从WEB API Controller返回异步方法中的Void

Question

我从这个博客获得的ASP.NET MVC 4 WEB API Controller中有这个异步方法:http: //www.strathweb.com/2012/04/html5-drag-and-drop-asynchronous-multi-file-upload -with-ASP净的WebAPI /

  public async Task<IList<RecivedFile>> Post()
    {
        List<RecivedFile> result = new List<RecivedFile>();
        if (Request.Content.IsMimeMultipartContent())
        {
            try
            {
                MultipartFormDataStreamProvider stream = new MultipartFormDataStreamProvider(HostingEnvironment.MapPath("~/USER-UPLOADS"));

                IEnumerable<HttpContent> bodyparts = await Request.Content.ReadAsMultipartAsync(stream);
                IDictionary<string, string> bodyPartFiles = stream.BodyPartFileNames;
                IList<string> newFiles = new List<string>();

                foreach (var item in bodyPartFiles)
                {
                    var newName = string.Empty;
                    var file = new FileInfo(item.Value);

                    if (item.Key.Contains("\""))
                        newName = Path.Combine(file.Directory.ToString(), item.Key.Substring(1, item.Key.Length - 2));
                    else
                        newName = Path.Combine(file.Directory.ToString(), item.Key);

                    File.Move(file.FullName, newName);
                    newFiles.Add(newName);
                }

                var uploadedFiles = newFiles.Select(i =>
                {
                    var fi = new FileInfo(i);
                    return new RecivedFile(fi.Name, fi.FullName, fi.Length);
                }).ToList();

                result.AddRange(uploadedFiles);
            }
            catch (Exception e)
            {
            }
        }
        return result;
    }

我的问题是为什么这个方法确实有一个返回类型的任务？它还不清楚"在哪里"或"向谁"返回此任务？这就像没有人等待/接收返回的对象.

我想知道如果我像这样返回void会有什么影响:

编辑:

我已经尝试了下面的代码,它完全打破了程序.就像运行时失去了对代码的引用一样,代码本身也没有完成运行.

    public async void Post()
    {
        List<RecivedFile> result = new List<RecivedFile>();
        if (Request.Content.IsMimeMultipartContent())
        {
            try
            {
                MultipartFormDataStreamProvider stream = new MultipartFormDataStreamProvider(HostingEnvironment.MapPath("~/USER-UPLOADS"));

                IEnumerable<HttpContent> bodyparts = await Request.Content.ReadAsMultipartAsync(stream);
                IDictionary<string, string> bodyPartFiles = stream.BodyPartFileNames;
                IList<string> newFiles = new List<string>();

                foreach (var item in bodyPartFiles)
                {
                    var newName = string.Empty;
                    var file = new FileInfo(item.Value);

                    if (item.Key.Contains("\""))
                        newName = Path.Combine(file.Directory.ToString(), item.Key.Substring(1, item.Key.Length - 2));
                    else
                        newName = Path.Combine(file.Directory.ToString(), item.Key);

                    File.Move(file.FullName, newName);
                    newFiles.Add(newName);
                }

            }
            catch (Exception e)
            {
            }
        }

    }

Answer 1

ASP.NET运行时等待它.您可能会发现此视频很有用.

async假设UI上下文的大多数示例.ASP.NET 还提供了一个上下文,但它是一个"请求"上下文 - 每个HTTP请求一个上下文.我async/ await帖子概述了这个"背景",async/ awaitFAQ更详细.