Not able to upload image using httpwebrequest in c# - c#

I am trying to upload an image programmatically via an API to another server. The API expects me to upload image in a byte array to be sent in a field: "image_content".
My implementation and calling code is as below. The web request hits the server but the server responds that the image is not present in my web request.
When I run the below code, I am getting error that the image is not present in the request. What am I missing here?
public static class FormUpload
{
private static readonly Encoding encoding = Encoding.UTF8;
public static HttpWebResponse MultipartFormDataPost(string postUrl, string userAgent, Dictionary<string, object> postParameters)
{
string formDataBoundary = String.Format("----------{0:N}", Guid.NewGuid());
string contentType = "multipart/form-data; boundary=" + formDataBoundary;
byte[] formData = GetMultipartFormData(postParameters, formDataBoundary);
return PostForm(postUrl, userAgent, contentType, formData);
}
private static HttpWebResponse PostForm(string postUrl, string userAgent, string contentType, byte[] formData)
{
HttpWebRequest request = WebRequest.Create(postUrl) as HttpWebRequest;
if (request == null)
{
throw new NullReferenceException("request is not a http request");
}
// Set up the request properties.
request.Method = "POST";
request.ContentType = contentType;
request.UserAgent = userAgent;
request.ContentLength = formData.Length;
// Send the form data to the request.
using (Stream requestStream = request.GetRequestStream())
{
requestStream.Write(formData, 0, formData.Length);
requestStream.Close();
}
return request.GetResponse() as HttpWebResponse;
}
private static byte[] GetMultipartFormData(Dictionary<string, object> postParameters, string boundary)
{
Stream formDataStream = new System.IO.MemoryStream();
bool needsCLRF = false;
foreach (var param in postParameters)
{
if (param.Value is FileParameter)
{
FileParameter fileToUpload = (FileParameter)param.Value;
// Add just the first part of this param, since we will write the file data directly to the Stream
string header = string.Format("--{0}\r\nContent-Disposition: form-data; name=\"{1}\"; filename=\"{2}\"\r\nContent-Type: {3}\r\n\r\n",
boundary,
param.Key,
fileToUpload.FileName ?? param.Key,
fileToUpload.ContentType ?? "application/octet-stream");
formDataStream.Write(encoding.GetBytes(header), 0, encoding.GetByteCount(header));
// Write the file data directly to the Stream, rather than serializing it to a string.
formDataStream.Write(fileToUpload.File, 0, fileToUpload.File.Length);
}
else
{
string postData = string.Format("--{0}\r\nContent-Disposition: form-data; name=\"{1}\"\r\n\r\n{2}",
boundary,
param.Key,
param.Value);
formDataStream.Write(encoding.GetBytes(postData), 0, encoding.GetByteCount(postData));
}
}
// Add the end of the request. Start with a newline
string footer = "\r\n--" + boundary + "--\r\n";
formDataStream.Write(encoding.GetBytes(footer), 0, encoding.GetByteCount(footer));
// Dump the Stream into a byte[]
formDataStream.Position = 0;
byte[] formData = new byte[formDataStream.Length];
formDataStream.Read(formData, 0, formData.Length);
formDataStream.Close();
return formData;
}
public class FileParameter
{
public byte[] File { get; set; }
public string FileName { get; set; }
public string ContentType { get; set; }
public FileParameter(byte[] file) : this(file, null) { }
public FileParameter(byte[] file, string filename) : this(file, filename, null) { }
public FileParameter(byte[] file, string filename, string contenttype)
{
File = file;
FileName = filename;
ContentType = contenttype;
}
}
}
The code to call above function is:
// Read file data
FileStream fs = new FileStream("c:\\myimage.jpeg", FileMode.Open, FileAccess.Read);
byte[] data = new byte[fs.Length];
fs.Read(data, 0, data.Length);
fs.Close();
// Generate post objects
Dictionary<string, object> postParameters = new Dictionary<string, object>();
postParameters.Add("image_content",data);
// Create request and receive response
string postURL = "myurl";
string userAgent = "Mozilla";
HttpWebResponse webResponse = FormUpload.MultipartFormDataPost(postURL, userAgent, postParameters);
// Process response
StreamReader responseReader = new StreamReader(webResponse.GetResponseStream());
string fullResponse = responseReader.ReadToEnd();
webResponse.Close();
Response.Write(fullResponse);

In my opinion, you should use the MultipartFormDataContent class because it "Provides a container for content encoded using multipart/form-data MIME type.". Try this
public static HttpWebResponse MultipartFormDataPost(string postUrl, string userAgent, byte[] data)
{
string contentType;
byte[] formData = Program.GetMultipartFormData(data, out contentType);
return PostForm(postUrl, userAgent, contentType, formData);
}
public static byte[] GetMultipartFormData(byte[] data, out string contentType)
{
var byteArrayContent = new ByteArrayContent(data);
byteArrayContent.Headers.ContentType = new MediaTypeHeaderValue("image/jpeg");
byteArrayContent.Headers.Add("image_content", "myimage.jpeg");
var content = new MultipartFormDataContent(String.Format("----------{0:N}", Guid.NewGuid())) { byteArrayContent };
contentType = content.Headers.ContentType.ToString();
return content.ReadAsByteArrayAsync().Result;
}

I was able to resolve the problem by the use of RestSharp Api mentioned in the stackoverflow question Upload file through c# using JSON request and RestSharp.

You all code is fine but you forgot To encode you parameter
try this
string postData = string.Format("--{0}\r\nContent-Disposition:
form-data; name=\"{1}\"\r\n\r\n{2}",
boundary,
HttpUtility.UrlEncode(param.Key),
HttpUtility.UrlEncode(param.Value));
In case of binary data
HttpUtility.UrlEncode(Convert.ToBase64String(byte[]))
try to use this code to add parameter in your request
NameValueCollection outgoingQueryString = HttpUtility.ParseQueryString(String.Empty);
outgoingQueryString.Add("uname", "username");
outgoingQueryString.Add("pname", "password");
string postdata = outgoingQueryString.ToString();
and write this postdata in you request

Related

Azure - C# - Error while uploading image using Rest API

I'm trying to upload an image to Azure storage container.
Using UploadBlobWithRestAPI, I'm trying to call the rest PUT API and using AuthorizationHeader, I'm trying to create the Authorization for the API call.
I'm getting the image created in the container. But, for some reason the image is not in the readable format. On downloading the image from container and trying to open in explorer, am getting "It appears that we don't support this file format". Any idea?
public void UploadBlobWithRestAPI(string MachineName,string ImageName)
{
string storageKey = "Storagekey";
string storageAccount = "Account";
string containerName = "test-container";
string blobName = "test2.jpg";
string method = "PUT";
Byte[] imageContentBytes = System.IO.File.ReadAllBytes(#"C:\\Test2.jpg");
int imageContentLength = (Encoding.UTF8.GetString(imageContentBytes)).Length;
string requestUri = $"https://{storageAccount}.blob.core.windows.net/{containerName}/{blobName}";
System.Net.HttpWebRequest request = (HttpWebRequest)WebRequest.Create(requestUri);
string now = DateTime.UtcNow.ToString("R");
request.Method = method;
request.ContentType = "application/octet-stream";
request.ContentLength = imageContentLength;
request.Headers.Add("x-ms-version", "2015-12-11");
request.Headers.Add("x-ms-date", now);
request.Headers.Add("x-ms-blob-type", "BlockBlob");
request.Headers.Add("Authorization", AuthorizationHeader(method, now, request, storageAccount, storageKey, containerName, blobName));
using (Stream requestStream = request.GetRequestStream())
{
requestStream.Write(Encoding.UTF8.GetBytes(Encoding.UTF8.GetString(imageContentBytes)), 0, imageContentLength);
}
using (HttpWebResponse resp = (HttpWebResponse)request.GetResponse())
{
MessageBox.Show(resp.StatusCode.ToString());
}
}
public string AuthorizationHeader(string method, string now, HttpWebRequest request, string storageAccount, string storageKey, string containerName, string blobName)
{
string headerResource = $"x-ms-blob-type:BlockBlob\nx-ms-date:{now}\nx-ms-version:2015-12-11";
string urlResource = $"/{storageAccount}/{containerName}/{blobName}";
string stringToSign = $"{method}\n\n\n{request.ContentLength}\n\n{request.ContentType}\n\n\n\n\n\n\n{headerResource}\n{urlResource}";
HMACSHA256 hmac = new HMACSHA256(Convert.FromBase64String(storageKey));
string signature = Convert.ToBase64String(hmac.ComputeHash(Encoding.UTF8.GetBytes(stringToSign)));
String AuthorizationHeader = String.Format("{0} {1}:{2}", "SharedKey", storageAccount, signature);
return AuthorizationHeader;
}
I believe the issue is coming because you are uploading an image (which is binary content) as a string which is corrupting the data.
Please try by changing the following:
int imageContentLength = (Encoding.UTF8.GetString(imageContentBytes)).Length;
to
int imageContentLength = imageContentBytes.Length;
and
using (Stream requestStream = request.GetRequestStream())
{
requestStream.Write(Encoding.UTF8.GetBytes(Encoding.UTF8.GetString(imageContentBytes)), 0, imageContentLength);
}
to
using (Stream requestStream = request.GetRequestStream())
{
requestStream.Write(imageContentBytes, 0, imageContentLength);
}
Also, please set the proper value for content type. Considering the file you are uploading is a JPEG image, please set the content type to image/jpeg instead of application/octet-stream. This will ensure that the image will be displayed correctly when loaded in a browser.

How to properly encode binary data to send over REST api PUT call

I'm trying to simply upload an image file to Azure blob store and the file is making it there but the file size is bigger than it should be and when I download the file via the browser it won't open as an image.
I manually uploaded the same file via the Azure interface and the file works and is 22k, uploading via my code is 29k and doesn't work.
NOTE: One thing to note is that all the examples used with this code only send text files. Maybe the Encoding.UTF8.GetBytes(requestBody); line is doing it?
This is where I'm Base64 encoding my data
HttpPostedFileBase hpf = Request.Files[file] as HttpPostedFileBase
byte[] binData;
using (BinaryReader b = new BinaryReader(hpf.InputStream))
binData = b.ReadBytes(hpf.ContentLength);
var result = System.Convert.ToBase64String(binData);
new BlobHelper("STORENAMEHERE", "SECRETKEY").PutBlob("images", "test.jpg", result);
This is what I'm using to PUT the data, it's from https://azurestoragesamples.codeplex.com/
public bool PutBlob(string container, string blob, string content)
{
return Retry<bool>(delegate()
{
HttpWebResponse response;
try
{
SortedList<string, string> headers = new SortedList<string, string>();
headers.Add("x-ms-blob-type", "BlockBlob");
response = CreateRESTRequest("PUT", container + "/" + blob, content, headers).GetResponse() as HttpWebResponse;
response.Close();
return true;
}
catch (WebException ex)
{
if (ex.Status == WebExceptionStatus.ProtocolError &&
ex.Response != null &&
(int)(ex.Response as HttpWebResponse).StatusCode == 409)
return false;
throw;
}
});
}
public HttpWebRequest CreateRESTRequest(string method, string resource, string requestBody = null, SortedList<string, string> headers = null,
string ifMatch = "", string md5 = "")
{
byte[] byteArray = null;
DateTime now = DateTime.UtcNow;
string uri = Endpoint + resource;
HttpWebRequest request = HttpWebRequest.Create(uri) as HttpWebRequest;
request.Method = method;
request.ContentLength = 0;
request.Headers.Add("x-ms-date", now.ToString("R", System.Globalization.CultureInfo.InvariantCulture));
request.Headers.Add("x-ms-version", "2009-09-19");
if (headers != null)
{
foreach (KeyValuePair<string, string> header in headers)
request.Headers.Add(header.Key, header.Value);
}
if (!String.IsNullOrEmpty(requestBody))
{
request.Headers.Add("Accept-Charset", "UTF-8");
byteArray = Encoding.UTF8.GetBytes(requestBody);
request.ContentLength = byteArray.Length;
}
request.Headers.Add("Authorization", AuthorizationHeader(method, now, request, ifMatch, md5));
if (!String.IsNullOrEmpty(requestBody))
request.GetRequestStream().Write(byteArray, 0, byteArray.Length);
return request;
}
I believe the problem is with how you're converting the binary data to string and then converting it back to byte array.
While converting binary data to string, you're using System.Convert.ToBase64String however while getting the byte array from string you're using Encoding.UTF8.GetBytes. Most likely this is causing the mismatch.
Please change the following line of code:
byteArray = Encoding.UTF8.GetBytes(requestBody);
to
byteArray = System.Convert.FromBase64String(requestBody);
and that should fix the problem.

How to passing variables to PHP Script [POST] using C#

I am using this code :
public void sendPostData(string url, string data)
{
WebRequest req = WebRequest.Create(url);
req.Method = "POST";
req.ContentType = "application/x-www-form-urlencoded";
byte[] bytes = Encoding.UTF8.GetBytes(data);
req.ContentLength = bytes.Length;
Stream dataStream = req.GetRequestStream();
dataStream.Write(bytes, 0, bytes.Length);
dataStream.Close();
}
PHP for POST processing :
<?php
if(#$_POST['filename'])
{
$data = filter_var($_POST['filename'], FILTER_SANITIZE_STRING);
$f = fopen($data.".txt", "w");
fclose($f);
}
?>
I am calling by this :
sendPostData("http://127.0.0.1/csharptest/index.php", "filename=myvariablehere");
So, all it does is create a file name "myvariablehere" on server instead of storing value in myvariablehere which holds the data.
I Want "myvariablehere" value stored on server.
Please lil help here !
Thanks
Try this code
public static class Upload
{
public static byte[] Post(string uri, NameValueCollection pairs)
{
byte[] response = null;
using (WebClient client = new WebClient())
{
response = client.UploadValues(uri, pairs);
}
return response;
}
}
and then simply call this
var response = Upload.Post("URL", new NameValueCollection() {
{ "fileName", "test" },
});

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());

httpmultipart request for this webservice getdata code in c# with android

HI i have a webservice IIS with this code:
// Implements multipart/form-data POST in C# http://www.ietf.org/rfc/rfc2388.txt
// http://www.briangrinstead.com/blog/multipart-form-post-in-c
public static class FormUpload
{
private static readonly Encoding encoding = Encoding.UTF8;
public static HttpWebResponse MultipartFormDataPost(string postUrl, string userAgent, Dictionary<string, object> postParameters)
{
string formDataBoundary = String.Format("----------{0:N}", Guid.NewGuid());
string contentType = "multipart/form-data; boundary=" + formDataBoundary;
byte[] formData = GetMultipartFormData(postParameters, formDataBoundary);
return PostForm(postUrl, userAgent, contentType, formData);
}
private static HttpWebResponse PostForm(string postUrl, string userAgent, string contentType, byte[] formData)
{
HttpWebRequest request = WebRequest.Create(postUrl) as HttpWebRequest;
if (request == null)
{
throw new NullReferenceException("request is not a http request");
}
// Set up the request properties.
request.Method = "POST";
request.ContentType = contentType;
request.UserAgent = userAgent;
request.CookieContainer = new CookieContainer();
request.ContentLength = formData.Length;
// You could add authentication here as well if needed:
// request.PreAuthenticate = true;
// request.AuthenticationLevel = System.Net.Security.AuthenticationLevel.MutualAuthRequested;
// request.Headers.Add("Authorization", "Basic " + Convert.ToBase64String(System.Text.Encoding.Default.GetBytes("username" + ":" + "password")));
// Send the form data to the request.
using (Stream requestStream = request.GetRequestStream())
{
requestStream.Write(formData, 0, formData.Length);
requestStream.Close();
}
return request.GetResponse() as HttpWebResponse;
}
private static byte[] GetMultipartFormData(Dictionary<string, object> postParameters, string boundary)
{
Stream formDataStream = new System.IO.MemoryStream();
bool needsCLRF = false;
foreach (var param in postParameters)
{
// Thanks to feedback from commenters, add a CRLF to allow multiple parameters to be added.
// Skip it on the first parameter, add it to subsequent parameters.
if (needsCLRF)
formDataStream.Write(encoding.GetBytes("\r\n"), 0, encoding.GetByteCount("\r\n"));
needsCLRF = true;
if (param.Value is FileParameter)
{
FileParameter fileToUpload = (FileParameter)param.Value;
// Add just the first part of this param, since we will write the file data directly to the Stream
string header = string.Format("--{0}\r\nContent-Disposition: form-data; name=\"{1}\"; filename=\"{2}\";\r\nContent-Type: {3}\r\n\r\n",
boundary,
param.Key,
fileToUpload.FileName ?? param.Key,
fileToUpload.ContentType ?? "application/octet-stream");
formDataStream.Write(encoding.GetBytes(header), 0, encoding.GetByteCount(header));
// Write the file data directly to the Stream, rather than serializing it to a string.
formDataStream.Write(fileToUpload.File, 0, fileToUpload.File.Length);
}
else
{
string postData = string.Format("--{0}\r\nContent-Disposition: form-data; name=\"{1}\"\r\n\r\n{2}",
boundary,
param.Key,
param.Value);
formDataStream.Write(encoding.GetBytes(postData), 0, encoding.GetByteCount(postData));
}
}
// Add the end of the request. Start with a newline
string footer = "\r\n--" + boundary + "--\r\n";
formDataStream.Write(encoding.GetBytes(footer), 0, encoding.GetByteCount(footer));
// Dump the Stream into a byte[]
formDataStream.Position = 0;
byte[] formData = new byte[formDataStream.Length];
formDataStream.Read(formData, 0, formData.Length);
formDataStream.Close();
return formData;
}
public class FileParameter
{
public byte[] File { get; set; }
public string FileName { get; set; }
public string ContentType { get; set; }
public FileParameter(byte[] file) : this(file, null) { }
public FileParameter(byte[] file, string filename) : this(file, filename, null) { }
public FileParameter(byte[] file, string filename, string contenttype)
{
File = file;
FileName = filename;
ContentType = contenttype;
}
}
}
Obtain from here: Multipart forms from C# client
But now, in mi android aplicattion i have this:
HttpParams httpParameters = new BasicHttpParams();
int timeoutConnection = 300000;
HttpConnectionParams.setConnectionTimeout(httpParameters, timeoutConnection);
int timeoutSocket = 300000;
HttpConnectionParams.setSoTimeout(httpParameters, timeoutSocket);
HttpClient httpClient = new DefaultHttpClient(httpParameters);
HttpPost postRequest = new HttpPost("http://172.21.1.87:9999/Service1.svc");
ByteArrayBody bab = new ByteArrayBody(ficheroAEnviar, "prueba.jpg");
MultipartEntity reqEntity = new MultipartEntity();
postRequest.addHeader("Content-Type", " multipart/form-data");
reqEntity.addPart("Dictionary", new FileBody(new File(fileUri.toString(), "application/zip")));
reqEntity.addPart("boundary", new StringBody("envio"));
postRequest.setEntity(reqEntity);
HttpResponse response = httpClient.execute(postRequest);
BufferedReader reader = new BufferedReader(new InputStreamReader(response.getEntity().getContent(), "UTF-8"));
String sResponse;
StringBuilder s = new StringBuilder();
postRequest.getAllHeaders();
while ((sResponse = reader.readLine()) != null) {
s = s.append(sResponse);
}
System.out.println("Response: " + s);
}
catch (Exception e) {
// handle exception here
Log.e(e.getClass().getName(), e.getMessage());
}
}
But it return me Bad request 400 error. Can i use the c# code to upload a multipartentity file with android or a need to make any change in some code?
Anyone have a example en c# and android to make this? Is for uppload videos about 10 or 15 MB.
Thanks
dont use multipart entity instead use as here you define the bounderies it is easy
try it
HttpURLConnection connection = null;
DataOutputStream outputStream = null;
DataInputStream inputStream = null;
String pathToOurFile = "/data/file_to_send.mp3";
String urlServer = "http://192.168.1.1/handle_upload.php";
String lineEnd = "\r\n";
String twoHyphens = "--";
String boundary = "*****";
int bytesRead, bytesAvailable, bufferSize;
byte[] buffer;
int maxBufferSize = 1*1024*1024;
try
{
FileInputStream fileInputStream = new FileInputStream(new File(pathToOurFile) );
URL url = new URL(urlServer);
connection = (HttpURLConnection) url.openConnection();
// Allow Inputs & Outputs
connection.setDoInput(true);
connection.setDoOutput(true);
connection.setUseCaches(false);
// Enable POST method
connection.setRequestMethod("POST");
connection.setRequestProperty("Connection", "Keep-Alive");
connection.setRequestProperty("Content-Type", "multipart/form-data;boundary="+boundary);
outputStream = new DataOutputStream( connection.getOutputStream() );
outputStream.writeBytes(twoHyphens + boundary + lineEnd);
outputStream.writeBytes("Content-Disposition: form-data; name=\"uploadedfile\";filename=\""+pathToOurFile +"\"" + lineEnd);
outputStream.writeBytes(lineEnd);
bytesAvailable = fileInputStream.available();
bufferSize = Math.min(bytesAvailable, maxBufferSize);
buffer = new byte[bufferSize];
// Read file
bytesRead = fileInputStream.read(buffer, 0, bufferSize);
while (bytesRead > 0)
{
outputStream.write(buffer, 0, bufferSize);
bytesAvailable = fileInputStream.available();
bufferSize = Math.min(bytesAvailable, maxBufferSize);
bytesRead = fileInputStream.read(buffer, 0, bufferSize);
}
outputStream.writeBytes(lineEnd);
outputStream.writeBytes(twoHyphens + boundary + twoHyphens + lineEnd);
// Responses from the server (code and message)
serverResponseCode = connection.getResponseCode();
serverResponseMessage = connection.getResponseMessage();
fileInputStream.close();
outputStream.flush();
outputStream.close();
}
catch (Exception ex)
{
//Exception handling
}
courtesy rafel

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