Uploading File from console application to WebAPI

Question

I'm trying to post a file + some info to a WebApi I control. My problem is that I can't access the file on the WebAPI side, all other fields are OK.

Here is my Console Application code

using (HttpClient client = new HttpClient()) { using (MultipartFormDataContent content = new MultipartFormDataContent()) { string filename = "my_filename.png"; content.Add(new StringContent(DateTime.Now.ToString("yyyy-MM-dd")), "data"); byte[] file_bytes = webClient.DownloadData($"https://my_url/my_file.png"); content.Add( new ByteArrayContent(file_bytes), "file"); string requestUri = "http://localhost:51114/api/File"; HttpResponseMessage result = client.PostAsync(requestUri, content).Result; Console.WriteLine("Upload result {0}", result.StatusCode); } }

Here is my WebAPI Code

 [HttpPost] public void Post(IFormFile file, [FromForm] DateTime data) { if (file == null || file.Length == 0) { Response.StatusCode = StatusCodes.Status400BadRequest; return; } // Never reaches this point..... file is null }

Any pointers on what I might be missing?

DividedByZero · Accepted Answer · 2018-06-26 20:21:04Z

If i'm not mistaken, you can submit a file to a WebAPI endpoint sending it as FormData with a Content-Type : multipart/form-data, something like this.

[HttpPost] [Route("..."] public void ReceiveFile() { System.Web.HttpPostedFile file = HttpContext.Current.Request.Files["keyName"]; System.IO.MemoryStream mem = new System.IO.MemoryStream(); file.InputStream.CopyTo(mem); byte[] data = mem.ToArray(); // you can replace the MemoryStream with file.saveAs("path") if you want. }

Thank you. +1 but it didn't solve my problem. I think that the issue is with the sending part, because i can use the method good using insomnia.

johnny 5 · Accepted Answer · 2018-06-26 20:33:08Z

You can grab out the content and convert it into a byte array in 2 lines of code, assuming you are only sending a single file (Note) its a good idea to use async for file upload so you don't consume as much cpu time:

var provider = await Request.Content.ReadAsMultipartAsync(new MultipartMemoryStreamProvider()); var file = provider.Contents.Single();

Thank you. +1 but it didn't solve my problem. I think that the issue is with the sending part, because i can use the method good using insomnia.

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