如何“克隆”传入的ASP.NET MVC Multipart请求以发送到Web API控制器

Question

Without performing a Server.TransferRequest I need a way to send the multipart body of an ASP.NET MVC Controller Request to an ASP.NET Web API controller which parses that multipart data. I would like to do an HttpClient PostAsync to the API passing along the multipart form data.

In a nutshell, the purpose of the API controller is to handle generic models and serialize them into key/value pairs. I want to use the MVC controllers to validate various models before sending the request on to the local API Controller. I do not want to call the API directly from the form.

Here are the steps:

1. User enters data into a form on a page (/Home/Index)
2. Index controller validates the model being passed in.
3. If valid, the Index controller performs an HttpClient POST to /api/Forms/Submit
4. The API controller accepts the multipart data and converts it to key value pairs and stores the data in the DB.
5. The API controller returns HttpResponseMessage w/response model.

I cannot read a Request.Content object on the MVC controller like I can on the API controller. What is the best way to 're-create' the multipart request?

Answer 1

In your MVC controller action after validating the model you can access the raw underlying request stream then pass the stream directly into an HttpWebRequest like so...

[HttpPost]
public ActionResult Index(MyModel m)
{
    Request.InputStream.Position = 0;

    //the incoming request stream
    var requestStream = HttpContext.Request.InputStream;

    //the outgoing web request
    var webRequest = (HttpWebRequest)WebRequest.Create("http://yaddayadda/api/TargetApiController/Target");

    Stream webStream = null;

    try
    {
        //copy incoming request body to outgoing request
        if (requestStream != null && requestStream.Length > 0)
        {
            webRequest.Method = "POST";
            webRequest.ContentLength = requestStream.Length;
            webRequest.ContentType = HttpContext.Request.ContentType; // <- included for multipart form content
            webStream = webRequest.GetRequestStream();
            requestStream.CopyTo(webStream);
            webStream.Close();
        }
    }
    finally
    {
        if (null != webStream)
        {
            webStream.Flush();
            webStream.Close();    // might need additional exception handling here
        }
    }

    using (HttpWebResponse response = (HttpWebResponse)webRequest.GetResponse())
    {
        var result = response.StatusCode;
    }


    return View();
}