C# 如何使用 HTTP POST multipart/form-data 将文件上传到服务器？

How to upload file to server with HTTP POST multipart/form-data?

提问by Xyroid

I am developing Windows Phone 8 app. I want to upload SQLite database via PHP web service using HTTP POST request with MIME type multipart/form-data & a string data called "userid=SOME_ID".

我正在开发 Windows Phone 8 应用程序。我想使用带有 MIME 类型 multipart/form-data 和名为“userid=SOME_ID”的字符串数据的 HTTP POST 请求，通过 PHP Web 服务上传 SQLite 数据库。

I don't want to use 3rd party libs like HttpClient, RestSharp or MyToolkit. I tried the below code but it doesn't upload the file & also doesn't give me any errors. It's working fine in Android, PHP, etc so there's no issue in web service. Below is my given code (for WP8). what's wrong with it?

我不想使用 HttpClient、RestSharp 或 MyToolkit 等 3rd 方库。我尝试了下面的代码，但它没有上传文件，也没有给我任何错误。它在 Android、PHP 等中运行良好，因此在 Web 服务中没有问题。下面是我给定的代码（用于 WP8）。它出什么问题了？

I've googled and I'm not getting specific for WP8

我用谷歌搜索过，但我没有具体了解 WP8

async void MainPage_Loaded(object sender, RoutedEventArgs e)
{
    var file = await Windows.ApplicationModel.Package.Current.InstalledLocation.GetFileAsync(DBNAME);
    //Below line gives me file with 0 bytes, why? Should I use 
    //IsolatedStorageFile instead of StorageFile
    //var file = await ApplicationData.Current.LocalFolder.GetFileAsync(DBNAME);
    byte[] fileBytes = null;
    using (var stream = await file.OpenReadAsync())
    {
        fileBytes = new byte[stream.Size];
        using (var reader = new DataReader(stream))
        {
            await reader.LoadAsync((uint)stream.Size);
            reader.ReadBytes(fileBytes);
        }
    }

    //var res = await HttpPost(Util.UPLOAD_BACKUP, fileBytes);
    HttpPost(fileBytes);
}

private void HttpPost(byte[] file_bytes)
{
    HttpWebRequest httpWebRequest = (HttpWebRequest)WebRequest.Create("http://www.myserver.com/upload.php");
    httpWebRequest.ContentType = "multipart/form-data";
    httpWebRequest.Method = "POST";
    var asyncResult = httpWebRequest.BeginGetRequestStream((ar) => { GetRequestStreamCallback(ar, file_bytes); }, httpWebRequest);  
}

private void GetRequestStreamCallback(IAsyncResult asynchronousResult, byte[] postData)  
{
    //DON'T KNOW HOW TO PASS "userid=some_user_id"  
    HttpWebRequest request = (HttpWebRequest)asynchronousResult.AsyncState;  
    Stream postStream = request.EndGetRequestStream(asynchronousResult);  
    postStream.Write(postData, 0, postData.Length);  
    postStream.Close();  
    var asyncResult = request.BeginGetResponse(new AsyncCallback(GetResponseCallback), request);  
}  

private void GetResponseCallback(IAsyncResult asynchronousResult)  
{  
    HttpWebRequest request = (HttpWebRequest)asynchronousResult.AsyncState;  
    HttpWebResponse response = (HttpWebResponse)request.EndGetResponse(asynchronousResult);  
    Stream streamResponse = response.GetResponseStream();  
    StreamReader streamRead = new StreamReader(streamResponse);  
    string responseString = streamRead.ReadToEnd();  
    streamResponse.Close();  
    streamRead.Close();  
    response.Close();  
}  

I also tried to solve my problem in Windows 8 but it's also not working.

我也尝试在 Windows 8 中解决我的问题，但它也不起作用。

public async Task Upload(byte[] fileBytes)
{
    using (var client = new HttpClient())
    {
        using (var content = new MultipartFormDataContent("Upload----" + DateTime.Now.ToString(System.Globalization.CultureInfo.InvariantCulture)))
        {
            content.Add(new StreamContent(new MemoryStream(fileBytes)));
            //Not sure below line is true or not
            content.Add(new StringContent("userid=farhanW8"));
            using (var message = await client.PostAsync("http://www.myserver.com/upload.php", content))
            {
                var input = await message.Content.ReadAsStringAsync();
            }
        }
    }
}

采纳答案by Xyroid

Here's my final working code. My web service needed one file (POST parameter name was "file") & a string value (POST parameter name was "userid").

这是我的最终工作代码。我的 Web 服务需要一个文件（POST 参数名称为“file”）和一个字符串值（POST 参数名称为“userid”）。

/// <summary>
/// Occurs when upload backup application bar button is clicked. Author : Farhan Ghumra
 /// </summary>
private async void btnUploadBackup_Click(object sender, EventArgs e)
{
    var dbFile = await ApplicationData.Current.LocalFolder.GetFileAsync(Util.DBNAME);
    var fileBytes = await GetBytesAsync(dbFile);
    var Params = new Dictionary<string, string> { { "userid", "9" } };
    UploadFilesToServer(new Uri(Util.UPLOAD_BACKUP), Params, Path.GetFileName(dbFile.Path), "application/octet-stream", fileBytes);
}

/// <summary>
/// Creates HTTP POST request & uploads database to server. Author : Farhan Ghumra
/// </summary>
private void UploadFilesToServer(Uri uri, Dictionary<string, string> data, string fileName, string fileContentType, byte[] fileData)
{
    string boundary = "----------" + DateTime.Now.Ticks.ToString("x");
    HttpWebRequest httpWebRequest = (HttpWebRequest)WebRequest.Create(uri);
    httpWebRequest.ContentType = "multipart/form-data; boundary=" + boundary;
    httpWebRequest.Method = "POST";
    httpWebRequest.BeginGetRequestStream((result) =>
    {
        try
        {
            HttpWebRequest request = (HttpWebRequest)result.AsyncState;
            using (Stream requestStream = request.EndGetRequestStream(result))
            {
                WriteMultipartForm(requestStream, boundary, data, fileName, fileContentType, fileData);
            }
            request.BeginGetResponse(a =>
            {
                try
                {
                    var response = request.EndGetResponse(a);
                    var responseStream = response.GetResponseStream();
                    using (var sr = new StreamReader(responseStream))
                    {
                        using (StreamReader streamReader = new StreamReader(response.GetResponseStream()))
                        {
                            string responseString = streamReader.ReadToEnd();
                            //responseString is depend upon your web service.
                            if (responseString == "Success")
                            {
                                MessageBox.Show("Backup stored successfully on server.");
                            }
                            else
                            {
                                MessageBox.Show("Error occurred while uploading backup on server.");
                            } 
                        }
                    }
                }
                catch (Exception)
                {

                }
            }, null);
        }
        catch (Exception)
        {

        }
    }, httpWebRequest);
}

/// <summary>
/// Writes multi part HTTP POST request. Author : Farhan Ghumra
/// </summary>
private void WriteMultipartForm(Stream s, string boundary, Dictionary<string, string> data, string fileName, string fileContentType, byte[] fileData)
{
    /// The first boundary
    byte[] boundarybytes = Encoding.UTF8.GetBytes("--" + boundary + "\r\n");
    /// the last boundary.
    byte[] trailer = Encoding.UTF8.GetBytes("\r\n--" + boundary + "--\r\n");
    /// the form data, properly formatted
    string formdataTemplate = "Content-Dis-data; name=\"{0}\"\r\n\r\n{1}";
    /// the form-data file upload, properly formatted
    string fileheaderTemplate = "Content-Dis-data; name=\"{0}\"; filename=\"{1}\";\r\nContent-Type: {2}\r\n\r\n";

    /// Added to track if we need a CRLF or not.
    bool bNeedsCRLF = false;

    if (data != null)
    {
        foreach (string key in data.Keys)
        {
            /// if we need to drop a CRLF, do that.
            if (bNeedsCRLF)
                WriteToStream(s, "\r\n");

            /// Write the boundary.
            WriteToStream(s, boundarybytes);

            /// Write the key.
            WriteToStream(s, string.Format(formdataTemplate, key, data[key]));
            bNeedsCRLF = true;
        }
    }

    /// If we don't have keys, we don't need a crlf.
    if (bNeedsCRLF)
        WriteToStream(s, "\r\n");

    WriteToStream(s, boundarybytes);
    WriteToStream(s, string.Format(fileheaderTemplate, "file", fileName, fileContentType));
    /// Write the file data to the stream.
    WriteToStream(s, fileData);
    WriteToStream(s, trailer);
}

/// <summary>
/// Writes string to stream. Author : Farhan Ghumra
/// </summary>
private void WriteToStream(Stream s, string txt)
{
    byte[] bytes = Encoding.UTF8.GetBytes(txt);
    s.Write(bytes, 0, bytes.Length);
}

/// <summary>
/// Writes byte array to stream. Author : Farhan Ghumra
/// </summary>
private void WriteToStream(Stream s, byte[] bytes)
{
    s.Write(bytes, 0, bytes.Length);
}

/// <summary>
/// Returns byte array from StorageFile. Author : Farhan Ghumra
/// </summary>
private async Task<byte[]> GetBytesAsync(StorageFile file)
{
    byte[] fileBytes = null;
    using (var stream = await file.OpenReadAsync())
    {
        fileBytes = new byte[stream.Size];
        using (var reader = new DataReader(stream))
        {
            await reader.LoadAsync((uint)stream.Size);
            reader.ReadBytes(fileBytes);
        }
    }

    return fileBytes;
}

I am very much thankful to Darin Rousseaufor helping me.

我非常感谢Darin Rousseau帮助我。

回答by loop

I've done it using MultipartFormDataContent :-

我已经使用 MultipartFormDataContent 完成了它：-

HttpClient httpClient = new HttpClient();
MultipartFormDataContent form = new MultipartFormDataContent();

form.Add(new StringContent(username), "username");
form.Add(new StringContent(useremail), "email");
form.Add(new StringContent(password), "password");            
form.Add(new ByteArrayContent(file_bytes, 0, file_bytes.Length), "profile_pic", "hello1.jpg");
HttpResponseMessage response = await httpClient.PostAsync("PostUrl", form);

response.EnsureSuccessStatusCode();
httpClient.Dispose();
string sd = response.Content.ReadAsStringAsync().Result;

回答by woondark

You can use this class:

你可以使用这个类：

using System.Collections.Specialized;
class Post_File
{
    public static void HttpUploadFile(string url, string file, string paramName, string contentType, NameValueCollection nvc)
    {
        string boundary = "---------------------------" + DateTime.Now.Ticks.ToString("x");
        byte[] boundarybytes = System.Text.Encoding.ASCII.GetBytes("\r\n--" + boundary + "\r\n");
        byte[] boundarybytesF = System.Text.Encoding.ASCII.GetBytes("--" + boundary + "\r\n");  // the first time it itereates, you need to make sure it doesn't put too many new paragraphs down or it completely messes up poor webbrick.  


        HttpWebRequest wr = (HttpWebRequest)WebRequest.Create(url);
        wr.Method = "POST";
        wr.KeepAlive = true;
        wr.Credentials = System.Net.CredentialCache.DefaultCredentials;
        wr.Accept = "text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8";
        var nvc2 = new NameValueCollection();
        nvc2.Add("Accepts-Language", "en-us,en;q=0.5");
        wr.Headers.Add(nvc2);
        wr.ContentType = "multipart/form-data; boundary=" + boundary;


        Stream rs = wr.GetRequestStream();

        bool firstLoop = true;
        string formdataTemplate = "Content-Disposition: form-data; name=\"{0}\"\r\n\r\n{1}";
        foreach (string key in nvc.Keys)
        {
            if (firstLoop)
            {
                rs.Write(boundarybytesF, 0, boundarybytesF.Length);
                firstLoop = false;
            }
            else
            {
                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, new FileInfo(file).Name, 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);
        }
        catch (Exception ex)
        {
            if (wresp != null)
            {
                wresp.Close();
                wresp = null;
            }
        }
        finally
        {
            wr = null;
        }
    }
}

use it:

用它：

NameValueCollection nvc = new NameValueCollection();
//nvc.Add("id", "TTR");
nvc.Add("table_name", "uploadfile");
nvc.Add("commit", "uploadfile");
Post_File.HttpUploadFile("http://example/upload_file.php", @"C:\user\yourfile.docx", "uploadfile", "application/vnd.ms-excel", nvc);

example server upload_file.php:

示例服务器upload_file.php：

m('File upload '.(@copy($_FILES['uploadfile']['tmp_name'],getcwd().'\'.'/'.$_FILES['uploadfile']['name']) ? 'success' : 'failed'));
function m($msg) {
    echo '<div style="background:#f1f1f1;border:1px solid #ddd;padding:15px;font:14px;text-align:center;font-weight:bold;">';
    echo $msg;
    echo '</div>';
}

回答by Hockic

I've been playing around a little bit and came up with a simplified, more generic solution:

我一直在玩，想出了一个简化的、更通用的解决方案：

private static string sendHttpRequest(string url, NameValueCollection values, NameValueCollection files = null)
{
    string boundary = "----------------------------" + DateTime.Now.Ticks.ToString("x");
    // The first boundary
    byte[] boundaryBytes = System.Text.Encoding.UTF8.GetBytes("\r\n--" + boundary + "\r\n");
    // The last boundary
    byte[] trailer = System.Text.Encoding.UTF8.GetBytes("\r\n--" + boundary + "--\r\n");
    // The first time it itereates, we need to make sure it doesn't put too many new paragraphs down or it completely messes up poor webbrick
    byte[] boundaryBytesF = System.Text.Encoding.ASCII.GetBytes("--" + boundary + "\r\n");

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

    // Get request stream
    Stream requestStream = request.GetRequestStream();

    foreach (string key in values.Keys)
    {
        // Write item to stream
        byte[] formItemBytes = System.Text.Encoding.UTF8.GetBytes(string.Format("Content-Disposition: form-data; name=\"{0}\";\r\n\r\n{1}", key, values[key]));
        requestStream.Write(boundaryBytes, 0, boundaryBytes.Length);
        requestStream.Write(formItemBytes, 0, formItemBytes.Length);
    }

    if (files != null)
    { 
        foreach(string key in files.Keys)
        {
            if(File.Exists(files[key]))
            {
                int bytesRead = 0;
                byte[] buffer = new byte[2048];
                byte[] formItemBytes = System.Text.Encoding.UTF8.GetBytes(string.Format("Content-Disposition: form-data; name=\"{0}\"; filename=\"{1}\"\r\nContent-Type: application/octet-stream\r\n\r\n", key, files[key]));
                requestStream.Write(boundaryBytes, 0, boundaryBytes.Length);
                requestStream.Write(formItemBytes, 0, formItemBytes.Length);

                using (FileStream fileStream = new FileStream(files[key], FileMode.Open, FileAccess.Read))
                {
                    while ((bytesRead = fileStream.Read(buffer, 0, buffer.Length)) != 0)
                    {
                        // Write file content to stream, byte by byte
                        requestStream.Write(buffer, 0, bytesRead);
                    }

                    fileStream.Close();
                }
            }
        }
    }

    // Write trailer and close stream
    requestStream.Write(trailer, 0, trailer.Length);
    requestStream.Close();

    using (StreamReader reader = new StreamReader(request.GetResponse().GetResponseStream()))
    {
        return reader.ReadToEnd();
    };
}

You can use it like this:

你可以这样使用它：

string fileLocation = Environment.GetFolderPath(Environment.SpecialFolder.MyDocuments) + Path.DirectorySeparatorChar + "somefile.jpg";
NameValueCollection values = new NameValueCollection();
NameValueCollection files = new NameValueCollection();
values.Add("firstName", "Alan");
files.Add("profilePicture", fileLocation);
sendHttpRequest("http://example.com/handler.php", values, files);

And in the PHP script you could handle data like this:

在 PHP 脚本中，您可以像这样处理数据：

echo $_POST['firstName'];
$name = $_POST['firstName'];
$image = $_FILES['profilePicture'];
$ds = DIRECTORY_SEPARATOR;
move_uploaded_file($image['tmp_name'], realpath(dirname(__FILE__)) . $ds . "uploads" . $ds . $image['name']);

回答by Wolf5

This simplistic version also works.

这个简单的版本也有效。

public void UploadMultipart(byte[] file, string filename, string contentType, string url)
{
    var webClient = new WebClient();
    string boundary = "------------------------" + DateTime.Now.Ticks.ToString("x");
    webClient.Headers.Add("Content-Type", "multipart/form-data; boundary=" + boundary);
    var fileData = webClient.Encoding.GetString(file);
    var package = string.Format("--{0}\r\nContent-Disposition: form-data; name=\"file\"; filename=\"{1}\"\r\nContent-Type: {2}\r\n\r\n{3}\r\n--{0}--\r\n", boundary, filename, contentType, fileData);

    var nfile = webClient.Encoding.GetBytes(package);

    byte[] resp = webClient.UploadData(url, "POST", nfile);
}

Add any extra required headers if needed.

如果需要，添加任何额外的必需标题。

回答by reza.cse08

It work for window phone 8.1. You can try this.

它适用于窗口电话 8.1。你可以试试这个。

Dictionary<string, object> _headerContents = new Dictionary<string, object>();
const String _lineEnd = "\r\n";
const String _twoHyphens = "--";
const String _boundary = "*****";
private async void UploadFile_OnTap(object sender, System.Windows.Input.GestureEventArgs e)
{
   Uri serverUri = new Uri("http:www.myserver.com/Mp4UploadHandler", UriKind.Absolute);    
   string fileContentType = "multipart/form-data";       
   byte[] _boundarybytes = Encoding.UTF8.GetBytes(_twoHyphens + _boundary + _lineEnd);
   byte[] _trailerbytes = Encoding.UTF8.GetBytes(_twoHyphens + _boundary + _twoHyphens + _lineEnd);
   Dictionary<string, object> _headerContents = new Dictionary<string, object>();
   SetEndHeaders();  // to add some extra parameter if you need

   httpWebRequest = (HttpWebRequest)WebRequest.Create(serverUri);
   httpWebRequest.ContentType = fileContentType + "; boundary=" + _boundary;
   httpWebRequest.Method = "POST";
   httpWebRequest.AllowWriteStreamBuffering = false;  // get response after upload header part

   var fileName = Path.GetFileName(MediaStorageFile.Path);    
   Stream fStream = (await MediaStorageFile.OpenAsync(Windows.Storage.FileAccessMode.Read)).AsStream(); //MediaStorageFile is a storage file from where you want to upload the file of your device    
   string fileheaderTemplate = "Content-Disposition: form-data; name=\"{0}\"" + _lineEnd + _lineEnd + "{1}" + _lineEnd;    
   long httpLength = 0;
   foreach (var headerContent in _headerContents) // get the length of upload strem
   httpLength += _boundarybytes.Length + Encoding.UTF8.GetBytes(string.Format(fileheaderTemplate, headerContent.Key, headerContent.Value)).Length;

   httpLength += _boundarybytes.Length + Encoding.UTF8.GetBytes("Content-Disposition: form-data; name=\"uploadedFile\";filename=\"" + fileName + "\"" + _lineEnd).Length
                                       + Encoding.UTF8.GetBytes(_lineEnd).Length * 2 + _trailerbytes.Length;
   httpWebRequest.ContentLength = httpLength + fStream.Length;  // wait until you upload your total stream 

    httpWebRequest.BeginGetRequestStream((result) =>
    {
       try
       {
         HttpWebRequest request = (HttpWebRequest)result.AsyncState;
         using (Stream stream = request.EndGetRequestStream(result))
         {
            foreach (var headerContent in _headerContents)
            {
               WriteToStream(stream, _boundarybytes);
               WriteToStream(stream, string.Format(fileheaderTemplate, headerContent.Key, headerContent.Value));
             }

             WriteToStream(stream, _boundarybytes);
             WriteToStream(stream, "Content-Disposition: form-data; name=\"uploadedFile\";filename=\"" + fileName + "\"" + _lineEnd);
             WriteToStream(stream, _lineEnd);

             int bytesRead = 0;
             byte[] buffer = new byte[2048];  //upload 2K each time

             while ((bytesRead = fStream.Read(buffer, 0, buffer.Length)) != 0)
             {
               stream.Write(buffer, 0, bytesRead);
               Array.Clear(buffer, 0, 2048); // Clear the array.
              }

              WriteToStream(stream, _lineEnd);
              WriteToStream(stream, _trailerbytes);
              fStream.Close();
         }
         request.BeginGetResponse(a =>
         { //get response here
            try
            {
               var response = request.EndGetResponse(a);
               using (Stream streamResponse = response.GetResponseStream())
               using (var memoryStream = new MemoryStream())
               {   
                   streamResponse.CopyTo(memoryStream);
                   responseBytes = memoryStream.ToArray();  // here I get byte response from server. you can change depends on server response
               }    
              if (responseBytes.Length > 0 && responseBytes[0] == 1)
                 MessageBox.Show("Uploading Completed");
              else
                  MessageBox.Show("Uploading failed, please try again.");
            }
            catch (Exception ex)
            {}
          }, null);
      }
      catch (Exception ex)
      {
         fStream.Close();                             
      }
   }, httpWebRequest);
}

private static void WriteToStream(Stream s, string txt)
{
   byte[] bytes = Encoding.UTF8.GetBytes(txt);
   s.Write(bytes, 0, bytes.Length);
 }

 private static void WriteToStream(Stream s, byte[] bytes)
 {
   s.Write(bytes, 0, bytes.Length);
 }

 private void SetEndHeaders()
 {
   _headerContents.Add("sId", LocalData.currentUser.SessionId);
   _headerContents.Add("uId", LocalData.currentUser.UserIdentity);
   _headerContents.Add("authServer", LocalData.currentUser.AuthServerIP);
   _headerContents.Add("comPort", LocalData.currentUser.ComPort);
 }

回答by Siddarth Kanted

The below code reads a file, converts it to a byte array and then makes a request to the server.

下面的代码读取文件，将其转换为字节数组，然后向服务器发出请求。

    public void PostImage()
    {
        HttpClient httpClient = new HttpClient();
        MultipartFormDataContent form = new MultipartFormDataContent();

        byte[] imagebytearraystring = ImageFileToByteArray(@"C:\Users\Downloads\icon.png");
        form.Add(new ByteArrayContent(imagebytearraystring, 0, imagebytearraystring.Count()), "profile_pic", "hello1.jpg");
        HttpResponseMessage response = httpClient.PostAsync("your url", form).Result;

        httpClient.Dispose();
        string sd = response.Content.ReadAsStringAsync().Result;
    }

    private byte[] ImageFileToByteArray(string fullFilePath)
    {
        FileStream fs = File.OpenRead(fullFilePath);
        byte[] bytes = new byte[fs.Length];
        fs.Read(bytes, 0, Convert.ToInt32(fs.Length));
        fs.Close();
        return bytes;
    }

回答by Brian K.

For people searching for 403 forbidden issue while trying to upload in multipart form the below might help as there is a case depending on the server configuration that you will get MULTIPART_STRICT_ERROR "!@eq 0" due to incorrect MultipartFormDataContent headers. Please note that both imagetag/filename variables include quotations (\") eg filename="\"myfile.png\"" .

对于在尝试以多部分形式上传时搜索 403 禁止问题的人来说，以下可能会有所帮助，因为根据服务器配置的情况，由于 MultipartFormDataContent 标头不正确，您将收到 MULTIPART_STRICT_ERROR "!@eq 0"。请注意，两个 imagetag/filename 变量都包含引号 (\")，例如 filename="\"myfile.png\"" 。

    MultipartFormDataContent form = new MultipartFormDataContent();
    ByteArrayContent imageContent = new ByteArrayContent(fileBytes, 0, fileBytes.Length);
    imageContent.Headers.TryAddWithoutValidation("Content-Disposition", "form-data; name="+imagetag+"; filename="+filename);
    imageContent.Headers.TryAddWithoutValidation("Content-Type", "image / png");
    form.Add(imageContent, imagetag, filename);

回答by Ali Sadri

hi guys after one day searching on web finally i solve problem with below source code hope to help you

大家好，在网上搜索了一天后，我终于用下面的源代码解决了问题，希望对您有所帮助

    public UploadResult UploadFile(string  fileAddress)
    {
        HttpClient client = new HttpClient();

        MultipartFormDataContent form = new MultipartFormDataContent();
        HttpContent content = new StringContent("fileToUpload");
        form.Add(content, "fileToUpload");       
        var stream = new FileStream(fileAddress, FileMode.Open);            
        content = new StreamContent(stream);
        var fileName = 
        content.Headers.ContentDisposition = new ContentDispositionHeaderValue("form-data")
        {
            Name = "name",
            FileName = Path.GetFileName(fileAddress),                 
        };
        form.Add(content);
        HttpResponseMessage response = null;          

        var url = new Uri("http://192.168.10.236:2000/api/Upload2");
        response = (client.PostAsync(url, form)).Result;          

    }

回答by Alex

I was also wanted to upload stuff to a Server and it was a Spring application i finally discovered that I needed to acctually set an content type for it to interpret it as a file. Just like this:

我还想将内容上传到服务器，这是一个 Spring 应用程序，我最终发现我需要实际设置内容类型以将其解释为文件。像这样：

...
MultipartFormDataContent form = new MultipartFormDataContent();
var fileStream = new FileStream(uniqueTempPathInProject, FileMode.Open);
var streamContent = new StreamContent(fileStream);
streamContent.Headers.ContentType=new MediaTypeHeaderValue("application/zip");
form.Add(streamContent, "file",fileName);
...