使用多部分实体将大文件从安卓上传到 c# asmx Web 服务

Question

我正在通过多部分实体请求从安卓获取请求数据。 如何接受该请求并将文件保存在服务器端。请检查已尝试的代码。来自安卓的文件是视频文件。

[WebMethod]
    [ScriptMethod(ResponseFormat = ResponseFormat.Json)]
    public UploadFileResponse FileUpload(FileStream stream)
    {        
        JavaScriptSerializer js = new JavaScriptSerializer();
        Context.Response.Clear();
        Context.Response.ContentType = "application/json";
        UploadFileResponse _response = null;
        bool IsFileUploaded = false;
        if (_response != null)
        {
            return _response;
        }
        else
        {
            _response = new UploadFileResponse();
        }
        try
        {
            MultipartParser parser = new MultipartParser(stream);

            if (parser.Success)
            {               
                string fileName = parser.Filename;
                string contentType = parser.ContentType;
                byte[] fileContent = parser.FileContents; 
                FileStream fileToupload = new FileStream("D:\FileUpload\" + fileName, FileMode.Create);
                fileToupload.Write(fileContent, 0, fileContent.Length);
                fileToupload.Close();
                fileToupload.Dispose();               
                _response.Result = true;
                _response.Message = "Success";
                stream.Close();
            }
            else
            {
                _response.Result = false;
                _response.Message = "Oops, something went wrong, please try again.";
            }
        }
        catch (Exception ex)
        {
            _response.Result = false;
            _response.Error = ex.Message;
            _response.Message = "Oops, something went wrong, please try again.";            
        }
        finally
        {
        }
        return _response;
    }

Answer 1

如果成功地将多部分数据发送到 Web 服务，则应该能够使用 HttpContext.Current.Request 捕获传入文件。

下面的代码会将文件保存到 Web 服务所在的当前目录。

[WebMethod]
    public void AttachFiles()
    {
        HttpPostedFile file = HttpContext.Current.Request.Files[0];
        using (var fileStream = new System.IO.FileStream(AppDomain.CurrentDomain.BaseDirectory+file.FileName, System.IO.FileMode.Create, System.IO.FileAccess.Write))
        {
            file.InputStream.CopyTo(fileStream);
        }
    }