Asp.net mvc 4 我可以使用MVC4'实现文件下载吗;什么是ApiController?

Asp.net mvc 4 我可以使用MVC4'实现文件下载吗;什么是ApiController?,asp.net-mvc-4,Asp.net Mvc 4,使用常规控制器,我可以通过返回FileResult来完成。这似乎不适用于ApiController。能做到吗?这样做对吗?试试这个 [HttpGet] public HttpResponseMessage Get() { var file = HttpContext.Current.Server.MapPath("~/Images/accent.png"); var stream = new FileStream(file, FileMode.Op

使用常规控制器,我可以通过返回FileResult来完成。这似乎不适用于ApiController。能做到吗?这样做对吗?

试试这个

[HttpGet]    
public HttpResponseMessage Get()
    {
        var file = HttpContext.Current.Server.MapPath("~/Images/accent.png");
        var stream = new FileStream(file, FileMode.Open);
        var content = new HttpResponseMessage(HttpStatusCode.OK)
                         {
                             Content = new StreamContent(stream)
                         };
        content.Content.Headers.ContentType = new MediaTypeHeaderValue("application/octet-stream");
        return content;
    }

多亏了你,我才有这个工作


Web API用于RESTful Web服务。因此,是的,这是一个不错的主意。我们得到了以下信息:
请求的资源不支持http方法“get”。
在方法上方添加一个
HttpGet
属性(
[HttpGet]
)@蒂努普
using System.Net;
using System.Net.Http;
using System.Net.Http.Headers;
using System.Web.Http;

namespace Web.Controllers
{
    //usage: /download/report

    [RoutePrefix("download")]
    public class DownloadController : ApiController
    {
        [HttpGet("report")]
        public HttpResponseMessage Report()
        {
            using (var service = new Client())
            {
                var report = service.BuildReport();

                return DownloadResponse(report, "Report.csv");
            }
        }

        private static HttpResponseMessage DownloadResponse(string content, string fileName)
        {
            var downloadContent = new StringContent(content);

            var mediaType = new MediaTypeHeaderValue("application/octet-stream");
            var disposition= new ContentDispositionHeaderValue("attachment") { FileName = fileName };

            downloadContent.Headers.ContentType = mediaType;
            downloadContent.Headers.ContentDisposition = disposition;

            var result = new HttpResponseMessage(HttpStatusCode.OK)
            {
                Content = downloadContent
            };

            return result;
        }
    }
}