Envoyer un fichier via POST HTTP avec C#

Question

J'ai cherché et lu à ce sujet, mais je n'ai rien trouvé de vraiment utile.

Je suis en train d'écrire une petite application C# Windows qui permet à l'utilisateur d'envoyer des fichiers à un serveur web, non pas par FTP, mais par HTTP en utilisant POST. Pensez à un formulaire web mais exécuté dans une application Windows.

J'ai créé mon objet HttpWebRequest en utilisant quelque chose comme ceci :

HttpWebRequest req = WebRequest.Create(uri) as HttpWebRequest 

et j'ai également défini les propriétés Method, ContentType et ContentLength. Mais c'est tout ce que je peux faire.

Voici mon morceau de code :

HttpWebRequest req = WebRequest.Create(uri) as HttpWebRequest;
req.KeepAlive = false;
req.Method = "POST";
req.Credentials = new NetworkCredential(user.UserName, user.UserPassword);
req.PreAuthenticate = true;
req.ContentType = file.ContentType;
req.ContentLength = file.Length;
HttpWebResponse response = null;

try
{
    response = req.GetResponse() as HttpWebResponse;
}
catch (Exception e) 
{
}

Ma question est donc essentiellement comment puis-je envoyer un fichier (fichier texte, image, audio, etc.) avec C# via HTTP POST.

Merci !

Answer 1

     public string SendFile(string filePath)
            {
                WebResponse response = null;
                try
                {
                    string sWebAddress = "Https://www.address.com";

                    string boundary = "---------------------------" + DateTime.Now.Ticks.ToString("x");
                    byte[] boundarybytes = System.Text.Encoding.ASCII.GetBytes("\r\n--" + boundary + "\r\n");
                    HttpWebRequest wr = (HttpWebRequest)WebRequest.Create(sWebAddress);
                    wr.ContentType = "multipart/form-data; boundary=" + boundary;
                    wr.Method = "POST";
                    wr.KeepAlive = true;
                    wr.Credentials = System.Net.CredentialCache.DefaultCredentials;
                    Stream stream = wr.GetRequestStream();
                    string formdataTemplate = "Content-Disposition: form-data; name=\"{0}\"\r\n\r\n{1}";

                    stream.Write(boundarybytes, 0, boundarybytes.Length);
                    byte[] formitembytes = System.Text.Encoding.UTF8.GetBytes(filePath);
                    stream.Write(formitembytes, 0, formitembytes.Length);
                    stream.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, "file", Path.GetFileName(filePath), Path.GetExtension(filePath));
                    byte[] headerbytes = System.Text.Encoding.UTF8.GetBytes(header);
                    stream.Write(headerbytes, 0, headerbytes.Length);

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

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

                    response = wr.GetResponse();
                    Stream responseStream = response.GetResponseStream();
                    StreamReader streamReader = new StreamReader(responseStream);
                    string responseData = streamReader.ReadToEnd();
                    return responseData;
                }
                catch (Exception ex)
                {
                    return ex.Message;
                }
                finally
                {
                    if (response != null)
                        response.Close();
                }
            }

Answer 2

Pour poster des fichiers à partir de tableaux d'octets :

private static string UploadFilesToRemoteUrl(string url, IList files, NameValueCollection nvc) {

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

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

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

        foreach (string key in nvc.Keys) {
            var formitem = string.Format(formdataTemplate, key, nvc[key]);
            var formitembytes = Encoding.UTF8.GetBytes(formitem);
            postQueue.Write(formitembytes);
        }

        var headerTemplate = "\r\n--" + boundary + "\r\n" +
            "Content-Disposition: form-data; name=\"{0}\"; filename=\"{1}\"\r\n" + 
            "Content-Type: application/zip\r\n\r\n";

        var i = 0;
        foreach (var file in files) {
            var header = string.Format(headerTemplate, "file" + i, "file" + i + ".zip");
            var headerbytes = Encoding.UTF8.GetBytes(header);
            postQueue.Write(headerbytes);
            postQueue.Write(file);
            i++;
        }

        postQueue.Write(Encoding.UTF8.GetBytes("\r\n--" + boundary + "--"));

        request.ContentLength = postQueue.Length;

        using (var requestStream = request.GetRequestStream()) {
            postQueue.CopyToStream(requestStream);
            requestStream.Close();
        }

        var webResponse2 = request.GetResponse();

        using (var stream2 = webResponse2.GetResponseStream())
        using (var reader2 = new StreamReader(stream2)) {

            var res =  reader2.ReadToEnd();
            webResponse2.Close();
            return res;
        }
    }

public class ByteArrayCustomQueue {

    private LinkedList arrays = new LinkedList();

    /// 
    /// Writes the specified data.
    /// 
    /// The data.
    public void Write(byte[] data) {
        arrays.AddLast(data);
    }

    /// 
    /// Gets the length.
    /// 
    /// 
    /// The length.
    /// 
    public int Length { get { return arrays.Sum(x => x.Length); } }

    /// 
    /// Copies to stream.
    /// 
    /// The request stream.
    /// 
    public void CopyToStream(Stream requestStream) {
        foreach (var array in arrays) {
            requestStream.Write(array, 0, array.Length);
        }
    }
}

Answer 3

Vous pouvez le faire directement avec HttpWebRequest/HttpWebResponse comme ceci.

        string serviceUrl = string.Format("{0}/upload?param={1}", "http://127.0.0.1:8080", HttpUtility.UrlEncode(parameter));
        HttpWebRequest request = (HttpWebRequest)WebRequest.Create(serviceUrl);
        request.Method = "POST";
        request.KeepAlive = true;

        FileStream file = File.OpenRead(pathToFile);
        request.ContentLength = file.Length;

        file.Seek(0, SeekOrigin.Begin);
        file.CopyTo(request.GetRequestStream());

        HttpWebResponse response = (request.GetResponse() as HttpWebResponse);
        StreamReader reader = new StreamReader(response.GetResponseStream(), Encoding.UTF8);
        string responseText = reader.ReadToEnd();

Answer 4

En utilisant .NET 4.5 en essayant d'effectuer un téléchargement de fichier POST de formulaire. J'ai essayé la plupart des méthodes ci-dessus mais en vain. J'ai trouvé la solution ici https://www.c-sharpcorner.com/article/upload-any-file-using-http-post-multipart-form-data

Mais je ne suis pas convaincu car je ne comprends pas pourquoi nous devons encore traiter de tels programmations de bas niveau dans ces utilisations courantes (devrait être géré agréablement par le cadre)