The answer provided by JPgrassi is what you would be doing to have MultiPart data. I think there are few more things that needs to be added, so I thought of writing my own answer.
MultiPart form data, as the name suggest, is not single type of data, but specifies that the form will be sent as a MultiPart MIME message, so you cannot have predefined formatter to read all the contents. You need to use ReadAsync function to read byte stream and get your different types of data, identify them and de-serialize them.
There are two ways to read the contents. First one is to read and keep everything in memory and the second way is to use a provider that will stream all the file contents into some randomly name files(with GUID) and providing handle in form of local path to access file (The example provided by jpgrassi is doing the second).
First Method: Keeping everything in-memory
//Async because this is asynchronous process and would read stream data in a buffer.
//If you don't make this async, you would be only reading a few KBs (buffer size)
//and you wont be able to know why it is not working
public async Task<HttpResponseMessage> Post()
{
if (!request.Content.IsMimeMultipartContent()) return null;
Dictionary<string, object> extractedMediaContents = new Dictionary<string, object>();
//Here I am going with assumption that I am sending data in two parts,
//JSON object, which will come to me as string and a file. You need to customize this in the way you want it to.
extractedMediaContents.Add(BASE64_FILE_CONTENTS, null);
extractedMediaContents.Add(SERIALIZED_JSON_CONTENTS, null);
request.Content.ReadAsMultipartAsync()
.ContinueWith(multiPart =>
{
if (multiPart.IsFaulted || multiPart.IsCanceled)
{
Request.CreateErrorResponse(HttpStatusCode.InternalServerError, multiPart.Exception);
}
foreach (var part in multiPart.Result.Contents)
{
using (var stream = part.ReadAsStreamAsync())
{
stream.Wait();
Stream requestStream = stream.Result;
using (var memoryStream = new MemoryStream())
{
requestStream.CopyTo(memoryStream);
//filename attribute is identifier for file vs other contents.
if (part.Headers.ToString().IndexOf("filename") > -1)
{
extractedMediaContents[BASE64_FILE_CONTENTS] = memoryStream.ToArray();
}
else
{
string jsonString = System.Text.Encoding.ASCII.GetString(memoryStream.ToArray());
//If you need just string, this is enough, otherwise you need to de-serialize based on the content type.
//Each content is identified by name in content headers.
extractedMediaContents[SERIALIZED_JSON_CONTENTS] = jsonString;
}
}
}
}
}).Wait();
//extractedMediaContents; This now has the contents of Request in-memory.
}
Second Method: Using a provider (as given by jpgrassi)
Point to note, this is only filename. If you want process file or store at different location, you need to stream read the file again.
public async Task<HttpResponseMessage> Post()
{
HttpResponseMessage response;
//Check if request is MultiPart
if (!Request.Content.IsMimeMultipartContent())
{
throw new HttpResponseException(HttpStatusCode.UnsupportedMediaType);
}
//This specifies local path on server where file will be created
string root = HttpContext.Current.Server.MapPath("~/App_Data");
var provider = new MultipartFormDataStreamProvider(root);
//This write the file in your App_Data with a random name
await Request.Content.ReadAsMultipartAsync(provider);
foreach (MultipartFileData file in provider.FileData)
{
//Here you can get the full file path on the server
//and other data regarding the file
//Point to note, this is only filename. If you want to keep / process file, you need to stream read the file again.
tempFileName = file.LocalFileName;
}
// You values are inside FormData. You can access them in this way
foreach (var key in provider.FormData.AllKeys)
{
foreach (var val in provider.FormData.GetValues(key))
{
Trace.WriteLine(string.Format("{0}: {1}", key, val));
}
}
//Or directly (not safe)
string name = provider.FormData.GetValues("name").FirstOrDefault();
response = Request.CreateResponse(HttpStatusCode.Ok);
return response;
}
