Question

我在这里要完成的是看看这个API能够处理多少每秒请求数。我试图在控制台应用程序中使用API，最终将是一次性代码。我的想法是创建一个for循环，尝试每2秒上传一个xml文档。我之前从未做过这样的事情，原谅我的无知。这是我的主要方法：

    static void Main()
    {            
       RunAsync().Wait();         
    }

RunAsync方法：

    static async Task RunAsync()
    {

        Uri apiUrl = new Uri("http://apiurl.com/upload/files/uploadfiles");
        const string file = @"C:\simple.xml";

        WebClient client = new WebClient();
        for (int i = 0; i <= 100; i++)
        {
            client.UploadFileCompleted += FileUploadSuccess;
            client.UploadFileAsync(apiUrl, file);
            await Task.Delay(2000);
            Console.WriteLine("Upload waiting 2 seconds...");
        }

        Console.WriteLine("Loop completed.");

    }

成功方法：

    private static void FileUploadSuccess(object sender, UploadFileCompletedEventArgs e)
    {
        string reply = System.Text.Encoding.UTF8.GetString(e.Result);
        Console.WriteLine("The file result was: {0}", reply);

    }

第一次通过e.Result引发异常。这是例外：

System.dll中出现未处理的“System.Reflection.TargetInvocationException”类型异常

在做了一些研究后，显然我不能在不等待它的情况下调用API方法（返回异步任务）。不幸的是，似乎UploadFileAsync并非“等待”。

以下是API方法：

public async Task<HttpResponseMessage> UploadFiles()
    {
        var pilotTokenObject = TokenHelper.CreatePilotTokenObject(Request);
        byte[] fileBuffer = null;

        HttpResponseMessage retVal = null;
        if (pilotTokenObject != null)
        {
            var content = Request.Content;

            if (content == null)
            {
                throw new PilotApiException("Empty request content", HttpStatusCode.NoContent);
            }
            if (!content.IsMimeMultipartContent())
            {
                throw new PilotApiException("Request does not contain not multi-part content");
            }
            var uploadModelController = new PilotUploadModelController();


            //*SAVE STREAMED FILE*
            string serverSavePath = ConfigurationManager.AppSettings["PilotUploadApiTempStoragePath"];
            if (!Directory.Exists(serverSavePath))
                Directory.CreateDirectory(serverSavePath);

            var provider = new MultipartFormDataStreamProvider(serverSavePath);

            await Request.Content.ReadAsMultipartAsync(provider);

            var fileData = provider.FileData;
            if (fileData == null || fileData.Count == 0)
            {
                throw new PilotApiException("No multipart/form file data present.");
            }

            bool uploaded = false;
            //Loop through each file
            fileData.ForEach((fileRequest) =>
            {
                if (RetryUntilFileReadable(Path.Combine(serverSavePath, fileRequest.LocalFileName), 1000, 5))
                {

                    var fileHeader = fileRequest.Headers;
                    if (fileHeader != null && fileHeader.ContentDisposition != null)
                    {
                        var fileName = fileHeader.ContentDisposition.FileName.Replace("\"", "");
                        var fileBytes = File.ReadAllBytes(Path.Combine(serverSavePath, fileRequest.LocalFileName));


                        //Save File to DB
                        var upload = uploadModelController.UploadHelper
                            .AddUploadFileToDb(pilotTokenObject.CentralUserDbUserId, pilotTokenObject.ClientIp, pilotTokenObject.UserAgentString,
                                UploadEnums.UploadKind.PilotUploadApi, fileName, fileBytes.Length, fileBytes,
                                UploadEnums.EncryptionType.None);

                        if (upload != null)
                            uploaded = true;
                    }
                }

            });



            if (uploaded)
            {
                retVal = Request.CreateResponse(HttpStatusCode.Accepted, new
                    {
                        Response = String.Format("file uploaded successfully.")
                    });
            }


        }

        return retVal;
    }

我是否以完全错误的方式解决这个问题？我想做什么甚至可行？

Answer 1

在我看来，以下内容在您的方案中会更好：

byte[] response = await Task.Run(() => client.UploadFile(apiUrl, file));
string reply = System.Text.Encoding.UTF8.GetString(response);
Console.WriteLine("The file result was: {0}", reply);
Console.WriteLine("Upload waiting 2 seconds...");
await Task.Delay(2000);

在这种情况下，尝试将旧的异步API与较新的异步/等待混合搭配似乎并不富有成效。最好只使用async / await兼容代码包装API的同步版本。

（请注意，在我看来，您也可以调用Thread.Sleep(2000)而不是创建一个新的延迟任务来等待，但上述情况也应该正常。）

使用UploadFileAsync为上传API创建负载测试器

1 个答案: