Web API中发送HTML表单数据：文件上传且带表单数据

参考资料

https://docs.microsoft.com/en-us/aspnet/web-api/overview/advanced/sending-html-form-data-part-2

代码实现

        [HttpPost]
        public async Task UploadExcel()
        {
            /*
             * 方法一：
             */
            var result = await Request.Content.ReadAsMultipartAsync();
            var requestJson1 = await result.Contents[0].ReadAsStringAsync();
            var requestJson2 = await result.Contents[1].ReadAsStringAsync();
            var requestJson3 = await result.Contents[2].ReadAsStringAsync();
            var requestJson4 = await result.Contents[3].ReadAsStringAsync();
            //接收转Json
            var request = JsonConvert.DeserializeObject>(requestJson4);

            /*
             * 方法二：
             */
            var requestJson4 = await result.Contents
            if (!Request.Content.IsMimeMultipartContent())
            {
                throw new HttpResponseException(HttpStatusCode.UnsupportedMediaType);
            }
            string root = HttpContext.Current.Server.MapPath("~/App_Data");
            var provider = new MultipartFormDataStreamProvider(root);
            try
            {
                await Request.Content.ReadAsMultipartAsync(provider);
                //获取表单数据
                foreach (var key in provider.FormData.AllKeys)
                {
                    foreach (var val in provider.FormData.GetValues(key))
                    {
                        //Trace.WriteLine(string.Format("{0}: {1}", key, val));
                    }
                }

                //获取文件数据
                if (HttpContext.Current.Request.Files.Count > 0)
                {
                    var file=HttpContext.Current.Request.Files[0];                    
                }
                return Request.CreateResponse(HttpStatusCode.OK);
            }
            catch (System.Exception e)
            {
                return Request.CreateErrorResponse(HttpStatusCode.InternalServerError, e);
            }
            return Request.CreateResponse(HttpStatusCode.OK);
        }