我正在尝试使用N-UNIT来测试我的Web API应用程序,但我无法找到一种正确的方法来测试我的文件上传方法。哪种方法是测试方法的最佳方法?
Web API Controller:
[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);
}
}
答案 0 :(得分:18)
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;
通过电汇的请求如下:
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--
答案 1 :(得分:3)
在花了一些时间研究WebClient后,我得出了这个:
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
}