Question

我有一个看起来像这样的C＃控制器

 public class UploadController : ApiController
{
    public async Task<HttpResponseMessage> Post()
    {
        // Check if the request contains multipart/form-data. 
        if (!Request.Content.IsMimeMultipartContent())
        {
            throw new HttpResponseException(HttpStatusCode.UnsupportedMediaType);
        }

        string root = HttpContext.Current.Server.MapPath("~/Uploads");
        var provider = new MultipartFormDataStreamProvider(root);

        try
        {
            StringBuilder sb = new StringBuilder(); // Holds the response body 

            // Read the form data and return an async task. 
            await Request.Content.ReadAsMultipartAsync(provider);

            // This illustrates how to get the form data. 
            foreach (var key in provider.FormData.AllKeys)
            {
                foreach (var val in provider.FormData.GetValues(key))
                {
                    sb.Append(string.Format("{0}: {1}\n", key, val));
                }
            }

            // This illustrates how to get the file names for uploaded files. 
            foreach (var file in provider.FileData)
            {
                FileInfo fileInfo = new FileInfo(file.LocalFileName);
                sb.Append(string.Format("Uploaded file: {0} ({1} bytes)\n", fileInfo.Name, fileInfo.Length));
            }
            return new HttpResponseMessage()
            {
                Content = new StringContent(sb.ToString())
            };
        }
        catch (System.Exception e)
        {
            return Request.CreateErrorResponse(HttpStatusCode.InternalServerError, e);
        }
    }


}

以及看起来像这样的Upload.html页面

**<form name="form1" method="post" enctype="multipart/form-data" action="api/Upload">
<div>
    <label for="caption">Image Caption</label>
    <input name="caption" type="text" />
</div>
<div>
    <label for="image1">Image File</label>
    <input name="image1" type="file" />
</div>
<div>
    <input ng-click="submit()" type="submit" value="Submit" />
</div>

**

它允许我选择文件点击上传，并将其保存到本地文件夹。但是当上传时，文件被称为“BodyPart_175b76cb-88f2-4ea4-bbc5-fd3038345e5e”，它的类型为file。我希望我的应用程序做的是上传文件并保留其名称和文件类型。这是这样做的正确方法吗？

修改

所以我发现文件实际上正在上传，如果我添加正确的扩展名，我可以打开它们。任何人都知道如何在上传后保留其文件名？

Answer 1

想出来

重载了一个MIME对象，返回正确的名称

    using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Net.Http;


namespace AngularFun.Controllers
{
    class MyMultipartFormDataStreamProvider : MultipartFormDataStreamProvider
    {
        public MyMultipartFormDataStreamProvider(string path)
            : base(path)
        {

        }

        public override string GetLocalFileName(System.Net.Http.Headers.HttpContentHeaders headers)
        {
            string fileName;
            if (!string.IsNullOrWhiteSpace(headers.ContentDisposition.FileName))
            {
                fileName = headers.ContentDisposition.FileName;
            }
            else
            {
                fileName = Guid.NewGuid().ToString() + ".data";
            }
            return fileName.Replace("\"", string.Empty);
        }
    }
}

正在上传Mime Multipart

修改

1 个答案: