Unggah file dengan HTTPWebrequest (multipart / form-data)


298

Apakah ada kelas, pustaka atau beberapa kode yang akan membantu saya mengunggah file dengan HTTPWebrequest ?

Edit 2:

Saya tidak ingin mengunggah ke folder WebDAV atau sesuatu seperti itu. Saya ingin mensimulasikan browser, jadi sama seperti Anda mengunggah avatar Anda ke forum atau mengunggah file melalui formulir di aplikasi web. Unggah ke formulir yang menggunakan multipart / formulir-data.

Edit:

WebClient tidak mencakup persyaratan saya, jadi saya mencari solusi dengan HTTPWebrequest .


10
Berikut adalah posting blog besar tentang ini - paraesthesia.com/archive/2009/12/16/...
hwiechers

1
@hwiechers: Itu benar-benar bekerja untuk saya tidak seperti jawaban lain di sini. Saya hanya perlu menambahkan cookie saya.
Tim Schmelter

1
Jika Anda menggunakan .NET> = 4.0 maka lihat jawaban ini untuk solusi yang tidak memerlukan objek khusus.
Joshcodes

Jawaban:


250

Mengambil kode di atas dan memperbaikinya karena melemparkan Internal Server Error 500. Ada beberapa masalah dengan \ r \ n posisi dan spasi yang buruk, dll. Menerapkan refactoring dengan aliran memori, menulis langsung ke aliran permintaan. Inilah hasilnya:

    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;
        }
    }

dan penggunaan sampel:

    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);

Itu dapat diperluas untuk menangani beberapa file atau cukup menyebutnya beberapa kali untuk setiap file. Namun itu sesuai dengan kebutuhan Anda.


5
Bekerja seperti pesona. Terima kasih banyak.
Yoo Matsuo

5
saya telah mencoba kode ini tetapi tidak mengunggah file jpeg dan tidak mendapatkan kesalahan? bagaimana ini mungkin.
Orhan Cinar

1
Ketika saya mencoba mengirim file lebih dari 1MB, maka mendapatkan 500 Server kesalahan, file di bawah 1MB berfungsi dengan baik, bagaimana mungkin?
David Horák

2
Saya menambahkan wr.CookieContainer untuk menjaga cookie panggilan sebelumnya.
JoaquinG

8
Jika Anda akan memperpanjang ini untuk melakukan banyak file, berhati-hatilah: hanya batas terakhir yang mendapatkan 2 tanda hubung ekstra: "\r\n--" + boundary + "--\r\n" Jika tidak, file tambahan akan terpotong.
Peter Drier

143

Saya mencari sesuatu seperti ini, Ditemukan di: http://bytes.com/groups/net-c/268661-how-upload-file-via-c-code (dimodifikasi untuk kebenaran):

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();
    }
}

9
FYI ... Anda dapat melakukan refactor MemoryStream perantara dan menulis langsung ke aliran permintaan. Kuncinya adalah memastikan untuk menutup aliran permintaan saat Anda selesai, yang menentukan panjang konten permintaan Anda!
John Clayton

5
Itu bekerja untuk saya setelah saya menghapus ruang ekstra. "r \ n Content-Type: application / octet-stream" harus "" r \ nContent-type: application / octet-stream ".
Karl B

1
Saya juga menemukan bahwa double \ r \ n di akhir header dapat menyebabkan masalah. Menghapus salah satunya memperbaiki masalah saya.
Hugo Estrada

2
Oke, kode ini tidak bekerja untuk saya, tetapi kode seorang Kristen bekerja dengan sempurna untuk saya pertama kali - stackoverflow.com/questions/566462/… - Saya menguji cgi-lib.berkeley.edu/ex/fup.html
CVertex

2
Saya tahu ini adalah pertanyaan lama "dijawab" tetapi saya baru minggu ini mencoba melakukan ini. Dengan kerangka .NET saat ini Anda dapat melakukan semua ini dalam 3 baris kode ... WebClient client = new WebClient (); byte [] responseBinary = client.UploadFile (url, file); string result = Encoding.UTF8.GetString (responseBinary);
soutarm

106

UPDATE: Menggunakan .NET 4.5 (atau .NET 4.0 dengan menambahkan paket Microsoft.Net.Http dari NuGet) ini dimungkinkan tanpa kode eksternal, ekstensi, dan manipulasi HTTP "level rendah". Berikut ini sebuah contoh:

// 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();
    }
}

2
Dapat digunakan dengan 4.0 menggunakan paket Microsoft.Net.Http NuGet. Lihat: stackoverflow.com/questions/11145053/… .
amolbk

2
Ini akhirnya menjadi cara yang sangat mudah untuk melakukan beberapa hal yang sangat kuat, termasuk mengatur header khusus untuk setiap bagian formulir.
JasonRShaver

10
@ php-jquery-programmer, ini adalah kode contoh umum sehingga parameter memiliki nama generik. Pikirkan "param1" sebagai "your_well_named_param_here" dan harap pertimbangkan kembali -1 Anda.
Joshcodes

3
Apa yang Anda sarankan daripada param1?
Joshcodes

2
Beri aku saran untuk mengubahnya ke kawan. Apakah "nama file" bekerja untuk Anda?
Joshcodes

16

FAQ Upload ASP.NET saya memiliki artikel tentang ini, dengan kode contoh: Unggah file menggunakan permintaan POST RFC 1867 dengan HttpWebRequest / WebClient . Kode ini tidak memuat file ke dalam memori (tidak seperti kode di atas), mendukung banyak file, dan mendukung nilai formulir, mengatur kredensial dan cookie, dll.

Sunting: sepertinya Axosoft menurunkan halaman. Terima kasih kawan

Itu masih dapat diakses melalui archive.org.


Terima kasih atas tautannya, Chris. Saya benar-benar mengimplementasikan yang lain ke dalam perpustakaan saya sendiri dan menambahkan dukungan tersebut (selain memori). Juga sudah dikonversi ke VB.NET :)
dr. evil

Terima kasih, Chris. Ini membantu satu ton!
flipdoubt

Solusi kelas satu! Terima kasih banyak.
Bob Denny

Terimakasih atas peringatannya! Sayangnya, saya tidak memiliki kendali atas situs itu. Saya menemukan halaman (termasuk kode sumber) di archive.org dan memperbarui tautan yang sesuai.
Chris Hynes

15

Berdasarkan kode yang diberikan di atas, saya menambahkan dukungan untuk banyak file dan juga mengunggah aliran secara langsung tanpa perlu memiliki file lokal.

Untuk mengunggah file ke url tertentu termasuk beberapa postingan lakukan hal berikut:

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 } },
    });

Untuk meningkatkan ini lebih banyak lagi yang bisa menentukan nama dan tipe pantomim dari file yang diberikan itu sendiri.

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();
        }


    }
}

Apakah ada yang akan berubah jika tipe konten multipart / terkait?
Som Bhattacharyya

12

sesuatu seperti ini dekat: (kode yang belum diuji)

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();

Terima kasih membeli, saya tidak mengejar WebDAV atau solusi serupa, saya mengklarifikasi jawaban saya. Silakan lihat hasil edit.
dr. evil


6

Mengambil di atas dan memodifikasinya menerima beberapa nilai header, dan beberapa file

    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;
    }
}

Ini tidak berfungsi untuk saya sampai saya mengubah entri batas terakhir. Pastikan batas setelah file terakhir memiliki dua tanda hubung di akhir\r\n--" + boundary + "--\r\n
Ben Ripley

5

Contoh VB (dikonversi dari contoh C # pada posting lain):

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

4

Saya harus berurusan dengan ini baru-baru ini - cara lain untuk mendekatinya adalah dengan menggunakan fakta bahwa WebClient adalah turunan, dan mengubah WebRequest yang mendasarinya dari sana:

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

Saya lebih suka C #, tetapi jika Anda terjebak dengan VB hasilnya akan terlihat seperti ini:

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

+1 Tetap webclient terlalu tidak dapat dikustomisasi sehingga mengimplementasikannya akan terasa canggung, tetapi ini adalah pendekatan yang sangat menarik, dan saya tidak tahu bahwa itu mungkin.
dr. evil

3

Ada contoh kerja lain dengan beberapa komentar saya:

        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;
        }

Dan ada contoh menggunakan:

            UploadFile[] files = new UploadFile[] 
            { 
                new UploadFile(@"C:\2.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!

2

Saya mencari untuk melakukan unggah file dan menambahkan beberapa parameter ke permintaan multipart / form-data di VB.NET dan tidak melalui pos formulir biasa. Terima kasih untuk jawaban @JoshCodes saya mendapatkan arahan yang saya cari. Saya memposting solusi saya untuk membantu orang lain menemukan cara untuk melakukan posting dengan file dan parameter yang setara dengan html dari apa yang saya coba capai adalah: 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>

Karena kenyataan bahwa saya harus memberikan apiKey dan tanda tangan (yang merupakan checksum dihitung dari parameter permintaan dan string kunci gabungan string api), saya perlu melakukannya di sisi server. Alasan lain yang perlu saya lakukan adalah sisi server adalah fakta bahwa postingan file dapat dilakukan kapan saja dengan menunjuk ke file yang sudah ada di server (menyediakan path), sehingga tidak akan ada file yang dipilih secara manual selama form posting sehingga bentuk file data tidak akan berisi aliran file. Sebaliknya saya bisa menghitung checksum melalui panggilan balik ajax dan mengirimkan file melalui posting html menggunakan JQuery. Saya menggunakan .net versi 4.0 dan tidak dapat memutakhirkan ke 4,5 dalam solusi aktual. Jadi saya harus menginstal Microsoft.Net.Http menggunakan 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

Kode @CristianRomanescu yang dimodifikasi untuk bekerja dengan aliran memori, menerima file sebagai array byte, memungkinkan null nvc, mengembalikan respons permintaan, dan bekerja dengan header Otorisasi. Menguji kode dengan Web Api 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;
    }
}

Kode uji:

[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();
    }
}

Bekerja dengan baik untuk saya - saya harus mengunggah dengan byte []. Terima kasih!
Pemain Dadu

1

Bagi saya, karya-karya berikut (sebagian besar diinspirasi dari semua jawaban berikut), saya mulai dari jawaban Elad dan memodifikasi / menyederhanakan hal-hal yang sesuai dengan kebutuhan saya (hapus bukan masukan file formulir, hanya satu file, ...).

Semoga bisa membantu seseorang :)

(PS: Saya tahu bahwa penanganan pengecualian tidak dilaksanakan dan mengasumsikan bahwa itu ditulis di dalam kelas, jadi saya mungkin perlu upaya integrasi ...)

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();
    }

ringkasan bagus dari jawaban yang lebih panjang. catatan: saya mendapat 400 permintaan buruk ketika menggunakan kode Anda dan perlu menghapus yang pertama sebelum batas pertama
jemand771

1

Tidak yakin apakah ini diposting sebelumnya, tetapi saya berhasil menggunakan WebClient ini. saya membaca dokumentasi untuk WebClient. Poin kunci yang mereka buat adalah

Jika properti BaseAddress bukan string kosong ("") dan alamat tidak mengandung URI absolut, alamat harus merupakan URI relatif yang dikombinasikan dengan BaseAddress untuk membentuk URI absolut dari data yang diminta. Jika properti QueryString bukan string kosong, itu ditambahkan ke alamat.

Jadi yang saya lakukan adalah wc.QueryString.Add ("source", generateImage) untuk menambahkan parameter kueri yang berbeda dan entah bagaimana itu cocok dengan nama properti dengan gambar yang saya unggah. Semoga ini bisa membantu

    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

Saya menulis sebuah kelas menggunakan WebClient kembali ketika melakukan unggahan formulir multi bagian.

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);
        }
    }
}

Ini akan bekerja dengan semua versi framework .NET.


0

Saya tidak pernah bisa mendapatkan contoh berfungsi dengan baik, saya selalu menerima 500 kesalahan saat mengirimnya ke server.

Namun saya menemukan metode yang sangat elegan untuk melakukannya di url ini

Ini mudah diperluas dan jelas berfungsi dengan file biner serta XML.

Anda menyebutnya menggunakan sesuatu yang mirip dengan ini

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

Lihat perpustakaan 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

Klien menggunakan Convert file to ToBase64String, setelah menggunakan Xml untuk menyebarluaskan panggilan Server, server ini menggunakan File.WriteAllBytes(path,Convert.FromBase64String(dataFile_Client_sent)).

Semoga beruntung!


0

Metode ini berfungsi untuk mengunggah banyak gambar secara bersamaan

        var flagResult = new viewModel();
        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 = method;
        wr.KeepAlive = true;
        wr.Credentials = System.Net.CredentialCache.DefaultCredentials;

        Stream rs = wr.GetRequestStream();


        string path = @filePath;
        System.IO.DirectoryInfo folderInfo = new DirectoryInfo(path);

        foreach (FileInfo file in folderInfo.GetFiles())
        {
            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.FullName, 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);
            var result = reader2.ReadToEnd();
            var cList = JsonConvert.DeserializeObject<HttpViewModel>(result);
            if (cList.message=="images uploaded!")
            {
                flagResult.success = true;
            }

        }
        catch (Exception ex)
        {
            //log.Error("Error uploading file", ex);
            if (wresp != null)
            {
                wresp.Close();
                wresp = null;
            }
        }
        finally
        {
            wr = null;
        }
        return flagResult;
    }

-1

Saya menyadari ini mungkin sangat terlambat, tetapi saya sedang mencari solusi yang sama. Saya menemukan respons berikut dari perwakilan Microsoft

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;

}

7
Jadi, intinya kode yang sama seperti dr. kejahatan di atas? stackoverflow.com/questions/566462/…
Travis Collins
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