补充Matt Frear的回答-这将是一个asp.net Core的选择，直接从流读取文件，而不需要保存和从磁盘读取:

public ActionResult OnPostUpload(List<IFormFile> files)
    {
        try
        {
            var file = files.FirstOrDefault();
            var inputstream = file.OpenReadStream();

            XSSFWorkbook workbook = new XSSFWorkbook(stream);

            var FIRST_ROW_NUMBER = {{firstRowWithValue}};

            ISheet sheet = workbook.GetSheetAt(0);
            // Example: var firstCellRow = (int)sheet.GetRow(0).GetCell(0).NumericCellValue;

            for (int rowIdx = 2; rowIdx <= sheet.LastRowNum; rowIdx++)
               {
                  IRow currentRow = sheet.GetRow(rowIdx);

                  if (currentRow == null || currentRow.Cells == null || currentRow.Cells.Count() < FIRST_ROW_NUMBER) break;

                  var df = new DataFormatter();                

                  for (int cellNumber = {{firstCellWithValue}}; cellNumber < {{lastCellWithValue}}; cellNumber++)
                      {
                         //business logic & saving data to DB                        
                      }               
                }
        }
        catch(Exception ex)
        {
            throw new FileFormatException($"Error on file processing - {ex.Message}");
        }
    }

这里有两种接受文件的方法。一个使用内存提供程序MultipartMemoryStreamProvider，另一个使用保存到磁盘的MultipartFormDataStreamProvider。注意，这一次只针对一个文件上传。您可以扩展它来保存多个文件。第二种方法可以支持大文件。我测试过超过200MB的文件，它工作得很好。使用内存方法不需要保存到磁盘，但如果超过一定的限制，将抛出内存外异常。

private async Task<Stream> ReadStream()
{
    Stream stream = null;
    var provider = new MultipartMemoryStreamProvider();
    await Request.Content.ReadAsMultipartAsync(provider);
    foreach (var file in provider.Contents)
    {
        var buffer = await file.ReadAsByteArrayAsync();
        stream = new MemoryStream(buffer);
    }

    return stream;
}

private async Task<Stream> ReadLargeStream()
{
    Stream stream = null;
    string root = Path.GetTempPath();
    var provider = new MultipartFormDataStreamProvider(root);
    await Request.Content.ReadAsMultipartAsync(provider);
    foreach (var file in provider.FileData)
    {
        var path = file.LocalFileName;
        byte[] content = File.ReadAllBytes(path);
        File.Delete(path);
        stream = new MemoryStream(content);
    }

    return stream;
}

下面是一个快速而简单的解决方案，它从HTTP正文中获取上传的文件内容并将其写入文件。我为文件上传包含了一个“骨架”HTML/JS片段。

Web API方法:

[Route("api/myfileupload")]        
[HttpPost]
public string MyFileUpload()
{
    var request = HttpContext.Current.Request;
    var filePath = "C:\\temp\\" + request.Headers["filename"];
    using (var fs = new System.IO.FileStream(filePath, System.IO.FileMode.Create))
    {
        request.InputStream.CopyTo(fs);
    }
    return "uploaded";
}

HTML文件上传:

<form>
    <input type="file" id="myfile"/>  
    <input type="button" onclick="uploadFile();" value="Upload" />
</form>
<script type="text/javascript">
    function uploadFile() {        
        var xhr = new XMLHttpRequest();                 
        var file = document.getElementById('myfile').files[0];
        xhr.open("POST", "api/myfileupload");
        xhr.setRequestHeader("filename", file.name);
        xhr.send(file);
    }
</script>

朝着同样的方向，我发布了一个客户端和服务器片段，使用WebApi, c# 4发送Excel文件:

public static void SetFile(String serviceUrl, byte[] fileArray, String fileName)
{
    try
    {
        using (var client = new HttpClient())
        {
                client.DefaultRequestHeaders.Accept.Add(new MediaTypeWithQualityHeaderValue("application/json"));
                using (var content = new MultipartFormDataContent())
                {
                    var fileContent = new ByteArrayContent(fileArray);//(System.IO.File.ReadAllBytes(fileName));
                    fileContent.Headers.ContentDisposition = new ContentDispositionHeaderValue("attachment")
                    {
                        FileName = fileName
                    };
                    content.Add(fileContent);
                    var result = client.PostAsync(serviceUrl, content).Result;
                }
        }
    }
    catch (Exception e)
    {
        //Log the exception
    }
}

服务器的webapi控制器:

public Task<IEnumerable<string>> Post()
{
    if (Request.Content.IsMimeMultipartContent())
    {
        string fullPath = HttpContext.Current.Server.MapPath("~/uploads");
        MyMultipartFormDataStreamProvider streamProvider = new MyMultipartFormDataStreamProvider(fullPath);
        var task = Request.Content.ReadAsMultipartAsync(streamProvider).ContinueWith(t =>
        {
            if (t.IsFaulted || t.IsCanceled)
                    throw new HttpResponseException(HttpStatusCode.InternalServerError);

            var fileInfo = streamProvider.FileData.Select(i =>
            {
                var info = new FileInfo(i.LocalFileName);
                return "File uploaded as " + info.FullName + " (" + info.Length + ")";
            });
            return fileInfo;

        });
        return task;
    }
    else
    {
        throw new HttpResponseException(Request.CreateResponse(HttpStatusCode.NotAcceptable, "Invalid Request!"));
    }
}

和自定义MyMultipartFormDataStreamProvider，需要自定义文件名: PS:我从另一个帖子http://www.codeguru.com/csharp/.net/uploading-files-asynchronously-using-asp.net-web-api.htm取了这个代码

public class MyMultipartFormDataStreamProvider : MultipartFormDataStreamProvider
{
    public MyMultipartFormDataStreamProvider(string path)
        : base(path)
    {

    }

    public override string GetLocalFileName(System.Net.Http.Headers.HttpContentHeaders headers)
    {
        string fileName;
        if (!string.IsNullOrWhiteSpace(headers.ContentDisposition.FileName))
        {
            fileName = headers.ContentDisposition.FileName;
        }
        else
        {
            fileName = Guid.NewGuid().ToString() + ".data";
        }
        return fileName.Replace("\"", string.Empty);
    }
}

这个问题甚至对于。net Core也有很多好的答案。我使用这两个框架提供的代码示例工作良好。我就不重复了。在我的例子中，重要的事情是如何使用Swagger的文件上传动作，就像这样:

以下是我的概述:

2 . asp.net WebAPI

上传文件使用:MultipartFormDataStreamProvider见答案在这里 如何使用它与Swagger

net核心

上传文件使用:IFormFile见答案在这里或MS文档 如何使用它与Swagger

[HttpPost]
public JsonResult PostImage(HttpPostedFileBase file)
{
    try
    {
        if (file != null && file.ContentLength > 0 && file.ContentLength<=10485760)
        {
            var fileName = Path.GetFileName(file.FileName);                                        

            var path = Path.Combine(Server.MapPath("~/") + "HisloImages" + "\\", fileName);

            file.SaveAs(path);
            #region MyRegion
            ////save imag in Db
            //using (MemoryStream ms = new MemoryStream())
            //{
            //    file.InputStream.CopyTo(ms);
            //    byte[] array = ms.GetBuffer();
            //} 
            #endregion
            return Json(JsonResponseFactory.SuccessResponse("Status:0 ,Message: OK"), JsonRequestBehavior.AllowGet);
        }
        else
        {
            return Json(JsonResponseFactory.ErrorResponse("Status:1 , Message: Upload Again and File Size Should be Less Than 10MB"), JsonRequestBehavior.AllowGet);
        }
    }
    catch (Exception ex)
    {

        return Json(JsonResponseFactory.ErrorResponse(ex.Message), JsonRequestBehavior.AllowGet);

    }
}