Conversion of Byte to PDF using ConvertAPI - c#

Our requirement is to convert file stored in database, array of byte to PDF using ConvertAPI. We have tried multiple options but getting different errors as mention below. Can someone please help us with function to convert Byte file format to PDF.
Error 1 - The remote server returned an error: (400) Bad Request.
using (var client = new WebClient())
{
client.Headers.Add("accept", "application/octet-stream");
client.Headers.Add("Content-Type", "application/x-www-form-urlencoded");
var resultFile = client.UploadData("https://v2.convertapi.com/doc/to/pdf?Secret=**********", byteTemplate);
}
 
Error 2 - The remote server returned an error: (400) Bad Request
var requestContent = new MultipartFormDataContent();
ByteArrayContent data = new ByteArrayContent(byteTemplate, 0, byteTemplate.Count());
requestContent.Add(data, "File", "Files");
requestContent.Add(new StringContent("**********"), "Secret");
var authParam = parameters.ContainsKey("secret") ? $"Secret={parameters["secret"]}" : $"Token={parameters["token"]}";
return new HttpClient().PostAsync($"https://v2.convertapi.com/{srcFormat}/to/{dstFormat}?{authParam}", requestContent).Result;
Error 3 - Code":5001,"Message":"Conversion failed."
or
Cannot access a disposed object.
Object name: 'System.Net.Http.MultipartFormDataContent'
public static HttpResponseMessage Convert(string srcFormat, string dstFormat, Dictionary<string, string> parameters, byte[] bytetemp, MemoryStream streamTemplate)
{
var requestContent = new MultipartFormDataContent();
streamTemplate.Position = 0;
requestContent.Add(new StreamContent(streamTemplate), "File", "ABC");
foreach (var parameter in parameters)
{
if (File.Exists(parameter.Value))
{
}
else
{
requestContent.Add(new StringContent(parameter.Value), parameter.Key);
}
}
var authParam = parameters.ContainsKey("secret") ? $"Secret={parameters["secret"]}" : $"Token={parameters["token"]}";
HttpContent rescont = new HttpClient().PostAsync($"https://v2.convertapi.com/{srcFormat}/to/{dstFormat}?{authParam}", requestContent).Result.Content;
String a = rescont.ReadAsStringAsync().Result;
}
Error 4 - The remote server returned an error: (400) Bad Request.
Old Code with New URL, Working with old URL
public static byte[] CovertWordtoPdf(byte[] response)
{
byte[] bufferDocxReport;
bufferDocxReport = response;
string reportName = "reportname.doc";
#region Convert DOCX report to PDF format
WebRequest convertToPdfRequest = WebRequest.Create("https://v2.convertapi.com/docx/to/pdf?Secret=************");
convertToPdfRequest.Method = "POST";
var boundary = "---------------------------" + DateTime.Now.Ticks.ToString("x", NumberFormatInfo.InvariantInfo);
convertToPdfRequest.ContentType = "multipart/form-data; boundary=" + boundary;
boundary = "--" + boundary;
using (var requestStream = convertToPdfRequest.GetRequestStream())
{
// Write the file
var buffer = Encoding.ASCII.GetBytes(boundary + Environment.NewLine);
requestStream.Write(buffer, 0, buffer.Length);
buffer = Encoding.UTF8.GetBytes(string.Format("Content-Disposition: form-data; name=\"{0}\"; filename=\"{1}\"{2}", "name", reportName, Environment.NewLine));
requestStream.Write(buffer, 0, buffer.Length);
buffer = Encoding.ASCII.GetBytes(string.Format("Content-Type: {0}{1}{1}", "application/octet-stream", Environment.NewLine));
requestStream.Write(buffer, 0, buffer.Length);
requestStream.Write(bufferDocxReport, 0, bufferDocxReport.Length);
buffer = Encoding.ASCII.GetBytes(Environment.NewLine);
requestStream.Write(buffer, 0, buffer.Length);
var boundaryBuffer = Encoding.ASCII.GetBytes(boundary + "--");
requestStream.Write(boundaryBuffer, 0, boundaryBuffer.Length);
}
using (var convertToPdfResponse = convertToPdfRequest.GetResponse())
using (Stream convertToPdfResponseStream = convertToPdfResponse.GetResponseStream())
{
bufferDocxReport = ReadToEnd(convertToPdfResponseStream);
}
return bufferDocxReport;
#endregion
}
Error 5 - Code:5001,Message:"Conversion failed.
var requestContent = new MultipartFormDataContent();
streamTemplate.Position = 0;
StreamContent data = new StreamContent(streamTemplate);
requestContent.Add(data, "File", "Files");
requestContent.Add(new StringContent("************"), "Secret");
var authParam = parameters.ContainsKey("secret") ? $"Secret={parameters["secret"]}" : $"Token={parameters["token"]}";
return new HttpClient().PostAsync($"https://v2.convertapi.com/{srcFormat}/to/{dstFormat}?{authParam}", requestContent).Result;

The working solution would be
const string fileToConvert = #"C:\Projects\_temp\test1.docx";
var bytesToConvert = File.ReadAllBytes(fileToConvert);
var url = new Uri("https://v2.convertapi.com/docx/to/pdf?secret=<YourSecret>");
var content = new ByteArrayContent(bytesToConvert);
content.Headers.Add("content-type", "application/octet-stream");
content.Headers.Add("content-disposition", "attachment; filename=\"test1.docx\"");
var fileBytes = new HttpClient().PostAsync(url, content).Result.Content.ReadAsByteArrayAsync().Result;
I would like to point several important things:
We should set content-type for request to application/octet-stream
because we send plain binary data in request body.
Also content-disposition should be provided to let Rest API
endpoint to know what data is sent, in this case it is file.

Related

Cant correct upload file it says no file found for upload

im trying to create method who can upload file from special path to API link, but when i run application its says no file for upload has been added, so i cant understand, where is problem, i think maybe this line is not working correct bellow
requestUploadFileToDocument.GetRequestStream().Write(data, 0, data.Length);
My full code for method
var ConKey = ConfigurationManager.AppSettings["ConKey"];
var AddFile = ConfigurationManager.AppSettings["AddFile"];
var Document_IDSTflow = ConfigurationManager.AppSettings["Document_IDSTflow"];
var file = ConfigurationManager.AppSettings["file"];
string UploadFileToDocumentUrl = String.Format(AddFile + "Document_ID=" + Document_IDSTflow + "&key=" + ConKey);
ASCIIEncoding encoder = new ASCIIEncoding();
byte[] data = encoder.GetBytes(file);
HttpWebRequest requestUploadFileToDocument = WebRequest.Create(UploadFileToDocumentUrl) as HttpWebRequest;
requestUploadFileToDocument.Method = "POST";
requestUploadFileToDocument.ContentType = "multipart/form-data";
requestUploadFileToDocument.ContentLength = data.Length;
requestUploadFileToDocument.GetRequestStream().Write(data, 0, data.Length);
var httpResponse = (HttpWebResponse)requestUploadFileToDocument.GetResponse();
using (var streamReader = new StreamReader(httpResponse.GetResponseStream()))
{
var result2 = streamReader.ReadToEnd();
DocumentInfo DocumentInfoJsonData = JsonConvert.DeserializeObject<DocumentInfo>(result2);
if (DocumentInfoJsonData.Status.IsRequestSuccessful == true)
{
Console.WriteLine("Metode (POST) - OK");
}
else
{
Console.WriteLine("Metode (POST) - ERROR");
Console.WriteLine("\n" + "Error info: " + result2 + "\n");
}
}
Okey, i resolve this by my self after some wasted days :D
HttpClient client = new HttpClient();
MultipartFormDataContent form = new MultipartFormDataContent();
HttpContent content = new StringContent("fileToUpload");
form.Add(content, "fileToUpload");
var stream = new FileStream(FileLocation, FileMode.Open);
content = new StreamContent(stream);
content.Headers.ContentDisposition = new ContentDispositionHeaderValue("form-data")
{
Name = DocName,
FileName = FileNameExtension
};
form.Add(content);
HttpResponseMessage response = null;
response = client.PostAsync(UploadFileToDocumentUrl, form).Result;

Upload Excel file to web server

I was trying to upload a file to a web server through UWP c# using filestream but it always gives me an error i.e 405 method not allowed when I try to upload on http://example.com/httpdocs/content. Even for testing purpose I tried uploading on my localhost but still no luck.
Any Help?
Code :
public async Task<bool> Upload(StorageFile fileName)
{
HttpMultipartFormDataContent form = new HttpMultipartFormDataContent();
cts = new CancellationTokenSource();
using (IInputStream fileStream = await fileName.OpenSequentialReadAsync())
{
HttpStreamContent content = new HttpStreamContent(fileStream);
form.Add(content, "premier", fileName.Name);
using (HttpClient client = new HttpClient())
{
using (HttpRequestMessage request = new HttpRequestMessage(HttpMethod.Post, new Uri("http://example.com/httpdocs/content")))
{
request.Content = form;
request.Headers.TryAppendWithoutValidation("Content-Type", "application/x-www-form-urlencoded");
HttpResponseMessage response = await client.SendRequestAsync(request).AsTask(cts.Token);
var result = response.Content.ReadAsStringAsync().GetResults();
}
}
}
return true;
}
Hye, after trying much I found this code and it works perfectly. With a little correction that the request will be now sent to an aspx page i.e. http://example.com/abc.aspx. If you wanna send high mb's of excel file data then just change 4096 in Math.Min(4096, (int)fileStream.Length) accordingly
Client Side Code-
public static async Task<string> UploadFileEx(string uploadfile, string
url, string fileFormName, string contenttype, NameValueCollection
querystring, CookieContainer cookies)
{
try
{
if ((fileFormName == null) ||
(fileFormName.Length == 0))
{
fileFormName = "file";
}
if ((contenttype == null) ||
(contenttype.Length == 0))
{
contenttype = "application/vnd.openxmlformats-officedocument.spreadsheetml.sheet";
}
string postdata;
postdata = "?";
if (querystring != null)
{
foreach (string key in querystring.Keys)
{
postdata += key + "=" + querystring.Get(key) + "&";
}
}
Uri uri = new Uri(url + postdata);
string boundary = "----------" + DateTime.Now.Ticks.ToString("x");
HttpWebRequest webrequest = (HttpWebRequest)WebRequest.Create(uri);
webrequest.CookieContainer = cookies;
webrequest.ContentType = "multipart/form-data; boundary=" + boundary;
webrequest.Method = "POST";
// Build up the post message header
StringBuilder sb = new StringBuilder();
sb.Append("--");
sb.Append(boundary);
sb.Append("\r\n");
sb.Append("Content-Disposition: form-data; name=\"");
sb.Append(fileFormName);
sb.Append("\"; filename=\"");
var sd = sb.Append(Path.GetFileName(uploadfile));
sb.Append("\"");
sb.Append("\r\n");
sb.Append("Content-Type: ");
sb.Append(contenttype);
sb.Append("\r\n");
sb.Append("\r\n");
string postHeader = sb.ToString();
byte[] postHeaderBytes = Encoding.UTF8.GetBytes(postHeader);
// Build the trailing boundary string as a byte array
// ensuring the boundary appears on a line by itself
byte[] boundaryBytes = Encoding.ASCII.GetBytes("\r\n--" + boundary + "\r\n");
FileStream fileStream = new FileStream(uploadfile, FileMode.Open, FileAccess.Read);
long length = postHeaderBytes.Length + fileStream.Length + boundaryBytes.Length;
// webrequest.ContentLength = length;
webrequest.Headers[HttpRequestHeader.ContentLength] = length.ToString();
//Stream requestStream = webrequest.GetRequestStream();
Stream requestStream = await webrequest.GetRequestStreamAsync();
// Write out our post header
requestStream.Write(postHeaderBytes, 0, postHeaderBytes.Length);
// Write out the file contents
byte[] buffer = new Byte[checked((uint)Math.Min(4096, (int)fileStream.Length))];
int bytesRead = 0;
while ((bytesRead = fileStream.Read(buffer, 0, buffer.Length)) != 0)
requestStream.Write(buffer, 0, bytesRead);
// Write out the trailing boundary
requestStream.Write(boundaryBytes, 0, boundaryBytes.Length);
WebResponse response = await webrequest.GetResponseAsync();
Stream s = response.GetResponseStream();
StreamReader sr = new StreamReader(s);
return sr.ReadToEnd();
}
catch (Exception e)
{
e.ToString();
}
return null;
}

I want to send a message and photo to a channel with my bot in c# 2013, but I'm getting The remote server returned an error: (400) Bad Request

I want to send a message and photo to a channel with my bot in c# 2013.
Message and photo should be sent in one box. Photo in above the message .
i can send a message to the channel successfully but there are 2 problems:
when send photo , this error shows:
The remote server returned an error: (400) Bad Request.
I cannot send a text and photo together in only 1 send.
code :
using System;
using System.Collections.Generic;
using System.ComponentModel;
using System.Data;
using System.Drawing;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
using System.Windows.Forms;
using Telegram.Bot;
using Telegram.Bot.Types;
using System.Net;
namespace SendTxt-Photo
{
public partial class Form1 : Form
{
public Form1()
{
InitializeComponent();
}
string Token = "adasdsadsadsadasds";
string channel_id = "#BestLaptop";
private void Form1_Load(object sender, EventArgs e)
{
}
private void button1_Click(object sender, EventArgs e)
{
using (var stream = File.Open(#"image/1.jpg", FileMode.Open))
{
WebRequest req = WebRequest.Create("https://api.telegram.org/bot" + Token + "/sendMessage?chat_id=" + channel_id + "&text=" + textbox1.text);
req.UseDefaultCredentials = true;
WebRequest req1 = WebRequest.Create("https://api.telegram.org/bot" + Token + "/sendPhoto?chat_id=" + channel_id + "&Photo=" + stream );
req.UseDefaultCredentials = true;
req1.UseDefaultCredentials = true;
var result = req.GetResponse();
req.Abort();
var result1 = req1.GetResponse();
req1.Abort();
}
}
}
}
At first you must know that you send ...&photo=Sistem.IO.FileStream instead your file. It's root cause of bad request. Take a look in debug.
At second as you can see in api documentation sendPhoto method provide three types of photo parameter:
Pass a file_id as String to send a photo that exists on the Telegram servers (recommended), pass an HTTP URL as a String for Telegram to get a photo from the Internet, or upload a new photo using multipart/form-data.
So using multipart/form-data required in your case. It can be simplified by using RestSharp for example like (code was generated by Postman, haven't tested):
var client = new RestClient("https://api.telegram.org/botadasdsadsadsadasds/sendPhoto");
var request = new RestRequest(Method.POST);
request.AddHeader("cache-control", "no-cache");
request.AddHeader("content-type", "multipart/form-data; boundary=----WebKitFormBoundary7MA4YWxkTrZu0gW");
request.AddParameter("multipart/form-data; boundary=----WebKitFormBoundary7MA4YWxkTrZu0gW", "------WebKitFormBoundary7MA4YWxkTrZu0gW\r\nContent-Disposition: form-data; name=\"chat_id\"\r\n\r\n#BestLaptop\r\n------WebKitFormBoundary7MA4YWxkTrZu0gW\r\nContent-Disposition: form-data; name=\"photo\"; filename=\"1.jpg\"\r\nContent-Type: image/jpeg\r\n\r\n\r\n------WebKitFormBoundary7MA4YWxkTrZu0gW--", ParameterType.RequestBody);
IRestResponse response = client.Execute(request);
At third you can't sent photo above text in one message. For sending text and photo in one message use caption parameter of sendPhoto method. It provide 200 symbols length.
PS Why not using the Telegram.Bot nuget package?
This function is helper to upload file:
public static string UploadFilesToRemoteUrl(HttpWebRequest request, string[] files, NameValueCollection formFields = null)
{
string boundary = "----------------------------" + DateTime.Now.Ticks.ToString("x");
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, "photo", 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();
}
}
Here the code to send file (proxy used for Roscomnadzor) Use your variables instead of msg.GetParam({""}) you'll need chat_id, bot_id,photo and caption. And "proxy" if you need it.
string filePath = msg.GetParam("photo");
string URL = "https://api.telegram.org/bot" + msg.GetParam("bot_id") + "/sendPhoto";
HttpWebRequest request = (HttpWebRequest)WebRequest.Create(URL);
WebProxy myProxy = new WebProxy(msg.GetParam("proxy"));
myProxy.UseDefaultCredentials = true;
request.Proxy = myProxy;
string[] variable_name = {filePath};
NameValueCollection form = new NameValueCollection();
form["chat_id"] = msg.GetParam("chat_id");
form["caption"] = msg.GetParam("caption");
UploadFilesToRemoteUrl(request, variable_name, form);

Not able to read the file name from the http request header

I'm not able to read the file name as well as other values that come from the client. I'm using HttpWebRequest to send Multi part data to the server. My client side code looks like so:
public string upload(string file, string url)
{
HttpWebRequest requestToServer = (HttpWebRequest)
WebRequest.Create(url);
// Define a boundary string
string boundaryString = "----";
// Turn off the buffering of data to be written, to prevent
// OutOfMemoryException when sending data
requestToServer.AllowWriteStreamBuffering = false;
// Specify that request is a HTTP post
requestToServer.Method = WebRequestMethods.Http.Post;
// Specify that the content type is a multipart request
requestToServer.ContentType
= "multipart/form-data; boundary=" + boundaryString;
// Turn off keep alive
requestToServer.KeepAlive = false;
ASCIIEncoding ascii = new ASCIIEncoding();
string boundaryStringLine = "\r\n--" + boundaryString + "\r\n";
byte[] boundaryStringLineBytes = ascii.GetBytes(boundaryStringLine);
string lastBoundaryStringLine = "\r\n--" + boundaryString + "--\r\n";
byte[] lastBoundaryStringLineBytes = ascii.GetBytes(lastBoundaryStringLine);
NameValueCollection nvc = new NameValueCollection();
nvc.Add("id", "TTR");
// Get the byte array of the myFileDescription content disposition
string myFileDescriptionContentDisposition = Java.Lang.String.Format(
"Content-Disposition: form-data; name=\"{0}\"\r\n\r\n{1}",
"myFileDescription",
"A sample file description");
byte[] myFileDescriptionContentDispositionBytes
= ascii.GetBytes(myFileDescriptionContentDisposition);
string fileUrl = file;
// Get the byte array of the string part of the myFile content
// disposition
string myFileContentDisposition = Java.Lang.String.Format(
"Content-Disposition: form-data;name=\"{0}\"; "
+ "filename=\"{1}\"\r\nContent-Type: {2}\r\n\r\n",
"myFile", Path.GetFileName(fileUrl), Path.GetExtension(fileUrl));
byte[] myFileContentDispositionBytes =
ascii.GetBytes(myFileContentDisposition);
var name = Path.GetFileName(fileUrl);
FileInfo fileInfo = new FileInfo(fileUrl);
// Calculate the total size of the HTTP request
long totalRequestBodySize = boundaryStringLineBytes.Length * 2
+ lastBoundaryStringLineBytes.Length
+ myFileDescriptionContentDispositionBytes.Length
+ myFileContentDispositionBytes.Length
+ fileInfo.Length;
// And indicate the value as the HTTP request content length
requestToServer.ContentLength = totalRequestBodySize;
string formdataTemplate = "Content-Disposition: form-data; name=\"{0}\"\r\n\r\n{1}";
// Write the http request body directly to the server
using (Stream s = requestToServer.GetRequestStream())
{
//foreach (string key in nvc.Keys)
//{
// s.Write(boundaryStringLineBytes, 0, boundaryStringLineBytes.Length);
// string formitem = string.Format(formdataTemplate, key, nvc[key]);
// byte[] formitembytes = System.Text.Encoding.UTF8.GetBytes(formitem);
// s.Write(formitembytes, 0, formitembytes.Length);
//}
// Send the file description content disposition over to the server
s.Write(boundaryStringLineBytes, 0, boundaryStringLineBytes.Length);
s.Write(myFileDescriptionContentDispositionBytes, 0,
myFileDescriptionContentDispositionBytes.Length);
// Send the file content disposition over to the server
s.Write(boundaryStringLineBytes, 0, boundaryStringLineBytes.Length);
s.Write(myFileContentDispositionBytes, 0,
myFileContentDispositionBytes.Length);
// Send the file binaries over to the server, in 1024 bytes chunk
FileStream fileStream = new FileStream(fileUrl, FileMode.Open,
FileAccess.Read);
byte[] buffer = new byte[1024];
int bytesRead = 0;
while ((bytesRead = fileStream.Read(buffer, 0, buffer.Length)) != 0)
{
s.Write(buffer, 0, bytesRead);
} // end while
fileStream.Close();
// Send the last part of the HTTP request body
s.Write(lastBoundaryStringLineBytes, 0, lastBoundaryStringLineBytes.Length);
WebResponse response = requestToServer.GetResponse();
StreamReader responseReader = new StreamReader(response.GetResponseStream());
string replyFromServer = responseReader.ReadToEnd();
return replyFromServer;
}
}
The content-disposition values that are written on the client side aren't being retrieved on the server side. On the server side, the file name is read as "{0}" and subsequently other values are read as "{1}" and "{2}".
My server side code looks like so:
public async Task<HttpResponseMessage> UploadFile()
{
if (!Request.Content.IsMimeMultipartContent())
{
throw new HttpResponseException(HttpStatusCode.UnsupportedMediaType);
}
string root = HttpContext.Current.Server.MapPath("~/App_Data");
var provider = new MultipartFormDataStreamProvider(root);
var httpRequest = HttpContext.Current.Request;
var id = httpRequest.Form["{0}"];
var id2 = httpRequest.Form[0];
var s = id;
var l = id2;
// Read the form data.
await Request.Content.ReadAsMultipartAsync(provider);
// This illustrates how to get the file names.
foreach (MultipartFileData file in provider.FileData)
{
Trace.WriteLine(file.Headers.ContentDisposition.FileName);
Trace.WriteLine("Server file path: " + file.LocalFileName);
}
if (httpRequest.Files.Count > 0)
{
foreach (string file in httpRequest.Files)
{
var postedFile = httpRequest.Files[file];
var filePath = HttpContext.Current.Server.MapPath("~/" + postedFile.FileName);
postedFile.SaveAs(filePath);
// NOTE: To store in memory use postedFile.InputStream
}
return Request.CreateResponse(HttpStatusCode.Created);
}
return Request.CreateResponse(HttpStatusCode.BadRequest);
}
I've been stuck on this for 2 days and it's driving me crazy. I have chopped and changed my code several time but each time I have different issues. This is the closest I have come to making my code work except reading the headers properly on the server.
I will forever be grateful to the person the person who will help me out.
Your client is most likely something like Android application. You use Java.Lang.String.Format there, and syntax of java format strings is different from .NET format strings, so your {0} {1} etc placesholders are not getting expanded. To fix, just use regular .NET String.Format.

How to upload file to server with HTTP POST multipart/form-data?

I am developing Windows Phone 8 app. I want to upload SQLite database via PHP web service using HTTP POST request with MIME type multipart/form-data & a string data called "userid=SOME_ID".
I don't want to use 3rd party libs like HttpClient, RestSharp or MyToolkit. I tried the below code but it doesn't upload the file & also doesn't give me any errors. It's working fine in Android, PHP, etc so there's no issue in web service. Below is my given code (for WP8). what's wrong with it?
I've googled and I'm not getting specific for WP8
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();
}
I also tried to solve my problem in Windows 8 but it's also not working.
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();
}
}
}
}
Basic implementation using MultipartFormDataContent :-
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;
Here's my final working code. My web service needed one file (POST parameter name was "file") & a string value (POST parameter name was "userid").
/// <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;
}
I am very much thankful to Darin Rousseau for helping me.
This simplistic version also works.
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);
}
Add any extra required headers if needed.
I've been playing around a little bit and came up with a simplified, more generic solution:
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();
};
}
You can use it like this:
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);
And in the PHP script you could handle data like this:
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']);
You can use this class:
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;
}
}
}
use it:
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);
example server 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>';
}
Here is what worked for me while sending the file as mult-form data:
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 used above can be:
private HttpClient GetAuthenticatedHttpClient()
{
HttpClient httpClient = new HttpClient();
httpClient.BaseAddress = new Uri(<yourBaseURL>));
httpClient.DefaultRequestHeaders.Add("Token, <yourToken>);
return httpClient;
}
I know this is and old thread, but I was fighting with this and I would like to share my solution.
This solution works with HttpClient and MultipartFormDataContent, from System.Net.Http. You can release it with .NET Core 1.0 or higher, or .NET Framework 4.5 or higher.
As a quick summary, it's an asynchronous method that receives as parameters the URL in which you want to perform the POST, a key/value collection for sending strings, and a key/value collection for sending files.
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;
}
You can prepare your POST like this (you can add so many strings and files as you need):
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");
And finally, call the method like this:
var result = Post(url, strings, files).GetAwaiter().GetResult();
If you want, you can check your status code, and show the reason as below:
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));
}
And if someone need it, here I let a small example of how to receive store a file with PHP (at the other side of our .Net app):
<?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);
}
I hope you find it useful, I am attentive to your questions.
The below code reads a file, converts it to a byte array and then makes a request to the server.
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;
}
hi guys after one day searching on web finally i solve problem with below source code
hope to help you
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;
}
Here is multipart data post with basic authentication C#
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);
}
For people searching for 403 forbidden issue while trying to upload in multipart form the below might help as there is a case depending on the server configuration that you will get MULTIPART_STRICT_ERROR "!#eq 0" due to incorrect MultipartFormDataContent headers.
Please note that both imagetag/filename variables include quotations (\")
eg filename="\"myfile.png\"" .
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);
I was also wanted to upload stuff to a Server and it was a Spring application i finally discovered that I needed to acctually set an content type for it to interpret it as a file. Just like this:
...
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);
...
I know this is an old post, but after spending an entire afternoon trying, I have to share what worked for me. My solution is for a Xamarin application but the code should still work as long as it is C#:
The issue I had is that I wanted to use the same end-point and payload as my Angular app. (Code Below).
public class ApiService<T> where T : class
{
private string _webServiceUrl = Settings.WebServiceUrl;
private HttpClient httpClient;
public ApiService(string path = null)
{
_webServiceUrl = $"{_webServiceUrl}{path}/";
httpClient = new HttpClient();
httpClient.BaseAddress = new Uri(_webServiceUrl);
httpClient.DefaultRequestHeaders.Authorization = new AuthenticationHeaderValue("Bearer", Settings.Token);
}
private void HandleException(Exception ex, string url = "")
{
Settings.TotalHttpRequests -= 1;
Crashes.TrackError(ex, new Dictionary<string, string>{ { "URL", url }, });
MessagingCenter.Send<Page, string>(new Page(), "Exception", ex.Message);
}
public async Task<T> PostFormDataAsync(string url, string filePath, T t)
{
using (var form = new MultipartFormDataContent())
{
using (var fs = File.OpenRead(filePath))
{
byte[] bytes = new byte[fs.Length];
fs.Read(bytes, 0, Convert.ToInt32(fs.Length));
fs.Close();
var json = JsonConvert.SerializeObject(t);
var data = new StringContent(json);
form.Add(new ByteArrayContent(bytes, 0, bytes.Length), "file", fs.Name);
form.Add(data, "data");
Settings.TotalHttpRequests += 1;
var result = await httpClient.PostAsync(url, form);
Settings.TotalHttpRequests -= 1;
if (result.IsSuccessStatusCode)
{
var content = await result.Content.ReadAsStringAsync();
var item = JsonConvert.DeserializeObject<T>(content);
return item;
}
else
{
await RaiseErrorAsync(result);
return default(T);
}
}
}
}
private async Task RaiseErrorAsync(HttpResponseMessage result)
{
var message = await result.Content.ReadAsStringAsync();
MessagingCenter.Send<Page, string>(new Page(), "Error", message);
}
}
I have included some code for the benefit of other Xamarin Developers.
My API endpoint is as follows:
[DisableRequestSizeLimit]
[Consumes("multipart/form-data")]
[HttpPost]
public async Task<ActionResult> Post()
{
try
{
var data = (Request.Form["data"]).ToString();
var myObject = JsonConvert.DeserializeObject<MyClass>(data);
var file = Request.Form.Files[0];
myObject = await _service.AddAsync(myObject, file);
var result = _mapper.Map<MyClassDto>(myObject);
return Ok(result);
}
catch (Exception ex)
{
return BadRequest(ex.Message);
}
}
From this, I get my file as an IFormFile instance. I am also able to deserialize the object I sent as t of type T.
The ApiService can be modified to implement IApiService so that you can benefit from Dependency Service (in Xamarin).
On the Angular side, this is what I have on the component:
onSubmit() {
this.submitted = true;
if (this.form.invalid) {return;}
if (this.fileToUpload == null) {
Swal.fire('File', 'Please attach the file', 'error');
return;
}
const formData = new FormData();
formData.append('file', this.fileToUpload, this.fileToUpload.name);
formData.append('data', JSON.stringify(this.form.value));
this.service.postEndpoint('', formData).subscribe((data) => {
//Do something here
});
}
For the service:
import { ApiService } from '../api.service';
#Injectable({
providedIn: 'root'
})
export class MyService extends ApiService<MyClass>{
constructor(protected httpClient: HttpClient) {
super(httpClient, 'MyApiController');
}
}
The ApiService:
const httpOptions = {
headers : new HttpHeaders({
Authorization: 'Bearer ' + localStorage.getItem('token')
})
};
export abstract class ApiService<T> {
basePath = environment.apiPath;
apiURL = `${this.basePath}api/`;
constructor(protected httpClient: HttpClient, protected actionUrl: string) { }
public postEndpoint(endPoint: string, model: any): Observable<any> {
return this.httpClient.post(this.apiURL + `${this.actionUrl}/${endPoint}`, model, httpOptions);
}
}
I hope this saves someone some time! Happy coding!
It work for window phone 8.1. You can try this.
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);
}
Top to #loop answer.
We got below error for Asp.Net MVC,
Unable to connect to the remote server
Fix:
After adding the below code in Web.Confing issue has been resolved for us
<system.net>
<defaultProxy useDefaultCredentials="true" >
</defaultProxy>
</system.net>
Base on #Wolf5 's answer This works for me
var client = new WebClient();
client.Encoding = Encoding.UTF8;
var boundary = $"--------------------------{DateTime.Now.Ticks:x}";
client.Headers.Add("Content-Type", "multipart/form-data; boundary=" + boundary);
client.Headers.Add("Cookie", cookie);
var start = $"--{boundary}\r\nContent-Disposition: form-data; name=\"file\"; filename=\"{Path.GetFileName(fileName)}\"\r\nContent-Type: image/jpeg\r\n\r\n";
var end = $"\r\n--{boundary}--\r\n";
var lst = new List<byte>();
lst.AddRange(client.Encoding.GetBytes(start));
lst.AddRange(File.ReadAllBytes(fileName));
lst.AddRange(client.Encoding.GetBytes(end));
var resp = client.UploadData($"{ApiUrl}/api/upload/image", "POST", lst.ToArray());

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