9

我正在尝试使用从http://valums.com/ajax-upload/找到的脚本

我的控制器如下

using System;
using System.IO;
using System.Text.RegularExpressions;
using System.Web;
using System.Web.Hosting;
using System.Web.Mvc;
using MHNHub.Areas.ViewModels;
using MHNHub.Models;
using MHNHub.ViewModels;

namespace MHNHub.Areas.Admin.Controllers
{
    [Authorize(Roles = "Administrator")]
    public class ImageController : Controller
    {

        private MHNHubEntities _entities = new MHNHubEntities();

        //
        // GET: /Image/
        [AcceptVerbs(HttpVerbs.Get)]
        public ActionResult ImageUploader()
        {
            var viewModel = new ImageViewModel()
            {
                Image = new Image()
            };

            return PartialView(viewModel);
        }

        [AcceptVerbs(HttpVerbs.Post)]
        public ActionResult ImageUploader(Image image)
        {
            try
            {
                _entities.Images.AddObject(image);
                _entities.SaveChanges();

                return RedirectToAction("Index", "Product");
            }
            catch (Exception ex)
            {
                var viewModel = new ImageViewModel()
                                {
                                    Image = image,
                                    HasError = true,
                                    ErrorMessage = ex.Message
                                };
                return PartialView(viewModel);

            }
        }

        private string _uploadsFolder = HostingEnvironment.MapPath("~/App_Data/Files");

        public Guid Upload(HttpPostedFileBase fileBase)
        {
            var identifier = Guid.NewGuid();
            fileBase.SaveAs(GetDiskLocation(identifier));
            return identifier;
        }

        private string GetDiskLocation(Guid identifier)
        {
            return Path.Combine(_uploadsFolder, identifier.ToString());
        }

    }

}

我有这样的部分观点

<%@ Control Language="C#" Inherits="System.Web.Mvc.ViewUserControl<MHNHub.ViewModels.ImageViewModel>" %>

<script type="text/javascript">
    $(function () {
        $("#imagedialog").dialog({
            bgiframe: true,
            height: 170,
            width: 430,
            modal: true,
            autoOpen: false,
            resizable: true
        })
    });

    $(document).ready(function createUploader() {
        var uploader = new qq.FileUploader({
            element: document.getElementById('fileuploader'),
            action: '/Image/Upload/',
            name: 'name'
        });

    });

</script>    

<div id="imagedialog" title="Upload Image">

                <div id="fileuploader">

                </div>
                <h6>Drag and drop files supported in Firefox and Google Chrome with javascript enabled.</h6> 
                    <noscript>
                         <form action="/image/upload" enctype="multipart/form-data" method="post">
                            Select a file: <input type="file" name="photo" id="photo" />   

                            <input type="submit" value="Upload" name="submit"/>
                        </form>
                    </noscript>


</div>

<div class="editor-field">
    <img src="<%: Model.Image.FileName %>" />
    <%: Html.TextBoxFor(model => model.Image.FileName) %>
    <%: Html.ValidationMessageFor(model => model.Image.FileName)%>
    <a href="#" onclick="jQuery('#imagedialog').dialog('open'); return false">Upload Image</a>
</div>

我在母版页上正确链接了 fileuploader.js 和 fileuploader.css，并且上传程序正确显示，它甚至调用了我的操作，但 HttpPostedFileBase 为空，上传操作引发异常。关于我应该做什么的任何见解？

编辑

所以我发现使用萤火虫它发送一个 XmlHttpRequest。如何在上传操作中处理此问题？

4

1 回答 1

9

您在控制器操作中获得空参数的原因是因为此插件不会向multipart/form-data服务器发送请求。相反，它发送application/octet-stream内容类型请求标头并将文件内容直接写入请求流，将参数附加?qqfile到包含文件名的 URL。因此，如果您想在控制器上检索它，您需要直接读取流：

[HttpPost]
public ActionResult Upload(string qqfile)
{
    using (var reader = new BinaryReader(Request.InputStream))
    {
        // This will contain the uploaded file data and the qqfile the name
        byte[] file = reader.ReadBytes((int)Request.InputStream.Length);
    }
    return View();
}

如果您选择多个文件，插件只需向服务器发送多个请求，这样就可以工作。

此外，如果您想要处理大于int.MaxValue您必须从请求流中读取块并直接写入输出流而不是将整个文件加载到内存缓冲区中的文件：

using (var outputStream = File.Create(qqfile))
{
    const int chunkSize = 2 * 1024; // 2KB
    byte[] buffer = new byte[chunkSize];
    int bytesRead;
    while ((bytesRead = Request.InputStream.Read(buffer, 0, buffer.Length)) > 0)
    {
        outputStream.Write(buffer, 0, bytesRead);
    }
}

备注：createUploader从您的document.ready. 它应该是一个匿名函数。您甚至可以将它与$(function() { ... });您已经设置的模态对话框合并。

于 2010-09-13T06:38:09.340 回答