WEB API 使用MultipartFormDataStreamProvider上传文件

在这简短的教程中，我们打算去看如何上传多个文件在 ASP.Net WebAPI 中使用 MultipartFormDataStreamProvider。这一概念基于多部分/格式数据我们可以在这里发布多个文件的内容不仅将 NameValueCollection 作为服务器端提供的常规表单字段。在本教程中我们还看到了如何重写默认行为的 MultipartFormDataStreamProvider，将名称存储在一个独特的 BodyPart_ {GUID} 格式中。

using System;
using System.Collections.Generic;
using System.Diagnostics;
using System.IO;
using System.Linq;
using System.Net;
using System.Net.Http;
using System.Net.Http.Headers;
using System.Threading.Tasks;
using System.Web;
using System.Web.Http;
using System.Web.Mvc;

namespace UploadApplication.Controllers
{
    public class UploadController : ApiController
    {
        public async Task<HttpResponseMessage> Post()
        {
            // Check whether the POST operation is MultiPart?
            if (!Request.Content.IsMimeMultipartContent())
            {
                throw new HttpResponseException(HttpStatusCode.UnsupportedMediaType);
            }

            // Prepare CustomMultipartFormDataStreamProvider in which our multipart form
            // data will be loaded.
            string fileSaveLocation = HttpContext.Current.Server.MapPath("~/App_Data");
            CustomMultipartFormDataStreamProvider provider = new CustomMultipartFormDataStreamProvider(fileSaveLocation);
            List<string> files = new List<string>();

            try
            {
                // Read all contents of multipart message into CustomMultipartFormDataStreamProvider.
                await Request.Content.ReadAsMultipartAsync(provider);

                foreach (MultipartFileData file in provider.FileData)
                {
                    files.Add(Path.GetFileName(file.LocalFileName));
                }

                // Send OK Response along with saved file names to the client.
                return Request.CreateResponse(HttpStatusCode.OK, files);
            }
            catch (System.Exception e)
            {
                return Request.CreateErrorResponse(HttpStatusCode.InternalServerError, e);
            }
        }
    }

    // We implement MultipartFormDataStreamProvider to override the filename of File which
    // will be stored on server, or else the default name will be of the format like Body-
    // Part_{GUID}. In the following implementation we simply get the FileName from 
    // ContentDisposition Header of the Request Body.
    public class CustomMultipartFormDataStreamProvider : MultipartFormDataStreamProvider
    {
        public CustomMultipartFormDataStreamProvider(string path) : base(path) { }

        public override string GetLocalFileName(HttpContentHeaders headers)
        {
            return headers.ContentDisposition.FileName.Replace("\"", string.Empty);
        }
    }
}

测试程序

using System;
using System.Collections.Generic;
using System.IO;
using System.Linq;
using System.Net.Http;
using System.Net.Http.Headers;
using System.Text;
using System.Threading.Tasks;

namespace ConsoleApplication1
{
    class Program
    {
        static void Main(string[] args)
        {
            using (var client = new HttpClient())
            using (var content = new MultipartFormDataContent())
            {
                // Make sure to change API address
                client.BaseAddress = new Uri("http://localhost:53798/");

                // Add first file content 
                var fileContent1 = new ByteArrayContent(File.ReadAllBytes(@"c:\Users\aisadmin\Desktop\Me\NF2202533167366.pdf"));
                fileContent1.Headers.ContentDisposition = new ContentDispositionHeaderValue("attachment")
                {
                    FileName = "Sample.pdf"
                };

                // Add Second file content
                var fileContent2 = new ByteArrayContent(File.ReadAllBytes(@"c:\Users\aisadmin\Desktop\Sample.txt"));
                fileContent2.Headers.ContentDisposition = new ContentDispositionHeaderValue("attachment")
                {
                    FileName = "Sample.txt"
                };

                content.Add(fileContent1);
                content.Add(fileContent2);

                // Make a call to Web API
                var result = client.PostAsync("/api/upload", content).Result;

                Console.WriteLine(result.StatusCode);
                Console.ReadLine();
            }
        }
    }
}