CODEWITHSUNDEEP
c#asp.net-web-api.net-core
Sergio
Sergio

Reputation: 8259

Uploading File from console application to WebAPI

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?

Upvotes: 2

Views: 5426

Answers (2)

johnny 5
johnny 5

Reputation: 20987

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();

Upvotes: 1

DividedByZero
DividedByZero

Reputation: 11

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.

}

Upvotes: 1

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