Multipart/form-data file upload in asp.net core web api

Update- .net core 2.0 +

With .net core you can leverage the new IFormFile interface to upload both the image and properties in the same post. For example:

[HttpPost("content/upload-image")]
public async Task<IActionResult> UploadImage(MyFile upload)

The MyFile class can look like:

public class MyFile
{
    public string userId { get; set; }        
    public IFormFile File { get; set; }
    // Other properties
}

You can access the properties and the file as follows:

var file = upload.File // This is the IFormFile file
var param = upload.userId // param

To persist/save the file to disk you can do the following:

using (var stream = new FileStream(path, FileMode.Create))
{
    await file.File.CopyToAsync(stream);
}

.NET Framework

Yes it is. Depending on the client Framework you're using, you can configure your Web API for Content Type-Multipart, then do something like:

[HttpPost]
[Route("content/upload-image")]       
public async Task<HttpResponseMessage> Post()
{
    if (!Request.Content.IsMimeMultipartContent())
    {
        throw new HttpResponseException(HttpStatusCode.UnsupportedMediaType);
    }           
    // enter code here
}

Define and setup the directory where your image will be saved.

var root = HttpContext.Current.Server.MapPath("~/Content/Images/");
if (!Directory.Exists(root))
{
    Directory.CreateDirectory(root);
}  

Setup the StreamProvider and attempt to get the model data, which is the JSON you mentioned.

var streamProvider = new MultipartFormDataStreamProvider(root);
var result =
    await Request.Content.ReadAsMultipartAsync(streamProvider);
if (result.FormData["model"] == null)
{
    throw new HttpResponseException(HttpStatusCode.BadRequest);
}

Now access the files in the request.

try
{
    // Deserialize model data to your own DTO
    var model = result.FormData["model"];
    var formDto = JsonConvert
        .DeserializeObject<MyDto>(model, new IsoDateTimeConverter());
    var files = result.FileData.ToList();                
    if (files != null)
    {
        foreach (var file in files)
        {
            // Do anything with the file(s)
        }
    }
}