requestStream);
        }
    }

    PreparePostBody(webRequest);
}

If you use the async version, you have to modify the code similar to the above in Http.Async.cs.

Now u can use RestSharp like this

IRestRequest request = new RestRequest("urlpath", Method.PUT);
request.AddHeader("Content-Type", "application/zip");
request.AddFile("Testfile", "C:\\File.zip");

Client.Execute(request);

AddFile also provides an overload for setting direct byte[] data instead of a path. Hope that helps.