Cargar archivos con HTTPWebrequest (multipart/form-data)

Resuelto dr. evil asked hace 15 años • 22 respuestas

¿Existe alguna clase, biblioteca o algún fragmento de código que me ayude a cargar archivos con HTTPWebrequest ?

Edición 2:

No quiero subir a una carpeta WebDAV o algo así. Quiero simular un navegador, así como subir su avatar a un foro o cargar un archivo a través de un formulario en una aplicación web. Subir a un formulario que utilice datos de formulario/multiparte.

Editar:

WebClient no cubre mis requisitos, así que estoy buscando una solución con HTTPWebrequest .

dr. evil avatar Feb 20 '09 01:02 dr. evil
Aceptado

Tomé el código anterior y lo solucioné porque arroja el error interno del servidor 500. Hay algunos problemas con \r\n mal posicionado y espacios, etc. Se aplicó la refactorización con el flujo de memoria, escribiendo directamente en el flujo de solicitud. Aquí está el resultado:

    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;
        }
    }

y uso de muestra:

    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);

Podría ampliarse para manejar varios archivos o simplemente llamarlo varias veces para cada archivo. Como sea se adapta a tus necesidades.

Cristian Romanescu avatar Jun 08 '2010 11:06 Cristian Romanescu

Estaba buscando algo como esto, encontrado en: http://bytes.com/groups/net-c/268661-how-upload-file-via-c-code (modificado para que sea correcto):

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();
    }
}
dr. evil avatar Feb 19 '2009 22:02 dr. evil

Esto es posible sin código externo, extensiones y manipulación HTTP de "bajo nivel" (solo necesita el paquete Microsoft.Net.Http de NuGet). Aquí hay un ejemplo:

// 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();
    }
}
Joshcodes avatar Jun 04 '2013 18:06 Joshcodes

Según el código proporcionado anteriormente, agregué soporte para múltiples archivos y también cargué una transmisión directamente sin la necesidad de tener un archivo local.

Para cargar archivos a una URL específica, incluidos algunos parámetros de publicación, haga lo siguiente:

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 } },
    });

Para mejorar esto aún más, se podría determinar el nombre y el tipo MIME del archivo dado.

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();
        }


    }
}
Stefan avatar Apr 20 '2013 18:04 Stefan

algo como esto está cerca: (código no probado)

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();
Moose avatar Feb 19 '2009 18:02 Moose