the model:
public class UploadFileModel
{
public int Id { get; set; }
public string FileName { get; set; }
public HttpPostedFileBase File { get; set; }
}
the controller:
public void Post(UploadFileModel model)
{
// never arrives...
}
I am getting an error
"No MediaTypeFormatter is available to read an object of type 'UploadFileModel' from content with media type 'multipart/form-data'."
Is there anyway around this?
Upload Single FileTo add view, right click on action method and click on add view. Then select View from left side filter and select Razor View – Empty. Then click on Add button. Create design for your view as per your requirements.
Easiest way is to add ModelBinder attribute to the parameter. In following example, I have added ModelBinder attribute to the "data" parameter, so web API can understand how to use custom model binder, while binding the data. Another way is to add ModelBinder attribute to the type.
It's not easily possible. Model binding in Web API is fundamentally different than in MVC and you would have to write a MediaTypeFormatter that would read the stream of files into your model and additionally bind primitives which can be considerably challenging.
The easiest solution is to grab the file stream off the request using some type of MultipartStreamProvider
and the other parameters using FormData
name value collection off that provider
Example - http://www.asp.net/web-api/overview/working-with-http/sending-html-form-data,-part-2:
public async Task<HttpResponseMessage> PostFormData()
{
if (!Request.Content.IsMimeMultipartContent())
{
throw new HttpResponseException(HttpStatusCode.UnsupportedMediaType);
}
string root = HttpContext.Current.Server.MapPath("~/App_Data");
var provider = new MultipartFormDataStreamProvider(root);
try
{
await Request.Content.ReadAsMultipartAsync(provider);
// Show all the key-value pairs.
foreach (var key in provider.FormData.AllKeys)
{
foreach (var val in provider.FormData.GetValues(key))
{
Trace.WriteLine(string.Format("{0}: {1}", key, val));
}
}
return Request.CreateResponse(HttpStatusCode.OK);
}
catch (System.Exception e)
{
return Request.CreateErrorResponse(HttpStatusCode.InternalServerError, e);
}
}
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