NHttp.HttpRequest.ParseMultiPartItems C# (CSharp) Method

        private void ParseMultiPartItems(HttpClient client)
        {
            Files = new HttpFileCollection();

            if (client.MultiPartItems == null)
                return;

            foreach (var item in client.MultiPartItems)
            {
                string contentType = null;
                string name = null;
                string fileName = null;

                string header;

                if (item.Headers.TryGetValue("Content-Disposition", out header))
                {
                    string[] parts = header.Split(';');

                    for (int i = 0; i < parts.Length; i++)
                    {
                        string part = parts[i].Trim();

                        if (part.StartsWith("name="))
                            name = ParseContentDispositionItem(part.Substring(5));
                        else if (part.StartsWith("filename="))
                            fileName = ParseContentDispositionItem(part.Substring(9));
                    }
                }

                if (item.Headers.TryGetValue("Content-Type", out header))
                    contentType = header;

                if (name == null)
                {
                    Log.Info("Received multipart item without name");
                    continue;
                }

                if (item.Value != null)
                {
                    Form[name] = item.Value;
                }
                else
                {
                    Files.AddFile(name, new HttpPostedFile((int)item.Stream.Length, contentType, fileName, item.Stream));
                }
            }
        }