Question

我被要求在C＃中执行以下操作：

/**

* 1. Create a MultipartPostMethod

* 2. Construct the web URL to connect to the SDP Server

* 3. Add the filename to be attached as a parameter to the MultipartPostMethod with parameter name "filename"

* 4. Execute the MultipartPostMethod

* 5. Receive and process the response as required

* /

我编写了一些没有错误的代码，但是没有附加文件。

有人可以查看我的C＃代码，看看我是否错误地编写了代码？

这是我的代码：

var client = new HttpClient();
const string weblinkUrl = "http://testserver.com/attach?";
var method = new MultipartFormDataContent();
const string fileName = "C:\file.txt";
var streamContent = new StreamContent(File.Open(fileName, FileMode.Open));
method.Add(streamContent, "filename");

var result = client.PostAsync(weblinkUrl, method);
MessageBox.Show(result.Result.ToString());

Answer 1

有人多次询问过这个问题。以下是一些可能的解决方案：

C＃HttpClient 4.5 multipart / form-data upload：C# HttpClient 4.5 multipart/form-data upload

C＃中的HttpClient Multipart Form Post：HttpClient Multipart Form Post in C#

在个人笔记中，检查请求中发送的帖子数据，并检查响应。 Fiddler非常适合这一点。

Answer 2

我知道这是一个老帖子但是对于那些寻求解决方案的人来说，提供一个更直接的答案，这就是我发现的：

using System.Diagnostics;
using System.Net;
using System.Net.Http;
using System.Threading.Tasks;
using System.Web;
using System.Web.Http;

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

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

        try
        {
            // Read the form data.
            await Request.Content.ReadAsMultipartAsync(provider);

            // This illustrates how to get the file names.
            foreach (MultipartFileData file in provider.FileData)
            {
                Trace.WriteLine(file.Headers.ContentDisposition.FileName);
                Trace.WriteLine("Server file path: " + file.LocalFileName);
            }
            return Request.CreateResponse(HttpStatusCode.OK);
        }
        catch (System.Exception e)
        {
            return Request.CreateErrorResponse(HttpStatusCode.InternalServerError, e);
        }
    }

}

这是我找到它的地方 http://www.asp.net/web-api/overview/advanced/sending-html-form-data,-part-2
为了更精细的实施 http://galratner.com/blogs/net/archive/2013/03/22/using-html-5-and-the-web-api-for-ajax-file-uploads-with-image-preview-and-a-progress-bar.aspx

Answer 3

在C＃中发布MultipartFormDataContent很简单，但第一次可能会令人困惑。这是在发布.png .txt等时适合我的代码。

// 2. Create the url 
string url = "https://myurl.com/api/...";
string filename = "myFile.png";
// In my case this is the JSON that will be returned from the post
string result = "";
// 1. Create a MultipartPostMethod
// "NKdKd9Yk" is the boundary parameter
using (var formContent = new MultipartFormDataContent("NKdKd9Yk")) {

    content.Headers.ContentType.MediaType = "multipart/form-data";
    // 3. Add the filename C:\\... + fileName is the path your file
    Stream fileStream = System.IO.File.OpenRead("C:\\Users\\username\\Pictures\\" + fileName);
    content.Add(new StreamContent(fileStream), fileName, fileName);

    using (var client = new HttpClient())
    {
        // Bearer Token header if needed
        client.DefaultRequestHeaders.Add("Authorization", "Bearer " + _bearerToken);
        client.DefaultRequestHeaders.Accept.Add(new MediaTypeWithQualityHeaderValue("multipart/form-data"));

        try
        {
            // 4.. Execute the MultipartPostMethod
            var message = await client.PostAsync(url, formContent);
            // 5.a Receive the response
            result = await message.Content.ReadAsStringAsync();                
        }
        catch (Exception ex)
        {
            // Do what you want if it fails.
            throw ex;
        }
    }    
}
// 5.b Process the reponse Get a usable object from the JSON that is returned
MyObject myObject = JsonConvert.DeserializeObject<MyObject>(result);

在我的情况下，我需要在发布后对对象做一些事情，所以我用JsonConvert将它转换为该对象。

Answer 4

我调试了这个问题就在这里：

method.Add(streamContent, "filename");

这个＆＃39;添加＆＃39;实际上并没有将文件放在Multipart Content的BODY中。

Http MultipartFormDataContent

4 个答案: