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Returning a zipfile from a web api

I have built an asp net web api. I need to return a zipfile, as a result of some inner logic. I'm using this code and it works, but the resulting zip file, when unzipped manually, gave me this error "There are data after the end of the payload"
using (ZipFile zip = new ZipFile())
{
...
zip.Save(di.FullName + "\\" + "Update.zip");
}
string path = Path.Combine(Properties.Settings.Default.PathDisposizioniHTML, "Update.zip");
var response = new HttpResponseMessage(HttpStatusCode.OK);
var stream = new System.IO.FileStream(path, System.IO.FileMode.Open);
response.Content = new StreamContent(stream);
response.Content.Headers.ContentType = new System.Net.Http.Headers.MediaTypeHeaderValue("application/octet-stream");
This is how i receive the data in a .net console application:
using (Stream output = File.OpenWrite(#"C:\prova\MyFile.zip"))
using (Stream input = httpResponse.GetResponseStream())
{
input.CopyTo(output);
}

If you already have the zip file on your system, you shouldn't need to do anything special before sending it as a response.
This should work:
string filePath = #"C:\myfolder\myfile.zip";
return File(filePath, "application/zip");
If you're making the file on the fly, i.e. getting other files and programatically putting them into a zip file for the user, the following should work:
public IActionResult GetZipFile(){
//location of the file you want to compress
string filePath = #"C:\myfolder\myfile.ext";
//name of the zip file you will be creating
string zipFileName = "zipFile.zip";
byte[] result;
using (MemoryStream zipArchiveMemoryStream = new MemoryStream())
{
using (ZipArchive zipArchive = new ZipArchive(zipArchiveMemoryStream, ZipArchiveMode.Create, true))
{
ZipArchiveEntry zipEntry = zipArchive.CreateEntry(zipFileName);
using (Stream entryStream = zipEntry.Open())
{
using (MemoryStream tmpMemory = new MemoryStream(System.IO.File.ReadAllBytes(filePath)))
{
tmpMemory.CopyTo(entryStream);
};
}
}
zipArchiveMemoryStream.Seek(0, SeekOrigin.Begin);
result = zipArchiveMemoryStream.ToArray();
}
return File(result, "application/zip", zipFileName);
}
This is taken from a recent ASP.NET project of my own.

stream.ReadTimeout' threw an exception of type 'System.InvalidOperationException'

I am trying to download a zip file from SFTP and unzip in the memory to process the file
I am using SSH.Net to download the file.
private static void processfilesfromftp(List<TSOracleMicrosDownLoadSetUp> list)
{
SftpClient sftp = HelperFunctions.GetClientConnection();
if(sftp.IsConnected)
{
var files = sftp.ListDirectory("/");
ZipFile zips = new ZipFile();
string path = string.Empty;
foreach(var file in files)
{
Stream unzippedEntryStream = new MemoryStream();
path = string.Format("/{0}", file.Name);
//byte[] arr = sftp.ReadAllBytes(file.FullName);
var stream = new BufferedStream(sftp.OpenRead(file.FullName));
//System.IO.TextReader textReader = new System.IO.StreamReader(stream);
//System.IO.MemoryStream mStream = new MemoryStream();
using (ZipFile zip = ZipFile.Read(stream))
{
ZipEntry e = zip[0];
e.Extract(unzippedEntryStream);
System.IO.TextReader textReader = new System.IO.StreamReader(unzippedEntryStream);
string data = textReader.ReadToEnd();
}
}
}
}
memorystream throw error System.InvalidOperationException exception at
var stream = new BufferedStream(sftp.OpenRead(file.FullName));
Update
It is not throwing any error, but the final output of the unzip file is empty.
Using Framework 4.5.2 and Visual studio 2017

This is more a SSH.Net question and not specific Acumatica.
It seems the problem is related to the SSH connection.
To change the timeout you can use SshClient.ConnectionInfo.Timeout. But you need to catch the exception and handle it gracefully.
Here is a post with a similar issue.
BTW, you could use the included Acumatica library to read the zip file.

I think you are not writing the file from FTP to the memory stream so it's empty.
Try using the DownloadFile method from SSH.Net to write file content in the stream.
Reference:
https://stackoverflow.com/a/46907346/7376238
SftpClient _sftpClient;
_sftpClient = new SftpClient("sftp.server.domain", "MyLoginHere", "MyPasswordHere");
Stream fileBody = new MemoryStream();
_sftpClient.DownloadFile(ftpFile.FullName, fileBody);
fileBody.Position = 0;

'System.OutOfMemoryException' was thrown in ZipArchive for large file from URL

I have an URL which contains a zip file. The files need to be unzipped from the URL. The URL is Opened and Read using webclient and then added to a Stream. It is then used in the ZipArchive object which will unzip the files and store them in the D:\ drive. When a file is around 400Mb I get the 'System.OutOfMemoryException'.
Stream has to be used since the webClient.OpenRead(Uri Address) returns a Stream. As well as the use ZipArchive(Stream stream).
How can I stop from getting this message?
string zipFileUrl = "https://www.dropbox.com/s/clersbjdcshpdy6/oversize_zip_test_0.zip?dl=0"
string output_path = #"D:\";
using (WebClient webClient = new WebClient())
{
using (Stream streamFile = webClient.OpenRead(zipFileUrl))
{
using (ZipArchive archive = new ZipArchive(streamFile))//ERROR HERE
{
var entries = archive.Entries;
//Loops thru each file in Zip and adds it to directory
foreach (var entry in entries)
{
if (entry.FullName != "/" && entry.Name != "")
{
string completeFileName = Path.Combine(output_path, entry.FullName);
string directory = Path.GetDirectoryName(completeFileName);
//If directory does not exist then we create it.
if (!Directory.Exists(directory))
{
Directory.CreateDirectory(directory);
}
//Extracts zip from URL to extract path, and overwrites if file exists.
entry.ExtractToFile(completeFileName, true);
}
}
}
}

I think here might be your problem, from the ZipArchive.Init method
private void Init(Stream stream, ZipArchiveMode mode, Boolean leaveOpen)
{
Stream extraTempStream = null;
try
{
_backingStream = null;
//check stream against mode
switch (mode)
{
case ZipArchiveMode.Create:
// (SNIP)
case ZipArchiveMode.Read:
if (!stream.CanRead)
throw new ArgumentException(SR.ReadModeCapabilities);
if (!stream.CanSeek)
{
_backingStream = stream;
extraTempStream = stream = new MemoryStream();
_backingStream.CopyTo(stream);
stream.Seek(0, SeekOrigin.Begin);
}
break;
case ZipArchiveMode.Update:
// (SNIP)
default:
// (SNIP)
}
// (SNIP)
}
if streamFile.CanSeek is false (which from a WebClient it will be) it copies the entire file in to memory then works on the file. This is what is using up all the memory.
Try to find a 3rd party library that handles Zip files and does not need a stream that supports seeking. If you can't, copy the file to disk first to the temp folder with a FileStream with the FileOptions.DeleteOnClose option passed in, then use that stream in your zip before you close the stream.
string zipFileUrl = "https://www.dropbox.com/s/clersbjdcshpdy6/oversize_zip_test_0.zip?dl=0";
string output_path = #"D:\";
using (var tempFileStream = new FileStream(Path.GetTempFileName(), FileMode.Create,
FileAccess.ReadWrite, FileShare.None,
4096, FileOptions.DeleteOnClose))
{
using (WebClient webClient = new WebClient())
{
using (Stream streamFile = webClient.OpenRead(zipFileUrl))
{
streamFile.CopyTo(tempFileStream);
}
}
tempFileStream.Position = 0;
using (ZipArchive archive = new ZipArchive(tempFileStream))
{
var entries = archive.Entries;
//Loops thru each file in Zip and adds it to directory
foreach (var entry in entries)
{
if (entry.FullName != "/" && entry.Name != "")
{
string completeFileName = Path.Combine(output_path, entry.FullName);
string directory = Path.GetDirectoryName(completeFileName);
//If directory does not exist then we create it.
if (!Directory.Exists(directory))
{
Directory.CreateDirectory(directory);
}
//Extracts zip from URL to extract path, and overwrites if file exists.
entry.ExtractToFile(completeFileName, true);
}
}
}
}

C# FTP The given path's format is not supported

This is my code
I have an FTP with many zip files. and each zip file has a XML with the same name.
I want to parse these xml files.
What i did is this:
get a list of all zip files in the FTP and save the names in this variable directories.
Now I want to open each zip file, which its name is in the directories list. I did this.
foreach (string fileNameInFTP in directories)
{
}
Now to read the content of that zip file, I tried this:.
string fileName = FTPAddress + fileNameInFTP;
using (var file = File.OpenRead(fileName))
using (var zip = new ZipArchive(file, ZipArchiveMode.Read))
{
foreach (var entry in zip.Entries)
{
using (var stream = entry.Open())
{
// do whatever we want with stream
// ...
}
}
}
I got this exception The given path's format is not supported. on this line:
using (var file = File.OpenRead("ftp://" +FTPAddress +"/" + fileNameInFTP)) could u help please

You should use something like this instead of trying to use File.OpenRead for remote FTP file download.
http://msdn.microsoft.com/en-us/library/ms229711%28v=vs.110%29.aspx
// Get the object used to communicate with the server.
FtpWebRequest request = (FtpWebRequest)WebRequest.Create("ftp://www.contoso.com/test.htm");
request.Method = WebRequestMethods.Ftp.DownloadFile;
// This example assumes the FTP site uses anonymous logon.
request.Credentials = new NetworkCredential ("anonymous","janeDoe#contoso.com");
FtpWebResponse response = (FtpWebResponse)request.GetResponse();
Stream responseStream = response.GetResponseStream();
using (var zip = new ZipArchive(responseStream , ZipArchiveMode.Read))
{
//Loops through each file in the zip that has the ".xml" extension
foreach (var entry in zip.Entries.Where(x=> (Path.GetExtension(x.Name) ?? "").ToLower() ==".xml"))
{
using (var stream = entry.Open())
{
//Load xml file and do whatever you like with it.
var xmlDocument = XDocument.Load(stream);
}
}
}
Console.WriteLine("Download Complete, status {0}", response.StatusDescription);
response.Close();

you can't use File IO to open FTP stream, here is a sample of how we can open FTP using WebRequest in .NET:
private static void Main(string[] args)
{
var ftp = WebRequest.Create(#"ftp://ftp.microsoft.com/softlib/MSLFILES/aspwebwiz2k.zip");
//ftp.Credentials=new NetworkCredential("anonymous","anonymous");
var response=ftp.GetResponse();
var stream=response.GetResponseStream();
var ms = ToMemoryStream(stream);
var archive = new ZipArchive(ms, ZipArchiveMode.Read);
var entry=archive.GetEntry("file name here");
var doc=XDocument.Load(entry.Open());
}
public static MemoryStream ToMemoryStream( Stream stream)
{
var memoryStream = new MemoryStream();
var buffer = new byte[4096];
while (true)
{
var readCount = stream.Read(buffer, 0, buffer.Length);
if (readCount == 0)
break;
memoryStream.Write(buffer, 0, readCount);
}
memoryStream.Position = 0;
return memoryStream;
}

File upload in asp.net c#

hey guys, m using an api of "Bits on the Run" following is the code of upload API
public string Upload(string uploadUrl, NameValueCollection args, string filePath)
{
_queryString = args; //no required args
WebClient client = createWebClient();
_queryString["api_format"] = APIFormat ?? "xml"; //xml if not specified - normally set in required args routine
queryStringToArgs();
string callUrl = _apiURL + uploadUrl + "?" + _args;
callUrl = uploadUrl + "?" + _args;
try {
byte[] response = client.UploadFile(callUrl, filePath);
return Encoding.UTF8.GetString(response);
} catch {
return "";
}
}
and below is my code to upload a file, m using FileUpload control to get the full path of a file(but m not succeeded in that)...
botr = new BotR.API.BotRAPI("key", "secret_code");
var response = doc.Descendants("link").FirstOrDefault();
string url = string.Format("{0}://{1}{2}", response.Element("protocol").Value, response.Element("address").Value, response.Element("path").Value);
//here i want fullpath of the file, how can i achieve that here
string filePath = fileUpload.PostedFile.FileName;//"C://Documents and Settings//rkrishna//My Documents//Visual Studio 2008//Projects//BitsOnTheRun//BitsOnTheRun//rough_test.mp4";
col = new NameValueCollection();
FileStream fs = new FileStream(filePath, FileMode.Open);
col["file_size"] = fs.Length.ToString();
col["file_md5"] = BitConverter.ToString(HashAlgorithm.Create("MD5").ComputeHash(fs)).Replace("-", "").ToLower();
col["key"] = response.Element("query").Element("key").Value;
col["token"] = response.Element("query").Element("token").Value;
fs.Dispose();
string uploadResponse = botr.Upload(url, col, filePath);
i read in some forums saying that for some security purpose you can't get fullpath of a file from client side. If it is true then how can i achieve file upload in my scenario ?

Yes, this is true, for security reason you cannot get the fullpath of the client machine, what you can do is, try the following,
Stream stream = fileUpload.PostedFile.InputStream;
stream.Read(bytes, 0, fileUpload.PostedFile.ContentLength);
instead of creating your own FileStream use the stream provided by the FileUploadControl. Hoep it shall help.