How to post file to ASP.NET Web Api 2

Question

I'd like to post a file to my webapi. It's not the problem but:

1. I don't want to use javascript
2. The file must be received and saved synchronously

3. I would like my action to look like this:

   public void Post(byte[] file)
   {
   
   }
   

   or:

   public void Post(Stream stream)
   {
   
   }
   

4. I want to post file from code similiar to this (of course, now it doesn't work):

   <form id="postFile" enctype="multipart/form-data" method="post">
   
       <input type="file" name="file" />
   
       <button value="post" type="submit" form="postFile"  formmethod="post" formaction="<%= Url.RouteUrl("WebApi", new { @httpRoute = "" }) %>" />
   
   </form>
   

Any suggestions will be appreciated

Answer 1

The simplest example would be something like this

[HttpPost]
[Route("")]
public async Task<HttpResponseMessage> Post()
{
    if (!Request.Content.IsMimeMultipartContent())
    {
        throw new HttpResponseException(HttpStatusCode.UnsupportedMediaType);
    }

    var provider = new MultipartFormDataStreamProvider(HostingEnvironment.MapPath("~/App_Data"));

    var files = await Request.Content.ReadAsMultipartAsync(provider);

    // Do something with the files if required, like saving in the DB the paths or whatever
    await DoStuff(files);

    return Request.CreateResponse(HttpStatusCode.OK);;
}

There is no synchronous version of ReadAsMultipartAsync so you are better off playing along.

UPDATE:

If you are using IIS server hosting, you can try the traditional way:

public HttpResponseMessage Post()
{
    var httpRequest = HttpContext.Current.Request;
    if (httpRequest.Files.Count > 0)
    {
        foreach (string fileName in httpRequest.Files.Keys)
        {
            var file = httpRequest.Files[fileName];
            var filePath = HttpContext.Current.Server.MapPath("~/" + file.FileName);
            file.SaveAs(filePath);
        }

        return Request.CreateResponse(HttpStatusCode.Created);
    }

    return Request.CreateResponse(HttpStatusCode.BadRequest);
}