How to upload files in the ASP.NET Web API using Http Client

Steps for Web Application (Client Application)

Here in this example, the client application is "WebApiClient.Web".

Step-1: Add an Action to the HomeController (in MVC Client application) for get view for upload file.

                        public ActionResult Part6()
                        {
                            return View();
                        }
                    

Step-2: Add view for the Action & design.

Right Click on Action Method (here right click on form action) > Add View... > Enter Name > Add.
View 

@{
    ViewBag.Title = "Part6";
}
 
<h2>Part6 - Upload file to web api using Http Client</h2>
<div class="container">
    <div>
        @if (ViewBag.Success != null)
        {
            <div class="alert alert-success" role="alert">
                <strong>Well done !</strong> Successfully uploaded. <a href="@ViewBag.Success" target="_blank">open file</a>
            </div>
        }
        else if(ViewBag.Failed != null)
        {
            <div class="alert alert-error" role="alert">
                <strong>Error !</strong> @ViewBag.Failed
            </div>
        }
    </div>
    @using (Html.BeginForm("Part6","Home", FormMethod.Post,new{ role = "form", enctype="multipart/form-data"}))
    {
        <div class="form-group">
            <input type="file" id="file" name="file" />
        </div>
        <input type="submit" value="Submit" class="btn btn-default" />
    }
</div>
                    

Step-3: Add an another action for POST action for upload file to web api using HttpClient

[HttpPost]
public ActionResult Part6(HttpPostedFileBase file)
{
    using (var client = new HttpClient())
    {
        using (var content = new MultipartFormDataContent())
        {
            byte[] Bytes = new byte[file.InputStream.Length + 1];
            file.InputStream.Read(Bytes, 0, Bytes.Length);
            var fileContent = new ByteArrayContent(Bytes);
            fileContent.Headers.ContentDisposition = new System.Net.Http.Headers.ContentDispositionHeaderValue("attachment") { FileName = file.FileName };
            content.Add(fileContent);
            var requestUri = "http://localhost:1963/api/upload";
            var result = client.PostAsync(requestUri, content).Result;
            if (result.StatusCode == System.Net.HttpStatusCode.Created)
            {
                List<string> m = result.Content.ReadAsAsync<List<string>>().Result;
                ViewBag.Success = m.FirstOrDefault();
 
            }
            else
            {
                ViewBag.Failed = "Failed !" + result.Content.ToString();
            }
        }
    }
    return View();
}
                    

Step-4: Run Application.

Here we need to start both application as the client application will consume services from web api application.