Как загрузить файл на сервер с помощью HTTP POST multipart / form-data


Я разрабатываю приложение для Windows Phone 8. Я хочу загрузить базу данных SQLite через веб-службу PHP с помощью HTTP-запроса POST с типом MIME multipart / form-data и строковыми данными под названием "userid=SOME_ID". Я не хочу использовать сторонние библиотеки, такие как HttpClient, RestSharp или MyToolkit. Я попробовал ниже код, но он не загружает файл и также не дает мне никаких ошибок. Он отлично работает в Android, PHP и т. д., Поэтому в веб-службе нет проблем. Ниже приведен мой код (для WP8). что плохого это?

я погуглил свою проблему, но ничего не работает для меня. Я не получаю конкретных для 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();  
}  

Я также пытался решить мою проблему в Windows 8, но это также не работает.

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();
            }
        }
    }
}
8   51  

8 ответов:

Я сделал это с помощью MultipartFormDataContent : -

        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;

вот мой последний рабочий код. Мой веб-сервиса нужен один файл (название поста параметр "файл") и строковое значение (наименование параметра пост был "идентификатор пользователя").

/// <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;
}

Я очень благодарен Дарин Руссо помогаешь мне.

эта упрощенная версия также работает.

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);
}

добавить любые дополнительные необходимые заголовки, если это необходимо.

Я немного поиграл и придумал упрощенное, более общее решение:

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();
    };
}

вы можете использовать его как это:

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);

и в PHP-скрипте вы можете обрабатывать такие данные:

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']);

вы можете использовать этот класс:

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;
        }
    }
}

использовать:

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);

пример сервера 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>';
}

приведенный ниже код считывает файл, преобразует его в массив байтов, а затем делает запрос к серверу.

    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;
    }

он работает для Windows phone 8.1. Вы можете попробовать это.

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);
 }

для людей, ищущих 403 запрещенный вопрос при попытке загрузить в составном виде ниже может помочь, как есть случай в зависимости от конфигурации сервера, что вы получите MULTIPART_STRICT_ERROR"[email protected] 0 " из-за неправильных заголовков MultipartFormDataContent. Обратите внимание, что обе переменные imagetag/filename включают цитаты (\") например filename="\ " myfile.формат 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);