如何使用Web上传文件和提交表单数据API?

How to upload files and submit form data using Web API?

背景

我目前正在使用 C# 代码从网页读取 HTTP POST 表单提交中的值。它看起来像这样:

public class CompanyDTO
{
    public string CompanyName { get; set; }
    public string Address1 { get; set; }
    public string Address2 { get; set; }
    public string City { get; set; }
    public string State { get; set; }
    public string ZIP { get; set; }
}

// POST: api/Submission
public HttpResponseMessage Post(CompanyDTO Company)
{
    return JsonToBrowser("{'location':'" + Company.City + "," + Company.State + "'}");
}

我还有从页面上的文件上传控件读取附件的工作代码:

// POST: api/Submission
public HttpResponseMessage Post()
{
    // Process and save attachments/uploaded files
    var folderName = "App_Data";
    var PATH = HttpContext.Current.Server.MapPath("~/" + folderName);
    var rootUrl = Request.RequestUri.AbsoluteUri.Replace(Request.RequestUri.AbsolutePath, String.Empty);
    if (Request.Content.IsMimeMultipartContent())
    {
        var streamProvider = new CustomMultipartFormDataStreamProvider(PATH);
        var task = Request.Content.ReadAsMultipartAsync(streamProvider).ContinueWith<IEnumerable<FileDesc>>(t =>
        {
            if (t.IsFaulted || t.IsCanceled)
            {
                throw new HttpResponseException(HttpStatusCode.InternalServerError);
            }

            var fileInfo = streamProvider.FileData.Select(i =>
            {
                var info = new FileInfo(i.LocalFileName);
                return new FileDesc(info.Name, rootUrl + "/" + folderName + "/" + info.Name, info.Length / 1024);
            });
            return fileInfo;
        });

        return JsonToBrowser("{'status':'success'}");
    } 
    else 
    {
        throw new HttpResponseException(Request.CreateResponse(HttpStatusCode.NotAcceptable, "This request is not properly formatted"));
    }
}

显然最后一段代码还有一些内容,但我希望它至少能说明我的问题。

我的问题

我现在正在尝试合并这两段代码。我认为这就像将数据传输对象放在 Post 函数中一样简单:

public HttpResponseMessage Post(CompanyDTO Company)

这会导致在我执行 HTTP POST 时向浏览器返回以下内容:

{ "Message":"The request entity's media type 'multipart/form-data' is not supported for this resource.", "ExceptionMessage":"No MediaTypeFormatter is available to read an object of type 'CompanyDTO' from content with media type 'multipart/form-data'.", "ExceptionType":"System.Net.Http.UnsupportedMediaTypeException", "StackTrace":" at System.Net.Http.HttpContentExtensions.ReadAsAsync[T](HttpContent content, Type type, IEnumerable1 formatters, IFormatterLogger formatterLogger, CancellationToken cancellationToken)\r\n at System.Net.Http.HttpContentExtensions.ReadAsAsync(HttpContent content, Type type, IEnumerable1 formatters, IFormatterLogger formatterLogger, CancellationToken cancellationToken)\r\n at System.Web.Http.ModelBinding.FormatterParameterBinding.ReadContentAsync(HttpRequestMessage request, Type type, IEnumerable1 formatters, IFormatterLogger formatterLogger, CancellationToken cancellationToken)" }

我的问题

那么我如何使用 C# 和 ASP.NET 上传一个或多个文件和 read/process 常规表单数据字段。我使用的是 Visual Studio 2013(社区版)。 另外,我是 C# 的初学者,所以我可能需要一些额外的 hints/help 如果你参考 function/class 关于在哪里可以找到它。

我发现与此问题相关的最接近的 SO 问题是 How to submit form-data with optional file data in asp.net web api 但它没有任何 response/answer...

这是我遇到这个问题时的解决方案:

客户:

 public async Task UploadImage(byte[] image, string url)
        {
            Stream stream = new System.IO.MemoryStream(image);
            HttpStreamContent streamContent = new HttpStreamContent(stream.AsInputStream());

            Uri resourceAddress = null;
            Uri.TryCreate(url.Trim(), UriKind.Absolute, out resourceAddress);
            Windows.Web.Http.HttpRequestMessage request = new Windows.Web.Http.HttpRequestMessage(Windows.Web.Http.HttpMethod.Post, resourceAddress);
            request.Content = streamContent;

            var httpClient = new Windows.Web.Http.HttpClient();
            var cts = new CancellationTokenSource();
            Windows.Web.Http.HttpResponseMessage response = await httpClient.SendRequestAsync(request).AsTask(cts.Token);
        }

控制器:

public async Task<HttpResponseMessage> Post()
{
    Stream requestStream = await this.Request.Content.ReadAsStreamAsync();
    byte[] byteArray = null;
    using (MemoryStream ms = new MemoryStream())
    {
        await requestStream.CopyToAsync(ms);
        byteArray = ms.ToArray();
    }
    .
    .
    .
    return Request.CreateResponse(HttpStatusCode.OK);
}