Question

I am using ZipOutputStream from SharpZipLib and I wish to upload the zipped contents it creates directly to my MVC post action. I am successfully getting it to post however the parameter of my action method has null as the posted data when it gets to my MVC action.

Here is my Test code I’m using to test this out:

    public void UploadController_CanUploadTest()
    {
        string xml = "<test>xml test</test>"
        string url = "http://localhost:49316/Api/DataUpload/Upload/";

        WebClient client = new WebClient();

        var cc= new CredentialCache();
        cc.Add(new Uri(url),
              "Basic", 
              new NetworkCredential("Testuser", "user"));

        client.Credentials = cc;

        string _UserAgent = "Mozilla/4.0 (compatible; MSIE 6.0; Windows NT 5.2; .NET CLR 1.0.3705;)";
        client.Headers.Add(HttpRequestHeader.UserAgent, _UserAgent);
        client.Headers["Content-type"] = "application/x-www-form-urlencoded";

        using (var stream = client.OpenWrite(url, "POST"))
        {
            Zipped zip = new Zipped(stream, Encoding.UTF8, false);

            FileContent content = new FileContent("Upload", xml);

            var uploads = new List<FileContent>();
            uploads.Add(content);

            zip.Compress(uploads);

            stream.Flush();
            stream.Close();
        }
    }

This is my zipped class wrapper:

    public class Zipped : ICompression, IDisposable
{
    private Stream _stream = null;
    private bool _closeStreamOnDispose = true;
    private Encoding _encoding;

    public Zipped()
        : this(new MemoryStream())
    {

    }

    public Zipped(Stream stream)
        : this(stream, Encoding.UTF8, true)
    {
    }

    public Zipped(Stream stream, Encoding encoding)
        : this(stream, encoding, true)
    {
    }

    public Zipped(Stream stream, Encoding encoding, bool closeStreamOnDispose)
    {
        _stream = stream;
        _closeStreamOnDispose = closeStreamOnDispose;
        _encoding = encoding;
    }

    public Stream Compress(IList<FileContent> dataList)
    {
        ZipOutputStream outputStream = new ZipOutputStream(_stream);
        outputStream.SetLevel(9);

        foreach (var data in dataList)
        {
            ZipEntry entry = new ZipEntry(data.Name);
            entry.CompressionMethod = CompressionMethod.Deflated;

            outputStream.PutNextEntry(entry);

            byte[] dataAsByteArray = _encoding.GetBytes(data.Content);

            outputStream.Write(dataAsByteArray, 0, dataAsByteArray.Length);
            outputStream.CloseEntry();
        }

        outputStream.IsStreamOwner = false;
        outputStream.Flush();
        outputStream.Close();

        return _stream;
    }

    public List<FileContent> DeCompress()
    {
        ZipInputStream inputStream = new ZipInputStream(_stream);
        ZipEntry entry = inputStream.GetNextEntry();

        List<FileContent> dataList = new List<FileContent>();

        while(entry != null)
        {
            string entryFileName = entry.Name;

            byte[] buffer = new byte[4096];     // 4K is optimum

            // Unzip file in buffered chunks. This is just as fast as unpacking to a buffer the full size
            // of the file, but does not waste memory.
            // The "using" will close the stream even if an exception occurs.                
            using (MemoryStream tempMemoryStream = new MemoryStream())
            {
                StreamUtils.Copy(inputStream, tempMemoryStream, buffer);

                string copied = _encoding.GetString(tempMemoryStream.ToArray());
                dataList.Add(new FileContent(entry.Name, copied));
            }

            entry = inputStream.GetNextEntry();
        }

        return dataList;

    }

    public void Dispose()
    {
        if(_closeStreamOnDispose)
            _stream.Dispose();
    }

Here is my simple MVC action:

    [HttpPost]
    public ActionResult Upload(HttpPostedFileBase uploaded)
    {
        // uploaded is null at this point
    }

Answer 1

If you want to use HttpPostedFileBase in your controller action you need to send a multipart/form-data request from the client and not application/x-www-form-urlencoded.

In fact you set the content type to application/x-www-form-urlencoded but you are not respecting this because you are directly writing the raw bytes to the request which is invalid. Well, in fact it’s not respecting the HTTP protocol standard but it could still work if you read the raw request stream from the controller instead of using HttpPostedFileBase. I wouldn’t recommend you going that route.

So the correct HTTP request that you are sending must look like this:

Content-Type: multipart/form-data; boundary=AaB03x

--AaB03x
Content-Disposition: form-data; name="uploaded"; filename="input.zip"
Content-Type: application/zip

... byte contents of the zip ...
--AaB03x--

The boundary must be chosen so that it doesn’t appear anywhere in the contents of the file.

I have blogged about an example of how you could upload multiple files.