Télécharger des fichiers avec HTTPWebrequest (multipart / form-data)

y a-t-il une classe, une bibliothèque ou un morceau de code qui m'aidera à télécharger des fichiers avec HTTPWebrequest ?

Edit 2:

Je ne veux pas télécharger dans un dossier WebDAV ou quelque chose comme ça. Je veux simuler un navigateur, donc tout comme vous téléchargez votre avatar sur un forum ou téléchargez un fichier via un formulaire dans une application web. Télécharger un formulaire qui utilise un multipart/form-data.

Edit:

WebClient n'est pas couvrir mes besoins, donc je suis à la recherche d'une solution avec HTTPWebrequest .

281
demandé sur dr. evil 2009-02-19 21:03:01

20 réponses

a pris le code ci-dessus et a corrigé parce qu'il lance L'erreur interne du serveur 500. Il y a quelques problèmes avec \r\n mal positionné et les espaces etc. Appliquez le remaniement avec le flux de mémoire, en écrivant directement sur le flux de requête. Voici le résultat:

    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;
        }
    }

et utilisation de l'échantillon:

    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);

il peut être étendu pour traiter plusieurs fichiers ou simplement l'appeler plusieurs fois pour chaque fichier. Toutefois, il s'adapte à vos besoins.

235
répondu Cristian Romanescu 2010-06-08 11:39:08

je cherchais quelque chose comme ceci, trouvé dans : http://bytes.com/groups/net-c/268661-how-upload-file-via-c-code (modifié pour l'exactitude):

public static string UploadFilesToRemoteUrl(string url, string[] files, NameValueCollection formFields = null)
{
    string boundary = "----------------------------" + DateTime.Now.Ticks.ToString("x");

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

    Stream memStream = new System.IO.MemoryStream();

    var boundarybytes = System.Text.Encoding.ASCII.GetBytes("\r\n--" +
                                                            boundary + "\r\n");
    var endBoundaryBytes = System.Text.Encoding.ASCII.GetBytes("\r\n--" +
                                                                boundary + "--");


    string formdataTemplate = "\r\n--" + boundary +
                                "\r\nContent-Disposition: form-data; name=\"{0}\";\r\n\r\n{1}";

    if (formFields != null)
    {
        foreach (string key in formFields.Keys)
        {
            string formitem = string.Format(formdataTemplate, key, formFields[key]);
            byte[] formitembytes = System.Text.Encoding.UTF8.GetBytes(formitem);
            memStream.Write(formitembytes, 0, formitembytes.Length);
        }
    }

    string headerTemplate =
        "Content-Disposition: form-data; name=\"{0}\"; filename=\"{1}\"\r\n" +
        "Content-Type: application/octet-stream\r\n\r\n";

    for (int i = 0; i < files.Length; i++)
    {
        memStream.Write(boundarybytes, 0, boundarybytes.Length);
        var header = string.Format(headerTemplate, "uplTheFile", files[i]);
        var headerbytes = System.Text.Encoding.UTF8.GetBytes(header);

        memStream.Write(headerbytes, 0, headerbytes.Length);

        using (var fileStream = new FileStream(files[i], FileMode.Open, FileAccess.Read))
        {
            var buffer = new byte[1024];
            var bytesRead = 0;
            while ((bytesRead = fileStream.Read(buffer, 0, buffer.Length)) != 0)
            {
                memStream.Write(buffer, 0, bytesRead);
            }
        }
    }

    memStream.Write(endBoundaryBytes, 0, endBoundaryBytes.Length);
    request.ContentLength = memStream.Length;

    using (Stream requestStream = request.GetRequestStream())
    {
        memStream.Position = 0;
        byte[] tempBuffer = new byte[memStream.Length];
        memStream.Read(tempBuffer, 0, tempBuffer.Length);
        memStream.Close();
        requestStream.Write(tempBuffer, 0, tempBuffer.Length);
    }

    using (var response = request.GetResponse())
    {
        Stream stream2 = response.GetResponseStream();
        StreamReader reader2 = new StreamReader(stream2);
        return reader2.ReadToEnd();
    }
}
135
répondu dr. evil 2016-06-14 11:51:19

mise à jour: en utilisant .NET 4.5 (ou .net 4.0 en ajoutant le Microsoft.Net.Http paquet de NuGet) ceci est possible sans code externe, extensions, et manipulation HTTP "de bas niveau". Voici un exemple:

// Perform the equivalent of posting a form with a filename and two files, in HTML:
// <form action="{url}" method="post" enctype="multipart/form-data">
//     <input type="text" name="filename" />
//     <input type="file" name="file1" />
//     <input type="file" name="file2" />
// </form>
private async Task<System.IO.Stream> UploadAsync(string url, string filename, Stream fileStream, byte [] fileBytes)
{
    // Convert each of the three inputs into HttpContent objects

    HttpContent stringContent = new StringContent(filename);
    // examples of converting both Stream and byte [] to HttpContent objects
    // representing input type file
    HttpContent fileStreamContent = new StreamContent(fileStream);
    HttpContent bytesContent = new ByteArrayContent(fileBytes);

    // Submit the form using HttpClient and 
    // create form data as Multipart (enctype="multipart/form-data")

    using (var client = new HttpClient())
    using (var formData = new MultipartFormDataContent()) 
    {
        // Add the HttpContent objects to the form data

        // <input type="text" name="filename" />
        formData.Add(stringContent, "filename", "filename");
        // <input type="file" name="file1" />
        formData.Add(fileStreamContent, "file1", "file1");
        // <input type="file" name="file2" />
        formData.Add(bytesContent, "file2", "file2");

        // Invoke the request to the server

        // equivalent to pressing the submit button on
        // a form with attributes (action="{url}" method="post")
        var response = await client.PostAsync(url, formData);

        // ensure the request was a success
        if (!response.IsSuccessStatusCode)
        {
            return null;
        }
        return await response.Content.ReadAsStreamAsync();
    }
}
93
répondu Joshcodes 2018-06-19 13:11:46

My ASP.NET Upload FAQ contient un article sur ce sujet, avec un code d'exemple: Upload files using an RFC 1867 POST request with HttpWebRequest/WebClient . Ce code ne charge pas les fichiers en mémoire (par opposition au code ci-dessus), prend en charge plusieurs fichiers, et prend en charge les valeurs de formulaire, la configuration des justificatifs d'identité et des cookies, etc.

Edit: on dirait Qu'Axosoft a descendu la page. Merci les gars.

C'est toujours accessible via archive.org.

16
répondu Chris Hynes 2016-07-06 22:55:18

basé sur le code fourni ci-dessus, j'ai ajouté le support pour plusieurs fichiers et aussi le téléchargement d'un flux directement sans avoir besoin d'un fichier local.

pour télécharger des fichiers vers une url spécifique, y compris certains params post faire ce qui suit:

RequestHelper.PostMultipart(
    "http://www.myserver.com/upload.php", 
    new Dictionary<string, object>() {
        { "testparam", "my value" },
        { "file", new FormFile() { Name = "image.jpg", ContentType = "image/jpeg", FilePath = "c:\temp\myniceimage.jpg" } },
        { "other_file", new FormFile() { Name = "image2.jpg", ContentType = "image/jpeg", Stream = imageDataStream } },
    });

pour améliorer cela encore plus on pourrait déterminer le nom et le type mime du fichier donné lui-même.

public class FormFile 
{
    public string Name { get; set; }

    public string ContentType { get; set; }

    public string FilePath { get; set; }

    public Stream Stream { get; set; }
}

public class RequestHelper
{

    public static string PostMultipart(string url, Dictionary<string, object> parameters) {

        string boundary = "---------------------------" + DateTime.Now.Ticks.ToString("x");
        byte[] boundaryBytes = System.Text.Encoding.ASCII.GetBytes("\r\n--" + boundary + "\r\n");

        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;

        if(parameters != null && parameters.Count > 0) {

            using(Stream requestStream = request.GetRequestStream()) {

                foreach(KeyValuePair<string, object> pair in parameters) {

                    requestStream.Write(boundaryBytes, 0, boundaryBytes.Length);
                    if(pair.Value is FormFile) {
                        FormFile file = pair.Value as FormFile;
                        string header = "Content-Disposition: form-data; name=\"" + pair.Key + "\"; filename=\"" + file.Name + "\"\r\nContent-Type: " + file.ContentType + "\r\n\r\n";
                        byte[] bytes = System.Text.Encoding.UTF8.GetBytes(header);
                        requestStream.Write(bytes, 0, bytes.Length);
                        byte[] buffer = new byte[32768];
                        int bytesRead;
                        if(file.Stream == null) {
                            // upload from file
                            using(FileStream fileStream = File.OpenRead(file.FilePath)) {
                                while((bytesRead = fileStream.Read(buffer, 0, buffer.Length)) != 0)
                                    requestStream.Write(buffer, 0, bytesRead);
                                fileStream.Close();
                            }
                        }
                        else {
                            // upload from given stream
                            while((bytesRead = file.Stream.Read(buffer, 0, buffer.Length)) != 0)
                                requestStream.Write(buffer, 0, bytesRead);
                        }
                    }
                    else {
                        string data = "Content-Disposition: form-data; name=\"" + pair.Key + "\"\r\n\r\n" + pair.Value;
                        byte[] bytes = System.Text.Encoding.UTF8.GetBytes(data);
                        requestStream.Write(bytes, 0, bytes.Length);
                    }
                }

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

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


    }
}
12
répondu Stefan 2013-04-20 18:39:59

quelque chose comme ceci est proche: (code non testé)

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();
10
répondu Moose 2009-02-19 18:06:55

a pris ce qui précède et l'a modifié accepter certaines valeurs d'en-tête ,et plusieurs fichiers""

    NameValueCollection headers = new NameValueCollection();
        headers.Add("Cookie", "name=value;");
        headers.Add("Referer", "http://google.com");
    NameValueCollection nvc = new NameValueCollection();
        nvc.Add("name", "value");

    HttpUploadFile(url, new string[] { "c:\file1.txt", "c:\file2.jpg" }, new string[] { "file", "image" }, new string[] { "application/octet-stream", "image/jpeg" }, nvc, headers);

public static void HttpUploadFile(string url, string[] file, string[] paramName, string[] contentType, NameValueCollection nvc, NameValueCollection headerItems)
{
    //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);

    foreach (string key in headerItems.Keys)
    {
        if (key == "Referer")
        {
            wr.Referer = headerItems[key];
        }
        else
        {
            wr.Headers.Add(key, headerItems[key]);
        }
    }

    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 = "";

    for(int i =0; i<file.Count();i++)
    {
        header = string.Format(headerTemplate, paramName[i], System.IO.Path.GetFileName(file[i]), contentType[i]);
        byte[] headerbytes = System.Text.Encoding.UTF8.GetBytes(header);
        rs.Write(headerbytes, 0, headerbytes.Length);

        FileStream fileStream = new FileStream(file[i], 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();
        rs.Write(boundarybytes, 0, boundarybytes.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);
            wresp.Close();
            wresp = null;
    }
    finally
    {
        wr = null;
    }
}
6
répondu mopenstein 2012-01-10 04:05:09

je pense que vous recherchez quelque chose de plus comme WebClient .

, UploadFile() .

5
répondu John T 2009-02-19 18:05:40

VB Example (converti de C# example sur un autre post):

Private Sub HttpUploadFile( _
    ByVal uri As String, _
    ByVal filePath As String, _
    ByVal fileParameterName As String, _
    ByVal contentType As String, _
    ByVal otherParameters As Specialized.NameValueCollection)

    Dim boundary As String = "---------------------------" & DateTime.Now.Ticks.ToString("x")
    Dim newLine As String = System.Environment.NewLine
    Dim boundaryBytes As Byte() = Text.Encoding.ASCII.GetBytes(newLine & "--" & boundary & newLine)
    Dim request As Net.HttpWebRequest = Net.WebRequest.Create(uri)

    request.ContentType = "multipart/form-data; boundary=" & boundary
    request.Method = "POST"
    request.KeepAlive = True
    request.Credentials = Net.CredentialCache.DefaultCredentials

    Using requestStream As IO.Stream = request.GetRequestStream()

        Dim formDataTemplate As String = "Content-Disposition: form-data; name=""{0}""{1}{1}{2}"

        For Each key As String In otherParameters.Keys

            requestStream.Write(boundaryBytes, 0, boundaryBytes.Length)
            Dim formItem As String = String.Format(formDataTemplate, key, newLine, otherParameters(key))
            Dim formItemBytes As Byte() = Text.Encoding.UTF8.GetBytes(formItem)
            requestStream.Write(formItemBytes, 0, formItemBytes.Length)

        Next key

        requestStream.Write(boundaryBytes, 0, boundaryBytes.Length)

        Dim headerTemplate As String = "Content-Disposition: form-data; name=""{0}""; filename=""{1}""{2}Content-Type: {3}{2}{2}"
        Dim header As String = String.Format(headerTemplate, fileParameterName, filePath, newLine, contentType)
        Dim headerBytes As Byte() = Text.Encoding.UTF8.GetBytes(header)
        requestStream.Write(headerBytes, 0, headerBytes.Length)

        Using fileStream As New IO.FileStream(filePath, IO.FileMode.Open, IO.FileAccess.Read)

            Dim buffer(4096) As Byte
            Dim bytesRead As Int32 = fileStream.Read(buffer, 0, buffer.Length)

            Do While (bytesRead > 0)

                requestStream.Write(buffer, 0, bytesRead)
                bytesRead = fileStream.Read(buffer, 0, buffer.Length)

            Loop

        End Using

        Dim trailer As Byte() = Text.Encoding.ASCII.GetBytes(newLine & "--" + boundary + "--" & newLine)
        requestStream.Write(trailer, 0, trailer.Length)

    End Using

    Dim response As Net.WebResponse = Nothing

    Try

        response = request.GetResponse()

        Using responseStream As IO.Stream = response.GetResponseStream()

            Using responseReader As New IO.StreamReader(responseStream)

                Dim responseText = responseReader.ReadToEnd()
                Diagnostics.Debug.Write(responseText)

            End Using

        End Using

    Catch exception As Net.WebException

        response = exception.Response

        If (response IsNot Nothing) Then

            Using reader As New IO.StreamReader(response.GetResponseStream())

                Dim responseText = reader.ReadToEnd()
                Diagnostics.Debug.Write(responseText)

            End Using

            response.Close()

        End If

    Finally

        request = Nothing

    End Try

End Sub
5
répondu Keith Walton 2016-02-27 15:16:28

j'ai dû faire face à ce récemment - une autre façon d'aborder il est d'utiliser le fait que WebClient est héréditaire, et de changer la WebRequest sous-jacente à partir de là:

http://msdn.microsoft.com/en-us/library/system.net.webclient.getwebrequest (VS.80).aspx

je préfère le C#, mais si vous êtes coincé avec VB le résultat devrait ressembler à quelque chose comme ceci:

Public Class BigWebClient
    Inherits WebClient
    Protected Overrides Function GetWebRequest(ByVal address As System.Uri) As System.Net.WebRequest
        Dim x As WebRequest = MyBase.GetWebRequest(address)
        x.Timeout = 60 * 60 * 1000
        Return x
    End Function
End Class

'Use BigWebClient here instead of WebClient
4
répondu Chris 2009-04-24 18:58:14

Il ya un autre exemple de travail avec certains de mes commentaires:

        List<MimePart> mimeParts = new List<MimePart>();

        try
        {
            foreach (string key in form.AllKeys)
            {
                StringMimePart part = new StringMimePart();

                part.Headers["Content-Disposition"] = "form-data; name=\"" + key + "\"";
                part.StringData = form[key];

                mimeParts.Add(part);
            }

            int nameIndex = 0;

            foreach (UploadFile file in files)
            {
                StreamMimePart part = new StreamMimePart();

                if (string.IsNullOrEmpty(file.FieldName))
                    file.FieldName = "file" + nameIndex++;

                part.Headers["Content-Disposition"] = "form-data; name=\"" + file.FieldName + "\"; filename=\"" + file.FileName + "\"";
                part.Headers["Content-Type"] = file.ContentType;

                part.SetStream(file.Data);

                mimeParts.Add(part);
            }

            string boundary = "----------" + DateTime.Now.Ticks.ToString("x");

            req.ContentType = "multipart/form-data; boundary=" + boundary;
            req.Method = "POST";

            long contentLength = 0;

            byte[] _footer = Encoding.UTF8.GetBytes("--" + boundary + "--\r\n");

            foreach (MimePart part in mimeParts)
            {
                contentLength += part.GenerateHeaderFooterData(boundary);
            }

            req.ContentLength = contentLength + _footer.Length;

            byte[] buffer = new byte[8192];
            byte[] afterFile = Encoding.UTF8.GetBytes("\r\n");
            int read;

            using (Stream s = req.GetRequestStream())
            {
                foreach (MimePart part in mimeParts)
                {
                    s.Write(part.Header, 0, part.Header.Length);

                    while ((read = part.Data.Read(buffer, 0, buffer.Length)) > 0)
                        s.Write(buffer, 0, read);

                    part.Data.Dispose();

                    s.Write(afterFile, 0, afterFile.Length);
                }

                s.Write(_footer, 0, _footer.Length);
            }

            return (HttpWebResponse)req.GetResponse();
        }
        catch
        {
            foreach (MimePart part in mimeParts)
                if (part.Data != null)
                    part.Data.Dispose();

            throw;
        }

et il y a un exemple d'utilisation:

            UploadFile[] files = new UploadFile[] 
            { 
                new UploadFile(@"C:.jpg","new_file","image/jpeg") //new_file is id of upload field
            };

            NameValueCollection form = new NameValueCollection();

            form["id_hidden_input"] = "value_hidden_inpu"; //there is additional param (hidden fields on page)


            HttpWebRequest req = (HttpWebRequest)WebRequest.Create(full URL of action);

            // set credentials/cookies etc. 
            req.CookieContainer = hrm.CookieContainer; //hrm is my class. i copied all cookies from last request to current (for auth)
            HttpWebResponse resp = HttpUploadHelper.Upload(req, files, form);

            using (Stream s = resp.GetResponseStream())
            using (StreamReader sr = new StreamReader(s))
            {
                string response = sr.ReadToEnd();
            }
             //profit!
3
répondu Nigrimmist 2012-01-16 11:31:39

je cherchais à faire le téléchargement de fichier et ajouter quelques paramètres à un multipart / form-data request in VB.NET et pas par l'intermédiaire d'un poste régulier. Grâce à la réponse de @JoshCodes j'ai eu la direction que je cherchais. Je poste ma solution pour aider les autres à trouver un moyen d'effectuer un poste avec à la fois le fichier et les paramètres l'équivalent html de ce que j'essaie de réaliser est : html

<form action="your-api-endpoint" enctype="multipart/form-data" method="post"> 
<input type="hidden" name="action" value="api-method-name"/> 
<input type="hidden" name="apiKey" value="gs1xxxxxxxxxxxxxex"/> 
<input type="hidden" name="access" value="protected"/> 
<input type="hidden" name="name" value="test"/> 
<input type="hidden" name="title" value="test"/> 
<input type="hidden" name="signature" value="cf1d4xxxxxxxxcd5"/> 
<input type="file" name="file"/> 
<input type="submit" name="_upload" value="Upload"/> 
</form>

en raison du fait que je dois fournir l'apiKey et la signature (qui est un checksum calculé des paramètres de requête et clé api chaîne concaténée), je devais le faire Côté Serveur. L'autre raison pour laquelle je devais le faire côté serveur est le fait que le post du fichier peut être effectué à tout moment en pointant vers un fichier déjà sur le serveur (fournissant le chemin), de sorte qu'il n'y aurait pas de fichier manuellement sélectionné pendant le post de formulaire donc le fichier de données de formulaire ne contiendrait pas le flux de fichier.Sinon j'aurais pu calculer la somme de contrôle via un rappel ajax et soumettre le fichier via le post html utilisant JQuery. J'utilise la version 4.0 de .net et je ne peux pas passer à la version 4.5 dans la solution actuelle. J'ai donc dû installer Microsoft.Net.Http utilisant nuget cmd

PM> install-package Microsoft.Net.Http

Private Function UploadFile(req As ApiRequest, filePath As String, fileName As String) As String
    Dim result = String.empty
    Try
        ''//Get file stream
        Dim paramFileStream As Stream = File.OpenRead(filePath)
        Dim fileStreamContent As HttpContent = New  StreamContent(paramFileStream)
        Using client = New HttpClient()
            Using formData = New MultipartFormDataContent()
                ''// This adds parameter name ("action")
                ''// parameter value (req.Action) to form data
                formData.Add(New StringContent(req.Action), "action")
                formData.Add(New StringContent(req.ApiKey), "apiKey")
                For Each param In req.Parameters
                    formData.Add(New StringContent(param.Value), param.Key)
                Next
                formData.Add(New StringContent(req.getRequestSignature.Qualifier), "signature")
                ''//This adds the file stream and file info to form data
                formData.Add(fileStreamContent, "file", fileName)
                ''//We are now sending the request
                Dim response = client.PostAsync(GetAPIEndpoint(), formData).Result
                ''//We are here reading the response
                Dim readR = New StreamReader(response.Content.ReadAsStreamAsync().Result, Encoding.UTF8)
                Dim respContent = readR.ReadToEnd()

                If Not response.IsSuccessStatusCode Then
                    result =  "Request Failed : Code = " & response.StatusCode & "Reason = " & response.ReasonPhrase & "Message = " & respContent
                End If
                result.Value = respContent
            End Using
        End Using
    Catch ex As Exception
        result = "An error occurred : " & ex.Message
    End Try

    Return result
End Function
2
répondu Halima 2015-03-27 12:19:08

Je ne sais pas si cela a été posté avant mais j'ai obtenu ce travail avec WebClient. j'ai lu la documentation pour le WebClient. Un point clé qu'ils font est

si la propriété BaseAddress n'est pas une chaîne ("") et une adresse vides ne contient pas D'URI absolu, l'adresse doit être un URI relatif qui est combiné avec BaseAddress pour former L'URI absolu du demandé données. Si la propriété QueryString n'est pas une chaîne vide, elle est ajouté à adresse.

donc je n'ai fait que des wc.QueryString.Ajouter ("source", generatedImage) pour ajouter les différents paramètres de requête et d'une manière ou d'une autre il apparie le nom de la propriété avec l'image que j'ai chargée. Espérons que cela aide

    public void postImageToFacebook(string generatedImage, string fbGraphUrl)
    {
        WebClient wc = new WebClient();
        byte[] bytes = System.IO.File.ReadAllBytes(generatedImage);

        wc.QueryString.Add("source", generatedImage);
        wc.QueryString.Add("message", "helloworld");

        wc.UploadFile(fbGraphUrl, generatedImage);

        wc.Dispose();

    }
1
répondu dellyjm 2014-02-04 04:48:15

a modifié le code @CristianRomanescu pour qu'il fonctionne avec le flux de mémoire, accepte le fichier comme un tableau d'octets, autorise la nvc nulle, renvoie la réponse à la requête et travaille avec L'en-tête Authorization. Testé le code avec L'Api Web 2.

private string HttpUploadFile(string url, byte[] file, string fileName, string paramName, string contentType, NameValueCollection nvc, string authorizationHeader)
{
    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.Headers.Add("Authorization", authorizationHeader);
    wr.KeepAlive = true;

    Stream rs = wr.GetRequestStream();

    string formdataTemplate = "Content-Disposition: form-data; name=\"{0}\"\r\n\r\n{1}";
    if (nvc != null)
    {
        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, fileName, contentType);
    byte[] headerbytes = System.Text.Encoding.UTF8.GetBytes(header);
    rs.Write(headerbytes, 0, headerbytes.Length);

    rs.Write(file, 0, file.Length);

    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);
        var response = reader2.ReadToEnd();
        return response;
    }
    catch (Exception ex)
    {
        if (wresp != null)
        {
            wresp.Close();
            wresp = null;
        }
        return null;
    }
    finally
    {
        wr = null;
    }
}

Testcode:

[HttpPost]
[Route("postformdata")]
public IHttpActionResult PostFormData()
{
    // Check if the request contains multipart/form-data.
    if (!Request.Content.IsMimeMultipartContent())
    {
        throw new HttpResponseException(HttpStatusCode.UnsupportedMediaType);
    }

    var provider = new MultipartMemoryStreamProvider();

    try
    {
        // Read the form data.
        var result = Request.Content.ReadAsMultipartAsync(provider).Result;
        string response = "";
        // This illustrates how to get the file names.
        foreach (var file in provider.Contents)
        {
            var fileName = file.Headers.ContentDisposition.FileName.Trim('\"');
            var buffer =  file.ReadAsByteArrayAsync().Result;
            response = HttpUploadFile("https://localhost/api/v1/createfromfile", buffer, fileName, "file", "application/pdf", null, "AuthorizationKey");
        }
        return Ok(response);
    }
    catch (System.Exception e)
    {
        return InternalServerError();
    }
}
1
répondu Ogglas 2016-04-26 06:15:54

j'ai écrit une classe en utilisant WebClient chemin du retour quand faire le téléchargement de formulaire multipart.

http://ferozedaud.blogspot.com/2010/03/multipart-form-upload-helper.html

/// 
/// MimePart
/// Abstract class for all MimeParts
/// 

abstract class MimePart
{
    public string Name { get; set; }

    public abstract string ContentDisposition { get; }

    public abstract string ContentType { get; }

    public abstract void CopyTo(Stream stream);

    public String Boundary
    {
        get;
        set;
    }
}

class NameValuePart : MimePart
{
    private NameValueCollection nameValues;

    public NameValuePart(NameValueCollection nameValues)
    {
        this.nameValues = nameValues;
    }

    public override void CopyTo(Stream stream)
    {
        string boundary = this.Boundary;
        StringBuilder sb = new StringBuilder();

        foreach (object element in this.nameValues.Keys)
        {
            sb.AppendFormat("--{0}", boundary);
            sb.Append("\r\n");
            sb.AppendFormat("Content-Disposition: form-data; name=\"{0}\";", element);
            sb.Append("\r\n");
            sb.Append("\r\n");
            sb.Append(this.nameValues[element.ToString()]);

            sb.Append("\r\n");

        }

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

        //Trace.WriteLine(sb.ToString());
        byte [] data = Encoding.ASCII.GetBytes(sb.ToString());
        stream.Write(data, 0, data.Length);
    }

    public override string ContentDisposition
    {
        get { return "form-data"; }
    }

    public override string ContentType
    {
        get { return String.Empty; }
    }
} 

class FilePart : MimePart

{

    private Stream input;

    private String contentType;



    public FilePart(Stream input, String name, String contentType)

    {

        this.input = input;

        this.contentType = contentType;

        this.Name = name;

    }



    public override void CopyTo(Stream stream)

    {

        StringBuilder sb = new StringBuilder();

        sb.AppendFormat("Content-Disposition: {0}", this.ContentDisposition);

        if (this.Name != null)

            sb.Append("; ").AppendFormat("name=\"{0}\"", this.Name);

        if (this.FileName != null)

            sb.Append("; ").AppendFormat("filename=\"{0}\"", this.FileName);

        sb.Append("\r\n");

        sb.AppendFormat(this.ContentType);

        sb.Append("\r\n");

        sb.Append("\r\n");



    // serialize the header data.

    byte[] buffer = Encoding.ASCII.GetBytes(sb.ToString());

    stream.Write(buffer, 0, buffer.Length);



    // send the stream.

    byte[] readBuffer = new byte[1024];

    int read = input.Read(readBuffer, 0, readBuffer.Length);

    while (read > 0)

    {

        stream.Write(readBuffer, 0, read);

        read = input.Read(readBuffer, 0, readBuffer.Length);

    }



    // write the terminating boundary

    sb.Length = 0;

    sb.Append("\r\n");

    sb.AppendFormat("--{0}", this.Boundary);

    sb.Append("\r\n");

    buffer = Encoding.ASCII.GetBytes(sb.ToString());

    stream.Write(buffer, 0, buffer.Length);



}

 public override string ContentDisposition
 {
      get { return "file"; }
 }



 public override string ContentType
 {
    get { 
       return String.Format("content-type: {0}", this.contentType); 
     }
 }

 public String FileName { get; set; }

}

    /// 
    /// Helper class that encapsulates all file uploads
    /// in a mime part.
    /// 

    class FilesCollection : MimePart
    {
        private List files;

        public FilesCollection()
        {
            this.files = new List();
            this.Boundary = MultipartHelper.GetBoundary();
        }

        public int Count
        {
            get { return this.files.Count; }
        }

        public override string ContentDisposition
        {
            get
            {
                return String.Format("form-data; name=\"{0}\"", this.Name);
            }
        }

        public override string ContentType
        {
            get { return String.Format("multipart/mixed; boundary={0}", this.Boundary); }
        }

        public override void CopyTo(Stream stream)
        {
            // serialize the headers
            StringBuilder sb = new StringBuilder(128);
            sb.Append("Content-Disposition: ").Append(this.ContentDisposition).Append("\r\n");
            sb.Append("Content-Type: ").Append(this.ContentType).Append("\r\n");
            sb.Append("\r\n");
            sb.AppendFormat("--{0}", this.Boundary).Append("\r\n");

            byte[] headerBytes = Encoding.ASCII.GetBytes(sb.ToString());
            stream.Write(headerBytes, 0, headerBytes.Length);
            foreach (FilePart part in files)
            {
                part.Boundary = this.Boundary;
                part.CopyTo(stream);
            }
        }

        public void Add(FilePart part)
        {
            this.files.Add(part);
        }
    }

/// 
/// Helper class to aid in uploading multipart
/// entities to HTTP web endpoints.
/// 

class MultipartHelper
{
    private static Random random = new Random(Environment.TickCount);

    private List formData = new List();
    private FilesCollection files = null;
    private MemoryStream bufferStream = new MemoryStream();
    private string boundary;

    public String Boundary { get { return boundary; } }

    public static String GetBoundary()
    {
        return Environment.TickCount.ToString("X");
    }

    public MultipartHelper()
    {
        this.boundary = MultipartHelper.GetBoundary();
    }

    public void Add(NameValuePart part)
    {
        this.formData.Add(part);
        part.Boundary = boundary;
    }

    public void Add(FilePart part)
    {
        if (files == null)
        {
            files = new FilesCollection();
        }
        this.files.Add(part);
    }

    public void Upload(WebClient client, string address, string method)
    {
        // set header
        client.Headers.Add(HttpRequestHeader.ContentType, "multipart/form-data; boundary=" + this.boundary);
        Trace.WriteLine("Content-Type: multipart/form-data; boundary=" + this.boundary + "\r\n");

        // first, serialize the form data
        foreach (NameValuePart part in this.formData)
        {
            part.CopyTo(bufferStream);
        }

        // serialize the files.
        this.files.CopyTo(bufferStream);

        if (this.files.Count > 0)
        {
            // add the terminating boundary.
            StringBuilder sb = new StringBuilder();
            sb.AppendFormat("--{0}", this.Boundary).Append("\r\n");
            byte [] buffer = Encoding.ASCII.GetBytes(sb.ToString());
            bufferStream.Write(buffer, 0, buffer.Length);
        }

        bufferStream.Seek(0, SeekOrigin.Begin);

        Trace.WriteLine(Encoding.ASCII.GetString(bufferStream.ToArray()));
        byte [] response = client.UploadData(address, method, bufferStream.ToArray());
        Trace.WriteLine("----- RESPONSE ------");
        Trace.WriteLine(Encoding.ASCII.GetString(response));
    }

    /// 
    /// Helper class that encapsulates all file uploads
    /// in a mime part.
    /// 

    class FilesCollection : MimePart
    {
        private List files;

        public FilesCollection()
        {
            this.files = new List();
            this.Boundary = MultipartHelper.GetBoundary();
        }

        public int Count
        {
            get { return this.files.Count; }
        }

        public override string ContentDisposition
        {
            get
            {
                return String.Format("form-data; name=\"{0}\"", this.Name);
            }
        }

        public override string ContentType
        {
            get { return String.Format("multipart/mixed; boundary={0}", this.Boundary); }
        }

        public override void CopyTo(Stream stream)
        {
            // serialize the headers
            StringBuilder sb = new StringBuilder(128);
            sb.Append("Content-Disposition: ").Append(this.ContentDisposition).Append("\r\n");
            sb.Append("Content-Type: ").Append(this.ContentType).Append("\r\n");
            sb.Append("\r\n");
            sb.AppendFormat("--{0}", this.Boundary).Append("\r\n");

            byte[] headerBytes = Encoding.ASCII.GetBytes(sb.ToString());
            stream.Write(headerBytes, 0, headerBytes.Length);
            foreach (FilePart part in files)
            {
                part.Boundary = this.Boundary;
                part.CopyTo(stream);
            }
        }

        public void Add(FilePart part)
        {
            this.files.Add(part);
        }
    }
}

class Program
{
    static void Main(string[] args)
    {
        Trace.Listeners.Add(new ConsoleTraceListener());
        try
        {
            using (StreamWriter sw = new StreamWriter("testfile.txt", false))
            {
                sw.Write("Hello there!");
            }

            using (Stream iniStream = File.OpenRead(@"c:\platform.ini"))
            using (Stream fileStream = File.OpenRead("testfile.txt"))
            using (WebClient client = new WebClient())
            {
                MultipartHelper helper = new MultipartHelper();

                NameValueCollection props = new NameValueCollection();
                props.Add("fname", "john");
                props.Add("id", "acme");
                helper.Add(new NameValuePart(props));

                FilePart filepart = new FilePart(fileStream, "pics1", "text/plain");
                filepart.FileName = "1.jpg";
                helper.Add(filepart);

                FilePart ini = new FilePart(iniStream, "pics2", "text/plain");
                ini.FileName = "inifile.ini";
                helper.Add(ini);

                helper.Upload(client, "http://localhost/form.aspx", "POST");
            }
        }
        catch (Exception e)
        {
            Trace.WriteLine(e);
        }
    }
}

cela fonctionnera avec toutes les versions du framework .NET.

1
répondu feroze 2016-09-18 05:20:03

Je ne peux jamais obtenir les exemples pour fonctionner correctement, je reçois toujours une erreur 500 en l'envoyant au serveur.

cependant je suis tombé sur une méthode très élégante de le faire dans cette url

il est facilement extensible et fonctionne évidemment avec des fichiers binaires ainsi que XML.

vous l'appelez en utilisant quelque chose de similaire à ce

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);
        }
    }
}
0
répondu Chris 2010-06-07 11:02:04

pour moi, les travaux suivants (principalement inspiré de toutes les réponses suivantes), j'ai commencé à partir de la réponse D'Elad et modifier/simplifier les choses pour correspondre à mon besoin (supprimer les entrées de formulaire ne pas de fichier, un seul fichier,...).

Espère que ça peut aider quelqu'un :)

(PS: je sais que la gestion des exceptions n'est pas implémentée et cela suppose qu'elle a été écrite dans une classe, donc je pourrais avoir besoin d'un effort d'intégration...)

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();
    }
0
répondu gyzpunk 2013-03-07 09:47:59

consultez la bibliothèque MyToolkit:

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

0
répondu Rico Suter 2014-10-01 13:46:45

utilisation par le Client convertir le fichier en ToBase64String , après utilisation Xml pour promulguer pour appeler le serveur, ce serveur utilise File.WriteAllBytes(path,Convert.FromBase64String(dataFile_Client_sent)) .

bonne chance!

0
répondu Tran Anh Hien 2017-02-14 09:20:02

je me rends compte que c'est probablement très tard, mais je cherchais la même solution. J'ai trouvé la réponse suivante d'un Microsoft rep

private void UploadFilesToRemoteUrl(string url, string[] files, string logpath, NameValueCollection nvc)
{

    long length = 0;
    string boundary = "----------------------------" +
    DateTime.Now.Ticks.ToString("x");


    HttpWebRequest httpWebRequest2 = (HttpWebRequest)WebRequest.Create(url);
    httpWebRequest2.ContentType = "multipart/form-data; boundary=" +
    boundary;
    httpWebRequest2.Method = "POST";
    httpWebRequest2.KeepAlive = true;
    httpWebRequest2.Credentials = System.Net.CredentialCache.DefaultCredentials;



    Stream memStream = new System.IO.MemoryStream();
    byte[] boundarybytes = System.Text.Encoding.ASCII.GetBytes("\r\n--" + boundary + "\r\n");


    string formdataTemplate = "\r\n--" + boundary + "\r\nContent-Disposition: form-data; name=\"{0}\";\r\n\r\n{1}";

    foreach(string key in nvc.Keys)
    {
        string formitem = string.Format(formdataTemplate, key, nvc[key]);
        byte[] formitembytes = System.Text.Encoding.UTF8.GetBytes(formitem);
        memStream.Write(formitembytes, 0, formitembytes.Length);
    }


    memStream.Write(boundarybytes,0,boundarybytes.Length);

    string headerTemplate = "Content-Disposition: form-data; name=\"{0}\"; filename=\"{1}\"\r\n Content-Type: application/octet-stream\r\n\r\n";

    for(int i=0;i<files.Length;i++)
    {

        string header = string.Format(headerTemplate,"file"+i,files[i]);
        byte[] headerbytes = System.Text.Encoding.UTF8.GetBytes(header);
        memStream.Write(headerbytes,0,headerbytes.Length);


        FileStream fileStream = new FileStream(files[i], FileMode.Open,
        FileAccess.Read);
        byte[] buffer = new byte[1024];

        int bytesRead = 0;

        while ( (bytesRead = fileStream.Read(buffer, 0, buffer.Length)) != 0 )
        {
            memStream.Write(buffer, 0, bytesRead);
        }


        memStream.Write(boundarybytes,0,boundarybytes.Length);


        fileStream.Close();
    }

    httpWebRequest2.ContentLength = memStream.Length;
    Stream requestStream = httpWebRequest2.GetRequestStream();

    memStream.Position = 0;
    byte[] tempBuffer = new byte[memStream.Length];
    memStream.Read(tempBuffer,0,tempBuffer.Length);
    memStream.Close();
    requestStream.Write(tempBuffer,0,tempBuffer.Length );
    requestStream.Close();


    WebResponse webResponse2 = httpWebRequest2.GetResponse();

    Stream stream2 = webResponse2.GetResponseStream();
    StreamReader reader2 = new StreamReader(stream2);

    webResponse2.Close();
    httpWebRequest2 = null;
    webResponse2 = null;

}
-1
répondu mhodgdon 2009-12-17 21:46:00