我如何获得一个MultipartMemoryStreamProvider的文件内容作为字节数组？ [英] How do I get the file contents of a MultipartMemoryStreamProvider as a byte array?

问题描述

我已经创建了文件上传使用FORMDATA一个Web API控制器的组成部分。

我如何从MultipartMemoryStreamProvider文件内容作为字节数组？

这是Web API方法

 公共任务＆LT;＆IEnumerable的LT; FileModel＆GT;＆GT;邮政（）
   {
     如果（Request.Content.IsMimeMultipartContent（））
        {
            VAR streamProvider =新MultipartMemoryStreamProvider（）;
            VAR任务= Request.Content.ReadAsMultipartAsync(streamProvider).ContinueWith<IEnumerable<FileModel>>(t = GT;
            {                如果（t.IsFaulted || t.IsCanceled）
                {
                    抛出新的Htt presponseException（的HTTPStatus code.InternalServerError）;
                }                FileDataBO FILEDATA;
                变种的fileInfo = streamProvider.Contents.Select（ⅰ= GT; {
                    //保存到数据库
                    FILEDATA =新FileDataBO（）;
                    filedata.FileName = i.Headers.ContentDisposition.FileName;
                    filedata.FileType =JPEG;                    //如何获得文件内容的位置？它应该的byte []
                    //filedata.FileContent =？                    //去做
                    //_fileDataService.SaveFile(filedata）;                    返回新FileModel（i.Headers.ContentDisposition.FileName，2048）;
                }）;
                返回的fileInfo;
            }）;            返回任务;
        }
        其他
        {
            抛出新的Htt presponseException（Request.CreateResponse（的HTTPStatus code.NotAcceptable，这个请求的格式不正确））;
        }
   }
 

解决方案

您应该能够通过做获取内容 i.ReadAsByteArrayAsync（）

I have created a component that uploads files to a Web API controller using FormData.

How do I get the file contents from the MultipartMemoryStreamProvider as a byte array?

Here is the Web Api method

   public Task<IEnumerable<FileModel>> Post()
   {             
     if (Request.Content.IsMimeMultipartContent())
        {            
            var streamProvider = new MultipartMemoryStreamProvider();
            var task = Request.Content.ReadAsMultipartAsync(streamProvider).ContinueWith<IEnumerable<FileModel>>(t =>
            {

                if (t.IsFaulted || t.IsCanceled)
                {
                    throw new HttpResponseException(HttpStatusCode.InternalServerError);
                }

                FileDataBO filedata;
                var fileInfo = streamProvider.Contents.Select(i => {
                    //save to db
                    filedata = new FileDataBO ();
                    filedata.FileName = i.Headers.ContentDisposition.FileName;
                    filedata.FileType = "jpeg";

                    // HOW TO GET FILE CONTENT HERE??? IT SHOULD BYTE[]
                    //filedata.FileContent = ???

                    //TODO
                    //_fileDataService.SaveFile(filedata);    

                    return new FileModel(i.Headers.ContentDisposition.FileName, 2048);
                });
                return fileInfo;
            });

            return task;
        }             
        else
        {
            throw new HttpResponseException(Request.CreateResponse(HttpStatusCode.NotAcceptable, "This request is not properly formatted"));
        }       
   }

解决方案

You should be able to get the content by doing i.ReadAsByteArrayAsync()

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