Accessing the Request.Content in the new ASP.NET vnext web api way of doing things? - asp.net-web-api2

Question

I have searched high and low for this one and can't seem to find a way of accessing the Request.Content in an MVC web api. I basically am trying to create a File Service to and from Azure Blob and Table storage (table for storing metadata about the file, blob for the actual file)....
I was converting the steps in the following link, but this is where I have come unstuck
the back end I have working but can't find a way of the new unified controller passing a fileobject from json post through to the service! Any ideas would be greatly appreciated as always... or am I just going about this the wrong way?
Article here....
UPDATE: so to clarify, what I am trying to do in the new MVC 6 (where you no longer have an apicontroller to inherit from) is to access a file that has been uploaded to the api from a JSON post. That is the long and short of what I am trying to achieve.
I am trying to use the article based on the old Web API which uses the Request.Content to access it, however even if I use the WebAPIShim which they provide I still come unstuck with other objects or properties that are no longer available so I'm wondering if I need to approach it a different way, but either way, all I am trying to do is to get a file from a JSON post to a MVC 6 Web api and pass that file to my back end service....
ANY IDEAS?

Answer 1

Here is an example without relying on model binding.
You can always find the request data in Request.Body, or use Request.Form to get the request body as a form.
[HttpPost]
public async Task<IActionResult> UploadFile()
{
if (Request.Form.Files != null && Request.Form.Files.Count > 0)
{
var file = Request.Form.Files[0];
var contentType = file.ContentType;
using (var fileStream = file.OpenReadStream())
{
using (var memoryStream = new MemoryStream())
{
await fileStream.CopyToAsync(memoryStream);
// do what you want with memoryStream.ToArray()
}
}
}
return new JsonResult(new { });
}

Answer 2

If the only thing in your request is a File you can use the IFormFile class in your action:
public FileDetails UploadSingle(IFormFile file)
{
FileDetails fileDetails;
using (var reader = new StreamReader(file.OpenReadStream()))
{
var fileContent = reader.ReadToEnd();
var parsedContentDisposition = ContentDispositionHeaderValue.Parse(file.ContentDisposition);
fileDetails = new FileDetails
{
Filename = parsedContentDisposition.FileName,
Content = fileContent
};
}
return fileDetails;
}