7

我正在尝试使用 N-UNIT 来测试我的 Web API 应用程序，但我找不到合适的方法来测试我的文件上传方法。哪种方法是测试该方法的最佳方法？

Web API 控制器：

[AcceptVerbs("post")]
public async Task<HttpResponseMessage> Validate()
    {
        // Check if the request contains multipart/form-data.
        if (!Request.Content.IsMimeMultipartContent())
        {
            return Request.CreateErrorResponse(HttpStatusCode.UnsupportedMediaType,"please submit a valid request");
        }
        var provider = new MultipartMemoryStreamProvider(); // this loads the file into memory for later on processing 
        try
        {
            await Request.Content.ReadAsMultipartAsync(provider);
            var resp = new HttpResponseMessage(HttpStatusCode.OK);
            foreach (var item in provider.Contents)
            {
                if (item.Headers.ContentDisposition.FileName != null)
                {
                    Stream stream = item.ReadAsStreamAsync().Result;
        // do some stuff and return response
                    resp.Content = new StringContent(result, Encoding.UTF8, "application/xml"); //text/plain "application/xml"
                    return resp;
                }
            }
               return resp;
        }
        catch (System.Exception e)
        {
            return Request.CreateErrorResponse(HttpStatusCode.InternalServerError, e);
        }
    }
4

2 回答 2

20

根据您的上述评论，以下是一个示例：

HttpClient client = new HttpClient();

MultipartFormDataContent formDataContent = new MultipartFormDataContent();
formDataContent.Add(new StringContent("Hello World!"),name: "greeting");
StreamContent file1 = new StreamContent(File.OpenRead(@"C:\Images\Image1.jpeg"));
file1.Headers.ContentType = new MediaTypeHeaderValue("image/jpeg");
file1.Headers.ContentDisposition = new ContentDispositionHeaderValue("form-data");
file1.Headers.ContentDisposition.FileName = "Image1.jpeg";
formDataContent.Add(file1);
StreamContent file2 = new StreamContent(File.OpenRead(@"C:\Images\Image2.jpeg"));
file2.Headers.ContentType = new MediaTypeHeaderValue("image/jpeg");
file2.Headers.ContentDisposition = new ContentDispositionHeaderValue("form-data");
file2.Headers.ContentDisposition.FileName = "Image1.jpeg";
formDataContent.Add(file2);

HttpResponseMessage response = client.PostAsync("http://loclhost:9095/api/fileuploads", formDataContent).Result;

通过网络发出的请求如下：

POST http://localhost:9095/api/fileuploads HTTP/1.1
Content-Type: multipart/form-data; boundary="34d56c28-919b-42ab-8462-076b400bd03f"
Host: localhost:9095
Content-Length: 486
Expect: 100-continue
Connection: Keep-Alive

--34d56c28-919b-42ab-8462-076b400bd03f
Content-Type: text/plain; charset=utf-8
Content-Disposition: form-data; name=greeting

Hello World!
--34d56c28-919b-42ab-8462-076b400bd03f
Content-Type: image/jpeg
Content-Disposition: form-data; filename=Image1.jpeg

----Your Image here-------
--34d56c28-919b-42ab-8462-076b400bd03f
Content-Type: image/jpeg
Content-Disposition: form-data; filename=Image2.jpeg

----Your Image here-------
--34d56c28-919b-42ab-8462-076b400bd03f--
于 2013-07-18T21:56:51.297 回答
4

在花了一些时间研究 WebClient 之后，我想出了这个：

     try
        {
            var imageFile = Path.Combine("dir", "fileName");
            WebClient webClient = new WebClient();
            byte[] rawResponse = webClient.UploadFile(string.Format("{0}/api/values/", "http://localhost:12345/"), imageFile);
            Console.WriteLine("Sever Response: {0}", System.Text.Encoding.ASCII.GetString(rawResponse)); // for debugging purposes
            Console.WriteLine("File Upload was successful"); 
        }
        catch (WebException wexc)
        {
           Console.WriteLine("Failed with an exception of " + wexc.Message);  
           // anything other than 200 will trigger the WebException

        }
于 2013-07-18T15:49:36.773 回答