HTTP POST मल्टीपार्ट / फॉर्म-डेटा के साथ सर्वर पर फ़ाइल कैसे अपलोड करें?

Question 1

मैं विंडोज फोन 8 ऐप विकसित कर रहा हूं। मैं MIME प्रकार मल्टीपार्ट / फॉर्म-डेटा और "उपयोगकर्ता आईडी = SOME_ID" नामक एक स्ट्रिंग डेटा के साथ HTTP POST अनुरोध का उपयोग करके PHP वेब सेवा के माध्यम से SQLite डेटाबेस अपलोड करना चाहता हूं।

मैं HttpClient, RestSharp या MyToolkit जैसी 3 पार्टी के कामों का उपयोग नहीं करना चाहता। मैंने नीचे दिए गए कोड की कोशिश की, लेकिन यह फ़ाइल अपलोड नहीं करता है और मुझे कोई त्रुटि नहीं देता है। यह Android, PHP आदि में ठीक काम कर रहा है, इसलिए वेब सेवा में कोई समस्या नहीं है। नीचे मेरा दिया गया कोड (WP8 के लिए) है। इसके साथ गलत क्या है?

मैंने गुगली की है और मैं WP8 के लिए विशिष्ट नहीं हो रहा हूं

async void MainPage_Loaded(object sender, RoutedEventArgs e)
{
    var file = await Windows.ApplicationModel.Package.Current.InstalledLocation.GetFileAsync(DBNAME);
    //Below line gives me file with 0 bytes, why? Should I use 
    //IsolatedStorageFile instead of StorageFile
    //var file = await ApplicationData.Current.LocalFolder.GetFileAsync(DBNAME);
    byte[] fileBytes = null;
    using (var stream = await file.OpenReadAsync())
    {
        fileBytes = new byte[stream.Size];
        using (var reader = new DataReader(stream))
        {
            await reader.LoadAsync((uint)stream.Size);
            reader.ReadBytes(fileBytes);
        }
    }

    //var res = await HttpPost(Util.UPLOAD_BACKUP, fileBytes);
    HttpPost(fileBytes);
}

private void HttpPost(byte[] file_bytes)
{
    HttpWebRequest httpWebRequest = (HttpWebRequest)WebRequest.Create("http://www.myserver.com/upload.php");
    httpWebRequest.ContentType = "multipart/form-data";
    httpWebRequest.Method = "POST";
    var asyncResult = httpWebRequest.BeginGetRequestStream((ar) => { GetRequestStreamCallback(ar, file_bytes); }, httpWebRequest);  
}

private void GetRequestStreamCallback(IAsyncResult asynchronousResult, byte[] postData)  
{
    //DON'T KNOW HOW TO PASS "userid=some_user_id"  
    HttpWebRequest request = (HttpWebRequest)asynchronousResult.AsyncState;  
    Stream postStream = request.EndGetRequestStream(asynchronousResult);  
    postStream.Write(postData, 0, postData.Length);  
    postStream.Close();  
    var asyncResult = request.BeginGetResponse(new AsyncCallback(GetResponseCallback), request);  
}  

private void GetResponseCallback(IAsyncResult asynchronousResult)  
{  
    HttpWebRequest request = (HttpWebRequest)asynchronousResult.AsyncState;  
    HttpWebResponse response = (HttpWebResponse)request.EndGetResponse(asynchronousResult);  
    Stream streamResponse = response.GetResponseStream();  
    StreamReader streamRead = new StreamReader(streamResponse);  
    string responseString = streamRead.ReadToEnd();  
    streamResponse.Close();  
    streamRead.Close();  
    response.Close();  
}

मैंने विंडोज 8 में भी अपनी समस्या को हल करने की कोशिश की लेकिन यह भी काम नहीं कर रहा है।

public async Task Upload(byte[] fileBytes)
{
    using (var client = new HttpClient())
    {
        using (var content = new MultipartFormDataContent("Upload----" + DateTime.Now.ToString(System.Globalization.CultureInfo.InvariantCulture)))
        {
            content.Add(new StreamContent(new MemoryStream(fileBytes)));
            //Not sure below line is true or not
            content.Add(new StringContent("userid=farhanW8"));
            using (var message = await client.PostAsync("http://www.myserver.com/upload.php", content))
            {
                var input = await message.Content.ReadAsStringAsync();
            }
        }
    }
}

Question 2

MultipartFormDataContent का उपयोग करके मूल कार्यान्वयन: -

HttpClient httpClient = new HttpClient();
MultipartFormDataContent form = new MultipartFormDataContent();

form.Add(new StringContent(username), "username");
form.Add(new StringContent(useremail), "email");
form.Add(new StringContent(password), "password");            
form.Add(new ByteArrayContent(file_bytes, 0, file_bytes.Length), "profile_pic", "hello1.jpg");
HttpResponseMessage response = await httpClient.PostAsync("PostUrl", form);

response.EnsureSuccessStatusCode();
httpClient.Dispose();
string sd = response.Content.ReadAsStringAsync().Result;

Question 3

यहाँ मेरा अंतिम काम कोड है। मेरी वेब सेवा को एक फ़ाइल की आवश्यकता थी (POST पैरामीटर नाम "फ़ाइल" था) और एक स्ट्रिंग मान (POST पैरामीटर नाम "उपयोगकर्ता नाम") था।

/// <summary>
/// Occurs when upload backup application bar button is clicked. Author : Farhan Ghumra
 /// </summary>
private async void btnUploadBackup_Click(object sender, EventArgs e)
{
    var dbFile = await ApplicationData.Current.LocalFolder.GetFileAsync(Util.DBNAME);
    var fileBytes = await GetBytesAsync(dbFile);
    var Params = new Dictionary<string, string> { { "userid", "9" } };
    UploadFilesToServer(new Uri(Util.UPLOAD_BACKUP), Params, Path.GetFileName(dbFile.Path), "application/octet-stream", fileBytes);
}

/// <summary>
/// Creates HTTP POST request & uploads database to server. Author : Farhan Ghumra
/// </summary>
private void UploadFilesToServer(Uri uri, Dictionary<string, string> data, string fileName, string fileContentType, byte[] fileData)
{
    string boundary = "----------" + DateTime.Now.Ticks.ToString("x");
    HttpWebRequest httpWebRequest = (HttpWebRequest)WebRequest.Create(uri);
    httpWebRequest.ContentType = "multipart/form-data; boundary=" + boundary;
    httpWebRequest.Method = "POST";
    httpWebRequest.BeginGetRequestStream((result) =>
    {
        try
        {
            HttpWebRequest request = (HttpWebRequest)result.AsyncState;
            using (Stream requestStream = request.EndGetRequestStream(result))
            {
                WriteMultipartForm(requestStream, boundary, data, fileName, fileContentType, fileData);
            }
            request.BeginGetResponse(a =>
            {
                try
                {
                    var response = request.EndGetResponse(a);
                    var responseStream = response.GetResponseStream();
                    using (var sr = new StreamReader(responseStream))
                    {
                        using (StreamReader streamReader = new StreamReader(response.GetResponseStream()))
                        {
                            string responseString = streamReader.ReadToEnd();
                            //responseString is depend upon your web service.
                            if (responseString == "Success")
                            {
                                MessageBox.Show("Backup stored successfully on server.");
                            }
                            else
                            {
                                MessageBox.Show("Error occurred while uploading backup on server.");
                            } 
                        }
                    }
                }
                catch (Exception)
                {

                }
            }, null);
        }
        catch (Exception)
        {

        }
    }, httpWebRequest);
}

/// <summary>
/// Writes multi part HTTP POST request. Author : Farhan Ghumra
/// </summary>
private void WriteMultipartForm(Stream s, string boundary, Dictionary<string, string> data, string fileName, string fileContentType, byte[] fileData)
{
    /// The first boundary
    byte[] boundarybytes = Encoding.UTF8.GetBytes("--" + boundary + "\r\n");
    /// the last boundary.
    byte[] trailer = Encoding.UTF8.GetBytes("\r\n--" + boundary + "--\r\n");
    /// the form data, properly formatted
    string formdataTemplate = "Content-Dis-data; name=\"{0}\"\r\n\r\n{1}";
    /// the form-data file upload, properly formatted
    string fileheaderTemplate = "Content-Dis-data; name=\"{0}\"; filename=\"{1}\";\r\nContent-Type: {2}\r\n\r\n";

    /// Added to track if we need a CRLF or not.
    bool bNeedsCRLF = false;

    if (data != null)
    {
        foreach (string key in data.Keys)
        {
            /// if we need to drop a CRLF, do that.
            if (bNeedsCRLF)
                WriteToStream(s, "\r\n");

            /// Write the boundary.
            WriteToStream(s, boundarybytes);

            /// Write the key.
            WriteToStream(s, string.Format(formdataTemplate, key, data[key]));
            bNeedsCRLF = true;
        }
    }

    /// If we don't have keys, we don't need a crlf.
    if (bNeedsCRLF)
        WriteToStream(s, "\r\n");

    WriteToStream(s, boundarybytes);
    WriteToStream(s, string.Format(fileheaderTemplate, "file", fileName, fileContentType));
    /// Write the file data to the stream.
    WriteToStream(s, fileData);
    WriteToStream(s, trailer);
}

/// <summary>
/// Writes string to stream. Author : Farhan Ghumra
/// </summary>
private void WriteToStream(Stream s, string txt)
{
    byte[] bytes = Encoding.UTF8.GetBytes(txt);
    s.Write(bytes, 0, bytes.Length);
}

/// <summary>
/// Writes byte array to stream. Author : Farhan Ghumra
/// </summary>
private void WriteToStream(Stream s, byte[] bytes)
{
    s.Write(bytes, 0, bytes.Length);
}

/// <summary>
/// Returns byte array from StorageFile. Author : Farhan Ghumra
/// </summary>
private async Task<byte[]> GetBytesAsync(StorageFile file)
{
    byte[] fileBytes = null;
    using (var stream = await file.OpenReadAsync())
    {
        fileBytes = new byte[stream.Size];
        using (var reader = new DataReader(stream))
        {
            await reader.LoadAsync((uint)stream.Size);
            reader.ReadBytes(fileBytes);
        }
    }

    return fileBytes;
}

मेरी मदद करने के लिए मैं डारिन रूसो का बहुत आभारी हूं।

Question 4

यह सरलीकृत संस्करण भी काम करता है।

public void UploadMultipart(byte[] file, string filename, string contentType, string url)
{
    var webClient = new WebClient();
    string boundary = "------------------------" + DateTime.Now.Ticks.ToString("x");
    webClient.Headers.Add("Content-Type", "multipart/form-data; boundary=" + boundary);
    var fileData = webClient.Encoding.GetString(file);
    var package = string.Format("--{0}\r\nContent-Disposition: form-data; name=\"file\"; filename=\"{1}\"\r\nContent-Type: {2}\r\n\r\n{3}\r\n--{0}--\r\n", boundary, filename, contentType, fileData);

    var nfile = webClient.Encoding.GetBytes(package);

    byte[] resp = webClient.UploadData(url, "POST", nfile);
}

यदि आवश्यक हो तो कोई अतिरिक्त आवश्यक हेडर जोड़ें।

Question 5

मैं थोड़ा बहुत खेल रहा हूं और एक सरलीकृत, अधिक सामान्य समाधान के साथ आया हूं:

private static string sendHttpRequest(string url, NameValueCollection values, NameValueCollection files = null)
{
    string boundary = "----------------------------" + DateTime.Now.Ticks.ToString("x");
    // The first boundary
    byte[] boundaryBytes = System.Text.Encoding.UTF8.GetBytes("\r\n--" + boundary + "\r\n");
    // The last boundary
    byte[] trailer = System.Text.Encoding.UTF8.GetBytes("\r\n--" + boundary + "--\r\n");
    // The first time it itereates, we need to make sure it doesn't put too many new paragraphs down or it completely messes up poor webbrick
    byte[] boundaryBytesF = System.Text.Encoding.ASCII.GetBytes("--" + boundary + "\r\n");

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

    // Get request stream
    Stream requestStream = request.GetRequestStream();

    foreach (string key in values.Keys)
    {
        // Write item to stream
        byte[] formItemBytes = System.Text.Encoding.UTF8.GetBytes(string.Format("Content-Disposition: form-data; name=\"{0}\";\r\n\r\n{1}", key, values[key]));
        requestStream.Write(boundaryBytes, 0, boundaryBytes.Length);
        requestStream.Write(formItemBytes, 0, formItemBytes.Length);
    }

    if (files != null)
    { 
        foreach(string key in files.Keys)
        {
            if(File.Exists(files[key]))
            {
                int bytesRead = 0;
                byte[] buffer = new byte[2048];
                byte[] formItemBytes = System.Text.Encoding.UTF8.GetBytes(string.Format("Content-Disposition: form-data; name=\"{0}\"; filename=\"{1}\"\r\nContent-Type: application/octet-stream\r\n\r\n", key, files[key]));
                requestStream.Write(boundaryBytes, 0, boundaryBytes.Length);
                requestStream.Write(formItemBytes, 0, formItemBytes.Length);

                using (FileStream fileStream = new FileStream(files[key], FileMode.Open, FileAccess.Read))
                {
                    while ((bytesRead = fileStream.Read(buffer, 0, buffer.Length)) != 0)
                    {
                        // Write file content to stream, byte by byte
                        requestStream.Write(buffer, 0, bytesRead);
                    }

                    fileStream.Close();
                }
            }
        }
    }

    // Write trailer and close stream
    requestStream.Write(trailer, 0, trailer.Length);
    requestStream.Close();

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

आप इसे इस तरह से उपयोग कर सकते हैं:

string fileLocation = Environment.GetFolderPath(Environment.SpecialFolder.MyDocuments) + Path.DirectorySeparatorChar + "somefile.jpg";
NameValueCollection values = new NameValueCollection();
NameValueCollection files = new NameValueCollection();
values.Add("firstName", "Alan");
files.Add("profilePicture", fileLocation);
sendHttpRequest("http://example.com/handler.php", values, files);

और PHP स्क्रिप्ट में आप इस तरह से डेटा को संभाल सकते हैं:

echo $_POST['firstName'];
$name = $_POST['firstName'];
$image = $_FILES['profilePicture'];
$ds = DIRECTORY_SEPARATOR;
move_uploaded_file($image['tmp_name'], realpath(dirname(__FILE__)) . $ds . "uploads" . $ds . $image['name']);

Question 6

आप इस वर्ग का उपयोग कर सकते हैं:

using System.Collections.Specialized;
class Post_File
{
    public static void HttpUploadFile(string url, string file, string paramName, string contentType, NameValueCollection nvc)
    {
        string boundary = "---------------------------" + DateTime.Now.Ticks.ToString("x");
        byte[] boundarybytes = System.Text.Encoding.ASCII.GetBytes("\r\n--" + boundary + "\r\n");
        byte[] boundarybytesF = System.Text.Encoding.ASCII.GetBytes("--" + boundary + "\r\n");  // the first time it itereates, you need to make sure it doesn't put too many new paragraphs down or it completely messes up poor webbrick.  


        HttpWebRequest wr = (HttpWebRequest)WebRequest.Create(url);
        wr.Method = "POST";
        wr.KeepAlive = true;
        wr.Credentials = System.Net.CredentialCache.DefaultCredentials;
        wr.Accept = "text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8";
        var nvc2 = new NameValueCollection();
        nvc2.Add("Accepts-Language", "en-us,en;q=0.5");
        wr.Headers.Add(nvc2);
        wr.ContentType = "multipart/form-data; boundary=" + boundary;


        Stream rs = wr.GetRequestStream();

        bool firstLoop = true;
        string formdataTemplate = "Content-Disposition: form-data; name=\"{0}\"\r\n\r\n{1}";
        foreach (string key in nvc.Keys)
        {
            if (firstLoop)
            {
                rs.Write(boundarybytesF, 0, boundarybytesF.Length);
                firstLoop = false;
            }
            else
            {
                rs.Write(boundarybytes, 0, boundarybytes.Length);
            }
            string formitem = string.Format(formdataTemplate, key, nvc[key]);
            byte[] formitembytes = System.Text.Encoding.UTF8.GetBytes(formitem);
            rs.Write(formitembytes, 0, formitembytes.Length);
        }
        rs.Write(boundarybytes, 0, boundarybytes.Length);

        string headerTemplate = "Content-Disposition: form-data; name=\"{0}\"; filename=\"{1}\"\r\nContent-Type: {2}\r\n\r\n";
        string header = string.Format(headerTemplate, paramName, new FileInfo(file).Name, contentType);
        byte[] headerbytes = System.Text.Encoding.UTF8.GetBytes(header);
        rs.Write(headerbytes, 0, headerbytes.Length);

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

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

        WebResponse wresp = null;
        try
        {
            wresp = wr.GetResponse();
            Stream stream2 = wresp.GetResponseStream();
            StreamReader reader2 = new StreamReader(stream2);
        }
        catch (Exception ex)
        {
            if (wresp != null)
            {
                wresp.Close();
                wresp = null;
            }
        }
        finally
        {
            wr = null;
        }
    }
}

इसका इस्तेमाल करें:

NameValueCollection nvc = new NameValueCollection();
//nvc.Add("id", "TTR");
nvc.Add("table_name", "uploadfile");
nvc.Add("commit", "uploadfile");
Post_File.HttpUploadFile("http://example/upload_file.php", @"C:\user\yourfile.docx", "uploadfile", "application/vnd.ms-excel", nvc);

उदाहरण सर्वर upload_file.php:

m('File upload '.(@copy($_FILES['uploadfile']['tmp_name'],getcwd().'\\'.'/'.$_FILES['uploadfile']['name']) ? 'success' : 'failed'));
function m($msg) {
    echo '<div style="background:#f1f1f1;border:1px solid #ddd;padding:15px;font:14px;text-align:center;font-weight:bold;">';
    echo $msg;
    echo '</div>';
}

Question 7

फ़ाइल को मल्टी-फॉर्म डेटा के रूप में भेजने के दौरान यहां मेरे लिए क्या काम किया गया है:

    public T HttpPostMultiPartFileStream<T>(string requestURL, string filePath, string fileName)
    {
        string content = null;

        using (MultipartFormDataContent form = new MultipartFormDataContent())
        {
            StreamContent streamContent;
            using (var fileStream = new FileStream(filePath, FileMode.Open))
            {
                streamContent = new StreamContent(fileStream);

                streamContent.Headers.Add("Content-Type", "application/octet-stream");
                streamContent.Headers.Add("Content-Disposition", string.Format("form-data; name=\"file\"; filename=\"{0}\"", fileName));
                form.Add(streamContent, "file", fileName);

                using (HttpClient client = GetAuthenticatedHttpClient())
                {
                    HttpResponseMessage response = client.PostAsync(requestURL, form).GetAwaiter().GetResult();
                    content = response.Content.ReadAsStringAsync().GetAwaiter().GetResult();



                    try
                    {
                        return JsonConvert.DeserializeObject<T>(content);
                    }
                    catch (Exception ex)
                    {
                        // Log the exception
                    }

                    return default(T);
                }
            }
        }
    }

GetAuthenticatedHttpClient ऊपर इस्तेमाल किया जा सकता है:

private HttpClient GetAuthenticatedHttpClient()
{
    HttpClient httpClient = new HttpClient();
    httpClient.BaseAddress = new Uri(<yourBaseURL>));
    httpClient.DefaultRequestHeaders.Add("Token, <yourToken>);
    return httpClient;
}

Question 8

नीचे दिया गया कोड एक फ़ाइल पढ़ता है, इसे बाइट सरणी में परिवर्तित करता है और फिर सर्वर से अनुरोध करता है।

    public void PostImage()
    {
        HttpClient httpClient = new HttpClient();
        MultipartFormDataContent form = new MultipartFormDataContent();

        byte[] imagebytearraystring = ImageFileToByteArray(@"C:\Users\Downloads\icon.png");
        form.Add(new ByteArrayContent(imagebytearraystring, 0, imagebytearraystring.Count()), "profile_pic", "hello1.jpg");
        HttpResponseMessage response = httpClient.PostAsync("your url", form).Result;

        httpClient.Dispose();
        string sd = response.Content.ReadAsStringAsync().Result;
    }

    private byte[] ImageFileToByteArray(string fullFilePath)
    {
        FileStream fs = File.OpenRead(fullFilePath);
        byte[] bytes = new byte[fs.Length];
        fs.Read(bytes, 0, Convert.ToInt32(fs.Length));
        fs.Close();
        return bytes;
    }

Question 9

हाय दोस्तों एक दिन के बाद वेब पर खोज आखिरकार मैं आपकी मदद करने के लिए नीचे स्रोत कोड आशा के साथ समस्या हल करता हूं

    public UploadResult UploadFile(string  fileAddress)
    {
        HttpClient client = new HttpClient();

        MultipartFormDataContent form = new MultipartFormDataContent();
        HttpContent content = new StringContent("fileToUpload");
        form.Add(content, "fileToUpload");       
        var stream = new FileStream(fileAddress, FileMode.Open);            
        content = new StreamContent(stream);
        var fileName = 
        content.Headers.ContentDisposition = new ContentDispositionHeaderValue("form-data")
        {
            Name = "name",
            FileName = Path.GetFileName(fileAddress),                 
        };
        form.Add(content);
        HttpResponseMessage response = null;          

        var url = new Uri("http://192.168.10.236:2000/api/Upload2");
        response = (client.PostAsync(url, form)).Result;          

    }

Question 10

यहां मूल प्रमाणीकरण C # के साथ मल्टीपार्ट डेटा पोस्ट है

public string UploadFilesToRemoteUrl(string url)
    {
        try
        {                             

            Dictionary<string, object> formFields = new Dictionary<string, object>();
            formFields.Add("requestid", "{\"id\":\"idvalue\"}");

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

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

            // basic authentication.
            var username = "userid";
            var password = "password";

            string credidentials = username + ":" + password;
            var authorization = Convert.ToBase64String(Encoding.Default.GetBytes(credidentials));
            request.Headers["Authorization"] = "Basic " + authorization;

            request.Method = "POST";
            request.KeepAlive = true;

            Stream memStream = new System.IO.MemoryStream();
            WriteFormData(formFields, memStream, boundary);

            FileInfo fileToUpload = new FileInfo(@"filelocation with name");
            string fileFormKey = "file";
            if (fileToUpload != null)
            {
                WritefileToUpload(fileToUpload, memStream, boundary, fileFormKey);
            }
            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 responseSReam = response.GetResponseStream();
                StreamReader streamReader = new StreamReader(responseSReam);
                return streamReader.ReadToEnd();
            }
        }
        catch (WebException ex)
        {
            using (WebResponse response = ex.Response)
            {
                HttpWebResponse httpResponse = (HttpWebResponse)response;
                using (var streamReader = new StreamReader(response.GetResponseStream()))
                    return streamReader.ReadToEnd();

            }
        }
    }

    // write form id.
    public static void WriteFormData(Dictionary<string, object> dictionary, Stream stream, string mimeBoundary)
    {
        string formdataTemplate = "\r\n--" + mimeBoundary +
                                    "\r\nContent-Disposition: form-data; name=\"{0}\";\r\n\r\n{1}";
        if (dictionary != null)
        {
            foreach (string key in dictionary.Keys)
            {
                string formitem = string.Format(formdataTemplate, key, dictionary[key]);
                byte[] formitembytes = System.Text.Encoding.UTF8.GetBytes(formitem);
                stream.Write(formitembytes, 0, formitembytes.Length);
            }
        }
    }

    // write file.
    public static void WritefileToUpload(FileInfo file, Stream stream, string mimeBoundary, string formkey)
    {
        var boundarybytes = System.Text.Encoding.ASCII.GetBytes("\r\n--" + mimeBoundary + "\r\n");
        var endBoundaryBytes = System.Text.Encoding.ASCII.GetBytes("\r\n--" + mimeBoundary + "--");

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

        stream.Write(boundarybytes, 0, boundarybytes.Length);
        var header = string.Format(headerTemplate, formkey, file.Name);
        var headerbytes = System.Text.Encoding.UTF8.GetBytes(header);

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

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

Question 11

यह विंडो फोन 8.1 के लिए काम करता है। आप यह कोशिश कर सकते हैं।

Dictionary<string, object> _headerContents = new Dictionary<string, object>();
const String _lineEnd = "\r\n";
const String _twoHyphens = "--";
const String _boundary = "*****";
private async void UploadFile_OnTap(object sender, System.Windows.Input.GestureEventArgs e)
{
   Uri serverUri = new Uri("http:www.myserver.com/Mp4UploadHandler", UriKind.Absolute);    
   string fileContentType = "multipart/form-data";       
   byte[] _boundarybytes = Encoding.UTF8.GetBytes(_twoHyphens + _boundary + _lineEnd);
   byte[] _trailerbytes = Encoding.UTF8.GetBytes(_twoHyphens + _boundary + _twoHyphens + _lineEnd);
   Dictionary<string, object> _headerContents = new Dictionary<string, object>();
   SetEndHeaders();  // to add some extra parameter if you need

   httpWebRequest = (HttpWebRequest)WebRequest.Create(serverUri);
   httpWebRequest.ContentType = fileContentType + "; boundary=" + _boundary;
   httpWebRequest.Method = "POST";
   httpWebRequest.AllowWriteStreamBuffering = false;  // get response after upload header part

   var fileName = Path.GetFileName(MediaStorageFile.Path);    
   Stream fStream = (await MediaStorageFile.OpenAsync(Windows.Storage.FileAccessMode.Read)).AsStream(); //MediaStorageFile is a storage file from where you want to upload the file of your device    
   string fileheaderTemplate = "Content-Disposition: form-data; name=\"{0}\"" + _lineEnd + _lineEnd + "{1}" + _lineEnd;    
   long httpLength = 0;
   foreach (var headerContent in _headerContents) // get the length of upload strem
   httpLength += _boundarybytes.Length + Encoding.UTF8.GetBytes(string.Format(fileheaderTemplate, headerContent.Key, headerContent.Value)).Length;

   httpLength += _boundarybytes.Length + Encoding.UTF8.GetBytes("Content-Disposition: form-data; name=\"uploadedFile\";filename=\"" + fileName + "\"" + _lineEnd).Length
                                       + Encoding.UTF8.GetBytes(_lineEnd).Length * 2 + _trailerbytes.Length;
   httpWebRequest.ContentLength = httpLength + fStream.Length;  // wait until you upload your total stream 

    httpWebRequest.BeginGetRequestStream((result) =>
    {
       try
       {
         HttpWebRequest request = (HttpWebRequest)result.AsyncState;
         using (Stream stream = request.EndGetRequestStream(result))
         {
            foreach (var headerContent in _headerContents)
            {
               WriteToStream(stream, _boundarybytes);
               WriteToStream(stream, string.Format(fileheaderTemplate, headerContent.Key, headerContent.Value));
             }

             WriteToStream(stream, _boundarybytes);
             WriteToStream(stream, "Content-Disposition: form-data; name=\"uploadedFile\";filename=\"" + fileName + "\"" + _lineEnd);
             WriteToStream(stream, _lineEnd);

             int bytesRead = 0;
             byte[] buffer = new byte[2048];  //upload 2K each time

             while ((bytesRead = fStream.Read(buffer, 0, buffer.Length)) != 0)
             {
               stream.Write(buffer, 0, bytesRead);
               Array.Clear(buffer, 0, 2048); // Clear the array.
              }

              WriteToStream(stream, _lineEnd);
              WriteToStream(stream, _trailerbytes);
              fStream.Close();
         }
         request.BeginGetResponse(a =>
         { //get response here
            try
            {
               var response = request.EndGetResponse(a);
               using (Stream streamResponse = response.GetResponseStream())
               using (var memoryStream = new MemoryStream())
               {   
                   streamResponse.CopyTo(memoryStream);
                   responseBytes = memoryStream.ToArray();  // here I get byte response from server. you can change depends on server response
               }    
              if (responseBytes.Length > 0 && responseBytes[0] == 1)
                 MessageBox.Show("Uploading Completed");
              else
                  MessageBox.Show("Uploading failed, please try again.");
            }
            catch (Exception ex)
            {}
          }, null);
      }
      catch (Exception ex)
      {
         fStream.Close();                             
      }
   }, httpWebRequest);
}

private static void WriteToStream(Stream s, string txt)
{
   byte[] bytes = Encoding.UTF8.GetBytes(txt);
   s.Write(bytes, 0, bytes.Length);
 }

 private static void WriteToStream(Stream s, byte[] bytes)
 {
   s.Write(bytes, 0, bytes.Length);
 }

 private void SetEndHeaders()
 {
   _headerContents.Add("sId", LocalData.currentUser.SessionId);
   _headerContents.Add("uId", LocalData.currentUser.UserIdentity);
   _headerContents.Add("authServer", LocalData.currentUser.AuthServerIP);
   _headerContents.Add("comPort", LocalData.currentUser.ComPort);
 }

Question 12

मल्टीपार्ट में अपलोड करने की कोशिश करते समय 403 निषिद्ध मुद्दे की खोज करने वाले लोगों के लिए नीचे दी गई मदद हो सकती है क्योंकि सर्वर कॉन्फ़िगरेशन के आधार पर एक मामला है कि आपको MULTIPART_STRICT_ERROR मिल जाएगा! @Eq 0 "गलत MultipartFormDataContent हेडर के कारण! कृपया ध्यान दें कि दोनों इमेजेटैग / फाइलनाम वेरिएबल्स में कोटेशन (\ ") जैसे फ़ाइल नाम =" \ "myfile.png \" शामिल हैं।

    MultipartFormDataContent form = new MultipartFormDataContent();
    ByteArrayContent imageContent = new ByteArrayContent(fileBytes, 0, fileBytes.Length);
    imageContent.Headers.TryAddWithoutValidation("Content-Disposition", "form-data; name="+imagetag+"; filename="+filename);
    imageContent.Headers.TryAddWithoutValidation("Content-Type", "image / png");
    form.Add(imageContent, imagetag, filename);

Question 13

मैं भी एक सर्वर पर सामान अपलोड करना चाहता था और यह एक स्प्रिंग एप्लिकेशन था जिसे मैंने अंततः पाया कि मुझे इसे फ़ाइल के रूप में व्याख्या करने के लिए सामग्री प्रकार को एकांत में सेट करने की आवश्यकता थी। सिर्फ इस तरह:

...
MultipartFormDataContent form = new MultipartFormDataContent();
var fileStream = new FileStream(uniqueTempPathInProject, FileMode.Open);
var streamContent = new StreamContent(fileStream);
streamContent.Headers.ContentType=new MediaTypeHeaderValue("application/zip");
form.Add(streamContent, "file",fileName);
...

Question 14

शीर्ष पर @loop उत्तर।

हमें Asp.Net MVC के लिए त्रुटि मिली, दूरस्थ सर्वर से कनेक्ट करने में असमर्थ

फिक्स: वेब में नीचे दिए गए कोड को जोड़ने के बाद। हमारे लिए समस्या का समाधान किया गया है

  <system.net>
    <defaultProxy useDefaultCredentials="true" >
    </defaultProxy>
  </system.net>

Question 15

मुझे पता है कि यह पुराना धागा है, लेकिन मैं इससे लड़ रहा था और मैं अपने समाधान को साझा करना चाहूंगा।

इस समाधान के साथ HttpClientऔर MultipartFormDataContent, से काम करता है System.Net.Http। आप इसे .NET Core 1.0या उच्चतर, या उच्चतर के साथ जारी कर सकते हैं .NET Framework 4.5।

एक त्वरित सारांश के रूप में, यह एक अतुल्यकालिक विधि है जो URL के मापदंडों के रूप में प्राप्त होती है जिसमें आप POST प्रदर्शन करना चाहते हैं, तार भेजने के लिए एक कुंजी / मान संग्रह और फ़ाइलों को भेजने के लिए एक कुंजी / मान संग्रह।

private static async Task<HttpResponseMessage> Post(string url, NameValueCollection strings, NameValueCollection files)
{
    var formContent = new MultipartFormDataContent(/* If you need a boundary, you can define it here */);

    // Strings
    foreach (string key in strings.Keys)
    {
        string inputName = key;
        string content = strings[key];

        formContent.Add(new StringContent(content), inputName);
    }

    // Files
    foreach (string key in files.Keys)
    {
        string inputName = key;
        string fullPathToFile = files[key];

        FileStream fileStream = File.OpenRead(fullPathToFile);
        var streamContent = new StreamContent(fileStream);
        var fileContent = new ByteArrayContent(streamContent.ReadAsByteArrayAsync().Result);
        formContent.Add(fileContent, inputName, Path.GetFileName(fullPathToFile));
    }

    var myHttpClient = new HttpClient();
    var response = await myHttpClient.PostAsync(url, formContent);
    //string stringContent = await response.Content.ReadAsStringAsync(); // If you need to read the content

    return response;
}

आप अपना POST इस तरह तैयार कर सकते हैं (आप अपनी जरूरत के अनुसार इतने तार और फाइलें जोड़ सकते हैं):

string url = @"http://yoursite.com/upload.php"

NameValueCollection strings = new NameValueCollection();
strings.Add("stringInputName1", "The content for input 1");
strings.Add("stringInputNameN", "The content for input N");

NameValueCollection files = new NameValueCollection();
files.Add("fileInputName1", @"FullPathToFile1"); // Path + filename
files.Add("fileInputNameN", @"FullPathToFileN");

और अंत में, इस तरह से विधि को बुलाओ:

var result = Post(url, strings, files).GetAwaiter().GetResult();

यदि आप चाहें, तो आप अपना स्टेटस कोड देख सकते हैं, और नीचे दिए गए कारण बता सकते हैं:

if (result.StatusCode == HttpStatusCode.OK)
{
    // Logic if all was OK
}
else
{
    // You can show a message like this:
    Console.WriteLine(string.Format("Error. StatusCode: {0} | ReasonPhrase: {1}", result.StatusCode, result.ReasonPhrase));
}

और अगर किसी को इसकी आवश्यकता होती है, तो यहां मैं एक छोटा सा उदाहरण देता हूं कि पीएचपी के साथ फाइल कैसे प्राप्त करें (हमारे नेटवर्क्स के दूसरी तरफ)।

<?php

if (isset($_FILES['fileInputName1']) && $_FILES['fileInputName1']['error'] === UPLOAD_ERR_OK)
{
  $fileTmpPath = $_FILES['fileInputName1']['tmp_name'];
  $fileName = $_FILES['fileInputName1']['name'];

  move_uploaded_file($fileTmpPath, '/the/final/path/you/want/' . $fileName);
}

मुझे आशा है कि आप इसे उपयोगी पाएंगे, मैं आपके प्रश्नों के प्रति चौकस हूं।