来自 C# 客户端的多部分表单

Multipart forms from C# client

提问by bash74

I am trying to fill a form in a php application from a C# client (Outlook addin). I used Fiddler to see the original request from within the php application and the form is transmitted as a multipart/form. Unfortunately .Net does not come with native support for this type of forms (WebClient has only a method for uploading a file). Does anybody know a library or has some code to achieve this? I want to post different values and additionally (but only sometimes) a file.

我正在尝试从 C# 客户端（Outlook 插件）填写 php 应用程序中的表单。我使用 Fiddler 从 php 应用程序中查看原始请求，并且表单作为多部分/表单传输。不幸的是，.Net 没有原生支持这种类型的表单（WebClient 只有一种上传文件的方法）。有人知道图书馆或有一些代码来实现这一目标吗？我想发布不同的值，另外（但只是有时）一个文件。

Thanks for your help, Sebastian

感谢您的帮助，塞巴斯蒂安

采纳答案by dnolan

This is cut and pasted from some sample code I wrote, hopefully it should give the basics. It only supports File data and form-data at the moment.

这是从我写的一些示例代码中剪切和粘贴的，希望它应该提供基础知识。目前只支持文件数据和表单数据。

public class PostData
{

    private List<PostDataParam> m_Params;

    public List<PostDataParam> Params
    {
        get { return m_Params; }
        set { m_Params = value; }
    }

    public PostData()
    {
        m_Params = new List<PostDataParam>();

        // Add sample param
        m_Params.Add(new PostDataParam("email", "MyEmail", PostDataParamType.Field));
    }


    /// <summary>
    /// Returns the parameters array formatted for multi-part/form data
    /// </summary>
    /// <returns></returns>
    public string GetPostData()
    {
        // Get boundary, default is --AaB03x
        string boundary = ConfigurationManager.AppSettings["ContentBoundary"].ToString();

        StringBuilder sb = new StringBuilder();
        foreach (PostDataParam p in m_Params)
        {
            sb.AppendLine(boundary);

            if (p.Type == PostDataParamType.File)
            {
                sb.AppendLine(string.Format("Content-Disposition: file; name=\"{0}\"; filename=\"{1}\"", p.Name, p.FileName));
                sb.AppendLine("Content-Type: text/plain");
                sb.AppendLine();
                sb.AppendLine(p.Value);                 
            }
            else
            {
                sb.AppendLine(string.Format("Content-Disposition: form-data; name=\"{0}\"", p.Name));
                sb.AppendLine();
                sb.AppendLine(p.Value);
            }
        }

        sb.AppendLine(boundary);

        return sb.ToString();           
    }
}

public enum PostDataParamType
{
    Field,
    File
}

public class PostDataParam
{


    public PostDataParam(string name, string value, PostDataParamType type)
    {
        Name = name;
        Value = value;
        Type = type;
    }

    public string Name;
    public string FileName;
    public string Value;
    public PostDataParamType Type;
}

To send the data you then need to:

要发送数据，您需要：

HttpWebRequest oRequest = null;
oRequest = (HttpWebRequest)HttpWebRequest.Create(oURL.URL);
oRequest.ContentType = "multipart/form-data";                       
oRequest.Method = "POST";
PostData pData = new PostData();

byte[] buffer = encoding.GetBytes(pData.GetPostData());

// Set content length of our data
oRequest.ContentLength = buffer.Length;

// Dump our buffered postdata to the stream, booyah
oStream = oRequest.GetRequestStream();
oStream.Write(buffer, 0, buffer.Length);
oStream.Close();

// get the response
oResponse = (HttpWebResponse)oRequest.GetResponse();

Hope thats clear, i've cut and pasted from a few sources to get that tidier.

希望那很清楚，我已经从几个来源剪切和粘贴，以使内容更整洁。

回答by jumoel

Building on dnolans example, this is the version I could actually get to work (there were some errors with the boundary, encoding wasn't set) :-)

基于 dnlans 示例，这是我实际可以开始工作的版本（边界有一些错误，未设置编码）:-)

To send the data:

发送数据：

HttpWebRequest oRequest = null;
oRequest = (HttpWebRequest)HttpWebRequest.Create("http://you.url.here");
oRequest.ContentType = "multipart/form-data; boundary=" + PostData.boundary;
oRequest.Method = "POST";
PostData pData = new PostData();
Encoding encoding = Encoding.UTF8;
Stream oStream = null;

/* ... set the parameters, read files, etc. IE:
   pData.Params.Add(new PostDataParam("email", "[email protected]", PostDataParamType.Field));
   pData.Params.Add(new PostDataParam("fileupload", "filename.txt", "filecontents" PostDataParamType.File));
*/

byte[] buffer = encoding.GetBytes(pData.GetPostData());

oRequest.ContentLength = buffer.Length;

oStream = oRequest.GetRequestStream();
oStream.Write(buffer, 0, buffer.Length);
oStream.Close();

HttpWebResponse oResponse = (HttpWebResponse)oRequest.GetResponse();

The PostData class should look like:

PostData 类应如下所示：

public class PostData
{
    // Change this if you need to, not necessary
    public static string boundary = "AaB03x";

    private List<PostDataParam> m_Params;

    public List<PostDataParam> Params
    {
        get { return m_Params; }
        set { m_Params = value; }
    }

    public PostData()
    {
        m_Params = new List<PostDataParam>();
    }

    /// <summary>
    /// Returns the parameters array formatted for multi-part/form data
    /// </summary>
    /// <returns></returns>
    public string GetPostData()
    {
        StringBuilder sb = new StringBuilder();
        foreach (PostDataParam p in m_Params)
        {
            sb.AppendLine("--" + boundary);

            if (p.Type == PostDataParamType.File)
            {
                sb.AppendLine(string.Format("Content-Disposition: file; name=\"{0}\"; filename=\"{1}\"", p.Name, p.FileName));
                sb.AppendLine("Content-Type: application/octet-stream");
                sb.AppendLine();
                sb.AppendLine(p.Value);
            }
            else
            {
                sb.AppendLine(string.Format("Content-Disposition: form-data; name=\"{0}\"", p.Name));
                sb.AppendLine();
                sb.AppendLine(p.Value);
            }
        }

        sb.AppendLine("--" + boundary + "--");

        return sb.ToString();
    }
}

public enum PostDataParamType
{
    Field,
    File
}

public class PostDataParam
{
    public PostDataParam(string name, string value, PostDataParamType type)
    {
        Name = name;
        Value = value;
        Type = type;
    }

    public PostDataParam(string name, string filename, string value, PostDataParamType type)
    {
        Name = name;
        Value = value;
        FileName = filename;
        Type = type;
    }

    public string Name;
    public string FileName;
    public string Value;
    public PostDataParamType Type;
}

回答by Anthony

I needed to simulate a browser login to a website to get a login cookie, and the login form was multipart/form-data.

我需要模拟浏览器登录网站来获取登录cookie，登录表单是multipart/form-data。

I took some clues from the other answers here, and then tried to get my own scenario working. It took a bit of frustrating trial and error before it worked right, but here is the code:

我从这里的其他答案中获取了一些线索，然后尝试让我自己的场景发挥作用。在它正常工作之前进行了一些令人沮丧的反复试验，但这里是代码：

    public static class WebHelpers
    {
        /// <summary>
        /// Post the data as a multipart form
        /// </summary>
       public static HttpWebResponse MultipartFormDataPost(string postUrl, string userAgent, Dictionary<string, string> values)
       {
           string formDataBoundary = "---------------------------" + WebHelpers.RandomHexDigits(12);
           string contentType = "multipart/form-data; boundary=" + formDataBoundary;

           string formData = WebHelpers.MakeMultipartForm(values, formDataBoundary);
           return WebHelpers.PostForm(postUrl, userAgent, contentType, formData);
       }

        /// <summary>
        /// Post a form
        /// </summary>
        public static HttpWebResponse PostForm(string postUrl, string userAgent, string contentType, string formData)
        {
            HttpWebRequest request = WebRequest.Create(postUrl) as HttpWebRequest;

            if (request == null)
            {
                throw new NullReferenceException("request is not a http request");
            }

            // Add these, as we're doing a POST
            request.Method = "POST";
            request.ContentType = contentType;
            request.UserAgent = userAgent;
            request.CookieContainer = new CookieContainer();

            // We need to count how many bytes we're sending. 
            byte[] postBytes = Encoding.UTF8.GetBytes(formData);
            request.ContentLength = postBytes.Length;

            using (Stream requestStream = request.GetRequestStream())
            {
                // Push it out there
                requestStream.Write(postBytes, 0, postBytes.Length);
                requestStream.Close();
            }

            return request.GetResponse() as HttpWebResponse;
        }

        /// <summary>
        /// Generate random hex digits 
        /// </summary>
        public static string RandomHexDigits(int count)
        {
            Random random = new Random();
            StringBuilder result = new StringBuilder();
            for (int i = 0; i < count; i++)
            {
                int digit = random.Next(16);
                result.AppendFormat("{0:x}", digit);
            }

            return result.ToString();
        }

        /// <summary>
        /// Turn the key and value pairs into a multipart form
        /// </summary>
        private static string MakeMultipartForm(Dictionary<string, string> values, string boundary)
        {
            StringBuilder sb = new StringBuilder();

            foreach (var pair in values)
            {
                sb.AppendFormat("--{0}\r\nContent-Disposition: form-data; name=\"{1}\"\r\n\r\n{2}\r\n", boundary, pair.Key, pair.Value);
            }

            sb.AppendFormat("--{0}--\r\n", boundary);

            return sb.ToString();    
        }
    }
}

It doesn't handle file data, just form since that's all that I needed. I called like this:

它不处理文件数据，只处理表单，因为这就是我所需要的。我是这样打电话的：

    try
    {
        using (HttpWebResponse response = WebHelpers.MultipartFormDataPost(postUrl, UserAgentString, this.loginForm)) 
        {
            if (response != null)
            {
                Cookie loginCookie = response.Cookies["logincookie"];
                .....

回答by Brian Grinstead

Thanks for the answers, everybody! I recently had to get this to work, and used your suggestions heavily. However, there were a couple of tricky parts that did not work as expected, mostly having to do with actually including the file (which was an important part of the question). There are a lot of answers here already, but I think this may be useful to someone in the future (I could not find many clear examples of this online). I wrote a blog postthat explains it a little more.

谢谢大家的解答！我最近不得不让它工作，并大量使用你的建议。但是，有几个棘手的部分没有按预期工作，主要与实际包含文件有关（这是问题的重要部分）。这里已经有很多答案了，但我认为这可能对将来的某个人有用（我在网上找不到很多明确的例子）。我写了一篇博文，对它进行了更多的解释。

Basically, I first tried to pass in the file data as a UTF8 encoded string, but I was having problems with encoding files (it worked fine for a plain text file, but when uploading a Word Document, for example, if I tried to save the file that was passed through to the posted form using Request.Files[0].SaveAs(), opening the file in Word did not work properly. I found that if you write the file data directly using a Stream (rather than a StringBuilder), it worked as expected. Also, I made a couple of modifications that made it easier for me to understand.

基本上，我首先尝试将文件数据作为 UTF8 编码的字符串传递，但是我在编码文件时遇到了问题（它对于纯文本文件工作正常，但是在上传 Word 文档时，例如，如果我尝试保存使用 Request.Files[0].SaveAs() 传递到已发布表单的文件，在 Word 中打开文件无法正常工作。我发现如果您直接使用 Stream（而不是 StringBuilder）写入文件数据)，它按预期工作。另外，我做了一些修改，使我更容易理解。

By the way, the Multipart Forms Request for Commentsand the W3C Recommendation for mulitpart/form-dataare a couple of useful resources in case anyone needs a reference for the specification.

顺便说一下，Multipart Forms Request for Comments和W3C 对 multipart/form-data 的建议是一些有用的资源，以防任何人需要规范的参考。

I changed the WebHelpers class to be a bit smaller and have simpler interfaces, it is now called FormUpload. If you pass a FormUpload.FileParameteryou can pass the byte[] contents along with a file name and content type, and if you pass a string, it will treat it as a standard name/value combination.

我将 WebHelpers 类更改为更小且具有更简单的接口，现在称为FormUpload. 如果您传递 aFormUpload.FileParameter您可以传递 byte[] 内容以及文件名和内容类型，如果您传递一个字符串，它会将其视为标准名称/值组合。

Here is the FormUpload class:

这是 FormUpload 类：

// Implements multipart/form-data POST in C# http://www.ietf.org/rfc/rfc2388.txt
// http://www.briangrinstead.com/blog/multipart-form-post-in-c
public static class FormUpload
{
    private static readonly Encoding encoding = Encoding.UTF8;
    public static HttpWebResponse MultipartFormDataPost(string postUrl, string userAgent, Dictionary<string, object> postParameters)
    {
        string formDataBoundary = String.Format("----------{0:N}", Guid.NewGuid());
        string contentType = "multipart/form-data; boundary=" + formDataBoundary;

        byte[] formData = GetMultipartFormData(postParameters, formDataBoundary);

        return PostForm(postUrl, userAgent, contentType, formData);
    }
    private static HttpWebResponse PostForm(string postUrl, string userAgent, string contentType, byte[] formData)
    {
        HttpWebRequest request = WebRequest.Create(postUrl) as HttpWebRequest;

        if (request == null)
        {
            throw new NullReferenceException("request is not a http request");
        }

        // Set up the request properties.
        request.Method = "POST";
        request.ContentType = contentType;
        request.UserAgent = userAgent;
        request.CookieContainer = new CookieContainer();
        request.ContentLength = formData.Length;

        // You could add authentication here as well if needed:
        // request.PreAuthenticate = true;
        // request.AuthenticationLevel = System.Net.Security.AuthenticationLevel.MutualAuthRequested;
        // request.Headers.Add("Authorization", "Basic " + Convert.ToBase64String(System.Text.Encoding.Default.GetBytes("username" + ":" + "password")));

        // Send the form data to the request.
        using (Stream requestStream = request.GetRequestStream())
        {
            requestStream.Write(formData, 0, formData.Length);
            requestStream.Close();
        }

        return request.GetResponse() as HttpWebResponse;
    }

    private static byte[] GetMultipartFormData(Dictionary<string, object> postParameters, string boundary)
    {
        Stream formDataStream = new System.IO.MemoryStream();
        bool needsCLRF = false;

        foreach (var param in postParameters)
        {
            // Thanks to feedback from commenters, add a CRLF to allow multiple parameters to be added.
            // Skip it on the first parameter, add it to subsequent parameters.
            if (needsCLRF)
                formDataStream.Write(encoding.GetBytes("\r\n"), 0, encoding.GetByteCount("\r\n"));

            needsCLRF = true;

            if (param.Value is FileParameter)
            {
                FileParameter fileToUpload = (FileParameter)param.Value;

                // Add just the first part of this param, since we will write the file data directly to the Stream
                string header = string.Format("--{0}\r\nContent-Disposition: form-data; name=\"{1}\"; filename=\"{2}\";\r\nContent-Type: {3}\r\n\r\n",
                    boundary,
                    param.Key,
                    fileToUpload.FileName ?? param.Key,
                    fileToUpload.ContentType ?? "application/octet-stream");

                formDataStream.Write(encoding.GetBytes(header), 0, encoding.GetByteCount(header));

                // Write the file data directly to the Stream, rather than serializing it to a string.
                formDataStream.Write(fileToUpload.File, 0, fileToUpload.File.Length);
            }
            else
            {
                string postData = string.Format("--{0}\r\nContent-Disposition: form-data; name=\"{1}\"\r\n\r\n{2}",
                    boundary,
                    param.Key,
                    param.Value);
                formDataStream.Write(encoding.GetBytes(postData), 0, encoding.GetByteCount(postData));
            }
        }

        // Add the end of the request.  Start with a newline
        string footer = "\r\n--" + boundary + "--\r\n";
        formDataStream.Write(encoding.GetBytes(footer), 0, encoding.GetByteCount(footer));

        // Dump the Stream into a byte[]
        formDataStream.Position = 0;
        byte[] formData = new byte[formDataStream.Length];
        formDataStream.Read(formData, 0, formData.Length);
        formDataStream.Close();

        return formData;
    }

    public class FileParameter
    {
        public byte[] File { get; set; }
        public string FileName { get; set; }
        public string ContentType { get; set; }
        public FileParameter(byte[] file) : this(file, null) { }
        public FileParameter(byte[] file, string filename) : this(file, filename, null) { }
        public FileParameter(byte[] file, string filename, string contenttype)
        {
            File = file;
            FileName = filename;
            ContentType = contenttype;
        }
    }
}

Here is the calling code, which uploads a file and a few normal post parameters:

这是调用代码，它上传一个文件和一些正常的 post 参数：

// Read file data
FileStream fs = new FileStream("c:\people.doc", FileMode.Open, FileAccess.Read);
byte[] data = new byte[fs.Length];
fs.Read(data, 0, data.Length);
fs.Close();

// Generate post objects
Dictionary<string, object> postParameters = new Dictionary<string, object>();
postParameters.Add("filename", "People.doc");
postParameters.Add("fileformat", "doc");
postParameters.Add("file", new FormUpload.FileParameter(data, "People.doc", "application/msword"));

// Create request and receive response
string postURL = "http://localhost";
string userAgent = "Someone";
HttpWebResponse webResponse = FormUpload.MultipartFormDataPost(postURL, userAgent, postParameters);

// Process response
StreamReader responseReader = new StreamReader(webResponse.GetResponseStream());
string fullResponse = responseReader.ReadToEnd();
webResponse.Close();
Response.Write(fullResponse);

回答by Brian Grinstead

Below is the code which I'm using

下面是我正在使用的代码

    //This URL not exist, it's only an example.
    string url = "http://myBox.s3.amazonaws.com/";
    //Instantiate new CustomWebRequest class
    CustomWebRequest wr = new CustomWebRequest(url);
    //Set values for parameters
    wr.ParamsCollection.Add(new ParamsStruct("key", "${filename}"));
    wr.ParamsCollection.Add(new ParamsStruct("acl", "public-read"));
    wr.ParamsCollection.Add(new ParamsStruct("success_action_redirect", "http://www.yahoo.com"));
    wr.ParamsCollection.Add(new ParamsStruct("x-amz-meta-uuid", "14365123651274"));
    wr.ParamsCollection.Add(new ParamsStruct("x-amz-meta-tag", ""));
    wr.ParamsCollection.Add(new ParamsStruct("AWSAccessKeyId", "zzzz"));            
    wr.ParamsCollection.Add(new ParamsStruct("Policy", "adsfadsf"));
    wr.ParamsCollection.Add(new ParamsStruct("Signature", "hH6lK6cA="));
    //For file type, send the inputstream of selected file
    StreamReader sr = new StreamReader(@"file.txt");
    wr.ParamsCollection.Add(new ParamsStruct("file", sr, ParamsStruct.ParamType.File, "file.txt"));

    wr.PostData();

from the following link I've downloaded the same code http://www.codeproject.com/KB/cs/multipart_request_C_.aspx

从以下链接我下载了相同的代码 http://www.codeproject.com/KB/cs/multipart_request_C_.aspx

Any Help

任何帮助

回答by eeeeaaii

In the version of .NET I am using you also have to do this:

在我使用的 .NET 版本中，您还必须执行以下操作：

System.Net.ServicePointManager.Expect100Continue = false;

If you don't, the HttpWebRequestclass will automatically add the Expect:100-continuerequest header which fouls everything up.

如果不这样做，HttpWebRequest该类将自动添加使Expect:100-continue所有内容都变得混乱的请求标头。

Also I learned the hard way that you have to have the right number of dashes. whatever you say is the "boundary" in the Content-Typeheader has to be preceded by two dashes

我也学到了你必须拥有正确数量的破折号的艰难方法。无论你说什么是Content-Type标题中的“边界”都必须以两个破折号开头

--THEBOUNDARY

and at the end

最后

--THEBOUNDARY--

exactly as it does in the example code. If your boundary is a lot of dashes followed by a number then this mistake won't be obvious by looking at the http request in a proxy server

与示例代码中的完全一样。如果您的边界是很多破折号后跟一个数字，那么通过查看代理服务器中的 http 请求，这个错误不会很明显

回答by TheQult

A little optimization of the class before. In this version the files are not totally loaded into memory.

之前对类的一点优化。在这个版本中，文件没有完全加载到内存中。

Security advice: a check for the boundary is missing, if the file contains the bounday it will crash.

安全建议：缺少边界检查，如果文件包含边界，它将崩溃。

namespace WindowsFormsApplication1
{
    public static class FormUpload
    {
        private static string NewDataBoundary()
        {
            Random rnd = new Random();
            string formDataBoundary = "";
            while (formDataBoundary.Length < 15)
            {
                formDataBoundary = formDataBoundary + rnd.Next();
            }
            formDataBoundary = formDataBoundary.Substring(0, 15);
            formDataBoundary = "-----------------------------" + formDataBoundary;
            return formDataBoundary;
        }

        public static HttpWebResponse MultipartFormDataPost(string postUrl, IEnumerable<Cookie> cookies, Dictionary<string, string> postParameters)
        {
            string boundary = NewDataBoundary();

            HttpWebRequest request = (HttpWebRequest)WebRequest.Create(postUrl);

            // Set up the request properties
            request.Method = "POST";
            request.ContentType = "multipart/form-data; boundary=" + boundary;
            request.UserAgent = "PhasDocAgent 1.0";
            request.CookieContainer = new CookieContainer();

            foreach (var cookie in cookies)
            {
                request.CookieContainer.Add(cookie);
            }

            #region WRITING STREAM
            using (Stream formDataStream = request.GetRequestStream())
            {
                foreach (var param in postParameters)
                {
                    if (param.Value.StartsWith("file://"))
                    {
                        string filepath = param.Value.Substring(7);

                        // Add just the first part of this param, since we will write the file data directly to the Stream
                        string header = string.Format("--{0}\r\nContent-Disposition: form-data; name=\"{1}\"; filename=\"{2}\";\r\nContent-Type: {3}\r\n\r\n",
                            boundary,
                            param.Key,
                            Path.GetFileName(filepath) ?? param.Key,
                            MimeTypes.GetMime(filepath));

                        formDataStream.Write(Encoding.UTF8.GetBytes(header), 0, header.Length);

                        // Write the file data directly to the Stream, rather than serializing it to a string.

                        byte[] buffer = new byte[2048];

                        FileStream fs = new FileStream(filepath, FileMode.Open);

                        for (int i = 0; i < fs.Length; )
                        {
                            int k = fs.Read(buffer, 0, buffer.Length);
                            if (k > 0)
                            {
                                formDataStream.Write(buffer, 0, k);
                            }
                            i = i + k;
                        }
                        fs.Close();
                    }
                    else
                    {
                        string postData = string.Format("--{0}\r\nContent-Disposition: form-data; name=\"{1}\"\r\n\r\n{2}\r\n",
                            boundary,
                            param.Key,
                            param.Value);
                        formDataStream.Write(Encoding.UTF8.GetBytes(postData), 0, postData.Length);
                    }
                }
                // Add the end of the request
                byte[] footer = Encoding.UTF8.GetBytes("\r\n--" + boundary + "--\r\n");
                formDataStream.Write(footer, 0, footer.Length);
                request.ContentLength = formDataStream.Length;
                formDataStream.Close();
            }
            #endregion

            return request.GetResponse() as HttpWebResponse;
        }
    }
}

回答by Luis Domingues

Thanks for the code, it saved me a lot of time (including the Except100 error!).

感谢您的代码，它为我节省了很多时间（包括 Except100 错误！）。

Anyway, I found a bug in the code, here:

无论如何，我在代码中发现了一个错误，在这里：

formDataStream.Write(encoding.GetBytes(postData), 0, postData.Length);

In case your POST data is utf-16, postData.Length, will return the number of characters and not the number of bytes. This will truncate the data being posted (for example, if you have 2 chars that are encoded as utf-16, they take 4 bytes, but postData.Length will say it takes 2 bytes, and you loose the 2 final bytes of the posted data).

如果您的 POST 数据是 utf-16，则 postData.Length 将返回字符数而不是字节数。这将截断发布的数据（例如，如果您有 2 个编码为 utf-16 的字符，它们占用 4 个字节，但 postData.Length 会说它占用 2 个字节，并且您丢失了已发布的最后 2 个字节数据）。

Solution - replace that line with:

解决方案 - 将该行替换为：

byte[] aPostData=encoding.GetBytes(postData);
formDataStream.Write(aPostData, 0, aPostData.Length);

Using this, the length is calculated by the size of the byte[], not the string size.

使用这个，长度是由字节[]的大小计算的，而不是字符串大小。

回答by Yavanosta

My implementation

我的实现

/// <summary>
/// Sending file via multipart\form-data
/// </summary>
/// <param name="url">URL for send</param>
/// <param name="file">Local file path</param>
/// <param name="paramName">Request file param</param>
/// <param name="contentType">Content-Type file headr</param>
/// <param name="nvc">Additional post params</param>
private static string httpUploadFile(string url, string file, string paramName, string contentType, NameValueCollection nvc)
{
    //delimeter
    var boundary = "---------------------------" + DateTime.Now.Ticks.ToString("x");

    //creating request
    var wr = (HttpWebRequest)WebRequest.Create(url);
    wr.ContentType = "multipart/form-data; boundary=" + boundary;
    wr.Method = "POST";
    wr.KeepAlive = true;

    //sending request
    using(var requestStream = wr.GetRequestStream())
    {
        using (var requestWriter = new StreamWriter(requestStream, Encoding.UTF8))
        {
            //params
            const string formdataTemplate = "Content-Disposition: form-data; name=\"{0}\"\r\n\r\n{1}";
            foreach (string key in nvc.Keys)
            {
                requestWriter.Write(boundary);
                requestWriter.Write(String.Format(formdataTemplate, key, nvc[key]));
            }
            requestWriter.Write(boundary);

            //file header
            const string headerTemplate = "Content-Disposition: form-data; name=\"{0}\"; filename=\"{1}\"\r\nContent-Type: {2}\r\n\r\n";
            requestWriter.Write(String.Format(headerTemplate, paramName, file, contentType));

            //file content
            using (var fileStream = new FileStream(file, FileMode.Open, FileAccess.Read))
            {
                fileStream.CopyTo(requestStream);
            }

            requestWriter.Write("\r\n--" + boundary + "--\r\n");
        }
    }

    //reading response
    try
    {
        using (var wresp = (HttpWebResponse)wr.GetResponse())
        {
            if (wresp.StatusCode == HttpStatusCode.OK)
            {
                using (var responseStream = wresp.GetResponseStream())
                {
                    if (responseStream == null)
                        return null;
                    using (var responseReader = new StreamReader(responseStream))
                    {
                        return responseReader.ReadToEnd();
                    }
                }
            }

            throw new ApplicationException("Error while upload files. Server status code: " + wresp.StatusCode.ToString());
        }
    }
    catch (Exception ex)
    {
        throw new ApplicationException("Error while uploading file", ex);
    }
}

回答by codevision

With .NET 4.5 you currently could use System.Net.Http namespace. Below the example for uploading single file using multipart form data.

使用 .NET 4.5，您目前可以使用 System.Net.Http 命名空间。下面是使用多部分表单数据上传单个文件的示例。

using System;
using System.IO;
using System.Net.Http;

namespace HttpClientTest
{
    class Program
    {
        static void Main(string[] args)
        {
            var client = new HttpClient();
            var content = new MultipartFormDataContent();
            content.Add(new StreamContent(File.Open("../../Image1.png", FileMode.Open)), "Image", "Image.png");
            content.Add(new StringContent("Place string content here"), "Content-Id in the HTTP"); 
            var result = client.PostAsync("https://hostname/api/Account/UploadAvatar", content);
            Console.WriteLine(result.Result.ToString());
        }
    }
}