对于我来说，下面的作品(主要是受到以下所有回答的启发)，我从Elad的回答开始，修改/简化事情以符合我的需要(删除不是文件形式输入，只有一个文件，……)

希望它能帮助到一些人:)

(PS:我知道异常处理没有实现，并且假设它是在一个类中编写的，所以我可能需要一些集成工作…)

private void uploadFile()
    {
        Random rand = new Random();
        string boundary = "----boundary" + rand.Next().ToString();
        Stream data_stream;
        byte[] header = System.Text.Encoding.ASCII.GetBytes("\r\n--" + boundary + "\r\nContent-Disposition: form-data; name=\"file_path\"; filename=\"" + System.IO.Path.GetFileName(this.file) + "\"\r\nContent-Type: application/octet-stream\r\n\r\n");
        byte[] trailer = System.Text.Encoding.ASCII.GetBytes("\r\n--" + boundary + "--\r\n");

        // Do the request
        HttpWebRequest request = (HttpWebRequest)WebRequest.Create(MBF_URL);
        request.UserAgent = "My Toolbox";
        request.Method = "POST";
        request.KeepAlive = true;
        request.ContentType = "multipart/form-data; boundary=" + boundary;
        data_stream = request.GetRequestStream();
        data_stream.Write(header, 0, header.Length);
        byte[] file_bytes = System.IO.File.ReadAllBytes(this.file);
        data_stream.Write(file_bytes, 0, file_bytes.Length);
        data_stream.Write(trailer, 0, trailer.Length);
        data_stream.Close();

        // Read the response
        WebResponse response = request.GetResponse();
        data_stream = response.GetResponseStream();
        StreamReader reader = new StreamReader(data_stream);
        this.url = reader.ReadToEnd();

        if (this.url == "") { this.url = "No response :("; }

        reader.Close();
        data_stream.Close();
        response.Close();
    }

我的ASP。NET上传常见问题解答中有一篇关于这方面的文章，有示例代码:使用HttpWebRequest/WebClient的RFC 1867 POST请求上传文件。此代码不将文件加载到内存中(与上面的代码相反)，支持多个文件，并支持表单值、设置凭据和cookie等。

编辑:看起来好像是Axosoft把这个页面删除了。谢谢你的家伙。

它仍然可以通过archive.org访问。

采用上面的代码并修复，因为它抛出内部服务器错误500。\r\n的位置和空格等存在一些问题。应用内存流重构，直接写入请求流。结果如下:

    public static void HttpUploadFile(string url, string file, string paramName, string contentType, NameValueCollection nvc) {
        log.Debug(string.Format("Uploading {0} to {1}", file, url));
        string boundary = "---------------------------" + DateTime.Now.Ticks.ToString("x");
        byte[] boundarybytes = System.Text.Encoding.ASCII.GetBytes("\r\n--" + boundary + "\r\n");

        HttpWebRequest wr = (HttpWebRequest)WebRequest.Create(url);
        wr.ContentType = "multipart/form-data; boundary=" + boundary;
        wr.Method = "POST";
        wr.KeepAlive = true;
        wr.Credentials = System.Net.CredentialCache.DefaultCredentials;

        Stream rs = wr.GetRequestStream();

        string formdataTemplate = "Content-Disposition: form-data; name=\"{0}\"\r\n\r\n{1}";
        foreach (string key in nvc.Keys)
        {
            rs.Write(boundarybytes, 0, boundarybytes.Length);
            string formitem = string.Format(formdataTemplate, key, nvc[key]);
            byte[] formitembytes = System.Text.Encoding.UTF8.GetBytes(formitem);
            rs.Write(formitembytes, 0, formitembytes.Length);
        }
        rs.Write(boundarybytes, 0, boundarybytes.Length);

        string headerTemplate = "Content-Disposition: form-data; name=\"{0}\"; filename=\"{1}\"\r\nContent-Type: {2}\r\n\r\n";
        string header = string.Format(headerTemplate, paramName, file, contentType);
        byte[] headerbytes = System.Text.Encoding.UTF8.GetBytes(header);
        rs.Write(headerbytes, 0, headerbytes.Length);

        FileStream fileStream = new FileStream(file, FileMode.Open, FileAccess.Read);
        byte[] buffer = new byte[4096];
        int bytesRead = 0;
        while ((bytesRead = fileStream.Read(buffer, 0, buffer.Length)) != 0) {
            rs.Write(buffer, 0, bytesRead);
        }
        fileStream.Close();

        byte[] trailer = System.Text.Encoding.ASCII.GetBytes("\r\n--" + boundary + "--\r\n");
        rs.Write(trailer, 0, trailer.Length);
        rs.Close();

        WebResponse wresp = null;
        try {
            wresp = wr.GetResponse();
            Stream stream2 = wresp.GetResponseStream();
            StreamReader reader2 = new StreamReader(stream2);
            log.Debug(string.Format("File uploaded, server response is: {0}", reader2.ReadToEnd()));
        } catch(Exception ex) {
            log.Error("Error uploading file", ex);
            if(wresp != null) {
                wresp.Close();
                wresp = null;
            }
        } finally {
            wr = null;
        }
    }

以及示例用法:

    NameValueCollection nvc = new NameValueCollection();
    nvc.Add("id", "TTR");
    nvc.Add("btn-submit-photo", "Upload");
    HttpUploadFile("http://your.server.com/upload", 
         @"C:\test\test.jpg", "file", "image/jpeg", nvc);

可以扩展它来处理多个文件，或者对每个文件多次调用它。但它适合你的需要。

查看MyToolkit库:

var request = new HttpPostRequest("http://www.server.com");
request.Data.Add("name", "value"); // POST data
request.Files.Add(new HttpPostFile("name", "file.jpg", "path/to/file.jpg")); 

await Http.PostAsync(request, OnRequestFinished);

http://mytoolkit.codeplex.com/wikipage?title=Http

我永远不能让例子正常工作，我总是收到一个500错误时，把它发送到服务器。

然而，我在这个url中遇到了一个非常优雅的方法

它很容易扩展，显然可以处理二进制文件和XML。

你可以用类似的方法来称呼它

class Program
{
    public static string gsaFeedURL = "http://yourGSA.domain.com:19900/xmlfeed";

    static void Main()
    {
        try
        {
            postWebData();
        }
        catch (Exception ex)
        {
        }
    }

    // new one I made from C# web service
    public static void postWebData()
    {
        StringDictionary dictionary = new StringDictionary();
        UploadSpec uploadSpecs = new UploadSpec();
        UTF8Encoding encoding = new UTF8Encoding();
        byte[] bytes;
        Uri gsaURI = new Uri(gsaFeedURL);  // Create new URI to GSA feeder gate
        string sourceURL = @"C:\FeedFile.xml"; // Location of the XML feed file
        // Two parameters to send
        string feedtype = "full";
        string datasource = "test";            

        try
        {
            // Add the parameter values to the dictionary
            dictionary.Add("feedtype", feedtype);
            dictionary.Add("datasource", datasource);

            // Load the feed file created and get its bytes
            XmlDocument xml = new XmlDocument();
            xml.Load(sourceURL);
            bytes = Encoding.UTF8.GetBytes(xml.OuterXml);

            // Add data to upload specs
            uploadSpecs.Contents = bytes;
            uploadSpecs.FileName = sourceURL;
            uploadSpecs.FieldName = "data";

            // Post the data
            if ((int)HttpUpload.Upload(gsaURI, dictionary, uploadSpecs).StatusCode == 200)
            {
                Console.WriteLine("Successful.");
            }
            else
            {
                // GSA POST not successful
                Console.WriteLine("Failure.");
            }
        }
        catch (Exception ex)
        {
            Console.WriteLine(ex.Message);
        }
    }
}

类似这样的代码很接近:(未测试的代码)

byte[] data; // data goes here.

HttpWebRequest request = (HttpWebRequest)WebRequest.Create(url);
request.Credentials = userNetworkCredentials;
request.Method = "PUT";
request.ContentType = "application/octet-stream";
request.ContentLength = data.Length;
Stream stream = request.GetRequestStream();
stream.Write(data,0,data.Length);
stream.Close();
response = (HttpWebResponse)request.GetResponse();
StreamReader reader = new StreamReader(response.GetResponseStream());
temp = reader.ReadToEnd();
reader.Close();