.Net WebAPI 高速下载文件接口实现

接触WebAPI一年多了，感觉是个承上启下，开创未来的技术。老一辈程序员写接口就像写方法一样，不需要了解太多网页的知识，却可以在浏览器中访问这些接口；由于是基于HTTP协议，因此不管是PC、手机还是嵌入式均可顺利访问。对于当下软件多终端的设计结构来说，这就像是量身订做的一样。

在开发的过程中遇到了一些问题，大部分都可通过百度找到解决方案，但是有一个问题却一直没有很好地解决，那就是文件下载速度的问题。

网上找到的WebAPI下载文件的代码大都是以下方式：

using System;
using System.Collections;
using System.Collections.Generic;
using System.IO;
using System.Linq;
using System.Net;
using System.Net.Http;
using System.Runtime.Remoting.Contexts;
using System.Text;
using System.Threading.Tasks;
using System.Web;
using System.Web.Http;
using System.Web.Http.Cors;
using System.Net.Http.Headers;namespace Service.Controller
{public partial class APIController : ApiController{public HttpResponseMessage GetFileResponse(string FilePath, string FileName = null, string ContentType = "application/octet-stream"){FileStream fs;HttpResponseMessage result;fs = new FileStream(FilePath, FileMode.Open, FileAccess.Read, FileShare.Read);result = new HttpResponseMessage(System.Net.HttpStatusCode.OK) { Content = new StreamContent(fs) };result.Content.Headers.ContentType = new System.Net.Http.Headers.MediaTypeHeaderValue(ContentType);result.Content.Headers.ContentDisposition = new System.Net.Http.Headers.ContentDispositionHeaderValue("attachment");result.Content.Headers.ContentDisposition.FileName = (null != FileName ? FileName : Path.GetFileName(FilePath));return result; ｝｝
｝

这种方式虽然可以实现文件下载，但是下载速度实在太慢(在局域网中下载速度只有50K)，稍大一点的文件就有可能导致客户端请求超时。

改为以下方式可实现高速下载的目的。在百兆局域网内实际测试可满速下载。

using System;
using System.Collections;
using System.Collections.Generic;
using System.IO;
using System.Linq;
using System.Net;
using System.Net.Http;
using System.Runtime.Remoting.Contexts;
using System.Text;
using System.Threading.Tasks;
using System.Web;
using System.Web.Http;
using System.Web.Http.Cors;
using System.Net.Http.Headers;namespace Service.Controller
{public partial class APIController : ApiController{#region 辅助函数public interface IWriteStreamToResponse{T Suorce { get; set; }void WriteToStream(Stream outputStream, HttpContent content, TransportContext context);}public class StreamFromFileName : IWriteStreamToResponse{private string _suorce;public string Suorce{get{return _suorce;}set{_suorce = value;}}public async void WriteToStream(Stream outputStream, HttpContent content, TransportContext context){try{var buffer = new byte[1024 * 1024 * 2];using (var video = File.Open(_suorce, FileMode.Open, FileAccess.Read, FileShare.Read)){var length = (int)video.Length;var bytesRead = 1;while (length > 0 && bytesRead > 0){bytesRead = video.Read(buffer, 0, Math.Min(length, buffer.Length));await outputStream.WriteAsync(buffer, 0, bytesRead);length -= bytesRead;}}}catch (Exception ex){return;}finally{outputStream.Close();}}}public class StreamFromBytes : IWriteStreamToResponse{private Stream _source;public Stream Suorce{get{return _source;}set{_source = value;}}public async void WriteToStream(Stream outputStream, HttpContent content, TransportContext context){try{var buffer = new byte[1024 * 1024 * 2];using (var video = _source){var length = (int)video.Length;var bytesRead = 1;while (length > 0 && bytesRead > 0){bytesRead = video.Read(buffer, 0, Math.Min(length, buffer.Length));await outputStream.WriteAsync(buffer, 0, bytesRead);length -= bytesRead;}}}catch (Exception ex){return;}finally{_source.Close();outputStream.Close();}}}#endregionpublic HttpResponseMessage GetFileResponse(string FilePath, string FileName = null, string ContentType = "application/octet-stream"){HttpResponseMessage response;IWriteStreamToResponse ResponseStreamWriter;Action sendMethod;ResponseStreamWriter = new StreamFromFileName() { Suorce = FilePath };sendMethod = ResponseStreamWriter.WriteToStream;response = Request.CreateResponse();response.Content = new System.Net.Http.PushStreamContent(sendMethod);response.Content.Headers.ContentType = new System.Net.Http.Headers.MediaTypeHeaderValue(ContentType);response.Content.Headers.ContentDisposition = new System.Net.Http.Headers.ContentDispositionHeaderValue("attachment");response.Content.Headers.ContentDisposition.FileName = (null != FileName ? FileName : Path.GetFileName(FilePath));return response;｝｝
｝

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