ZipArchive won't modify a memory stream - c#

I'm trying to download a zip file and modify it before returning it. I expect the stream to be modified after adding additional files with ZipArchive in Update mode. However, it stays the same. What am I doing wrong?
using (WebClient webClient = new WebClient())
{
string url = "http://www.dynaexamples.com/examples-manual/ls-dyna_example.zip/at_download/file";
byte[] downloadedData = webClient.DownloadData(url);
using (MemoryStream stream = new MemoryStream())
{
stream.Write(downloadedData, 0, downloadedData.Length);
Console.WriteLine(stream.Length.ToString()); //911616
ZipArchive archive = new ZipArchive(stream, ZipArchiveMode.Update);
ZipArchiveEntry testFile = archive.CreateEntry("test.txt");
using (StreamWriter writer = new StreamWriter(testFile.Open()))
{
writer.WriteLine("test");
writer.Flush();
}
Console.WriteLine(stream.Length.ToString()); //911616
}
}

Related

Invalid zip file reading from a Stream in C#

I have the following code:
private static byte[] ConverterStringToByte(Stream body)
{
string fileName = "data_" + DateTime.Now.ToString("yyyyMMddhhmmss") + ".zip";
// Take out the bytes from the memory stream and safely close the stream
using (var ms = new MemoryStream())
{
body.CopyTo(ms);
using (var zipArchive = new ZipArchive(ms, ZipArchiveMode.Create, false))
{
var zipEntry = zipArchive.CreateEntry(fileName, CompressionLevel.Optimal);
using (BinaryWriter writer = new BinaryWriter(zipEntry.Open()))
{
ms.Position = 0;
writer.Write(ms.ToArray());
}
}
return ms.ToArray();
}
}
I am downloading the file successfully, however I'm getting
invalid file
when trying to open
I think it should be something like this. Not sure about fileName though, because it's the name of the file being put into archive, so I don't think it should have *.zip extension. Unless you are creating a zip of zips.
static byte[] ConverterStringToByte(Stream body)
{
string fileName = #"data_" + DateTime.Now.ToString("yyyyMMddhhmmss") + ".zip";
using (var ms = new MemoryStream())
{
using (var zipArchive = new ZipArchive(ms, ZipArchiveMode.Create, false))
{
var zipEntry = zipArchive.CreateEntry(fileName, CompressionLevel.Optimal);
using (var destStream = zipEntry.Open())
{
body.CopyTo(destStream);
}
}
return ms.ToArray();
}
}

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.

Create zip file in memory from bytes (text with arbitrary encoding)

The application i'm developing needs to compress xml files into zip files and send them through http requests to a web service. As I dont need to keep the zip files, i'm just performing the compression in memory. The web service is denying my requests because the zip files are apparently malformed.
I know there is a solution in this question which works perfectly, but it uses a StreamWriter. My problem with that solution is that StreamWriter requires an encoding or assumes UTF-8, and I do not need to know the enconding of the xml files. I just need to read the bytes from those files, and store them inside a zip file, whatever encoding they use.
So, to be clear, this question has nothing to do with encodings, as I don't need to transform the bytes into text or the oposite. I just need to compress a byte[].
I'm using the next code to test how my zip file is malformed:
static void Main(string[] args)
{
Encoding encoding = Encoding.GetEncoding("ISO-8859-1");
string xmlDeclaration = "<?xml version=\"1.0\" encoding=\"" + encoding.WebName.ToUpperInvariant() + "\"?>";
string xmlBody = "<Test>ª!\"·$%/()=?¿\\|##~€¬'¡º</Test>";
string xmlContent = xmlDeclaration + xmlBody;
byte[] bytes = encoding.GetBytes(xmlContent);
string fileName = "test.xml";
string zipPath = #"C:\Users\dgarcia\test.zip";
Test(bytes, fileName, zipPath);
}
static void Test(byte[] bytes, string fileName, string zipPath)
{
byte[] zipBytes;
using (var memoryStream = new MemoryStream())
using (var zipArchive = new ZipArchive(memoryStream, ZipArchiveMode.Create, leaveOpen: false))
{
var zipEntry = zipArchive.CreateEntry(fileName);
using (Stream entryStream = zipEntry.Open())
{
entryStream.Write(bytes, 0, bytes.Length);
}
//Edit: as the accepted answer states, the problem is here, because i'm reading from the memoryStream before disposing the zipArchive.
zipBytes = memoryStream.ToArray();
}
using (var fileStream = new FileStream(zipPath, FileMode.OpenOrCreate))
{
fileStream.Write(zipBytes, 0, zipBytes.Length);
}
}
If I try to open that file, I get an "Unexpected end of file" error. So apparently, the web service is correctly reporting a malformed zip file. What I have tried so far:
Flushing the entryStream.
Closing the entryStream.
Both flushing and closing the entryStream.
Note that if I open the zipArchive directly from the fileStream the zip file is formed with no errors. However, the fileStream is just there as a test, and I need to create my zip file in memory.
You are trying to get bytes from MemoryStream too early, ZipArchive did not write them all yet. Instead, do like this:
using (var memoryStream = new MemoryStream()) {
// note "leaveOpen" true, to not dispose memoryStream too early
using (var zipArchive = new ZipArchive(memoryStream, ZipArchiveMode.Create, leaveOpen: true)) {
var zipEntry = zipArchive.CreateEntry(fileName);
using (Stream entryStream = zipEntry.Open()) {
entryStream.Write(bytes, 0, bytes.Length);
}
}
// now, after zipArchive is disposed - all is written to memory stream
zipBytes = memoryStream.ToArray();
}
If you use a memory stream to load your text you can control the encoding type and it works across a WCF service. This is the implementation i am using currently and it works on my WCF services
private byte[] Zip(string text)
{
var bytes = Encoding.UTF8.GetBytes(text);
using (var msi = new MemoryStream(bytes))
using (var mso = new MemoryStream())
{
using (var gs = new GZipStream(mso, CompressionMode.Compress))
{
CopyTo(msi, gs);
}
return mso.ToArray();
}
}
private string Unzip(byte[] bytes)
{
using (var msi = new MemoryStream(bytes))
using (var mso = new MemoryStream())
{
using (var gs = new GZipStream(msi, CompressionMode.Decompress))
{
CopyTo(gs, mso);
}
return Encoding.UTF8.GetString(mso.ToArray());
}
}

Using MemoryStream and DotNetZip to zip a json file

I have a JSON file created, and I am going to zip it using DotNetZip.
Using with StreamWriter to zip it is working, if I try to use MemoryStream it will not working.
StreamWriter :
sw = new StreamWriter(assetsFolder + #"manifest.json");
sw.Write(strManifest);
sw.Close();
zip.AddFile(Path.Combine(assetsFolder, "manifest.json"), "/");
zip.AddFile(Path.Combine(assetsFolder, "XXXXXXX"), "/");
zip.Save(outputStream);
MemoryStream :
var manifestStream = GenerateStreamFromString(strManifest);
public static Stream GenerateStreamFromString(string s)
{
MemoryStream stream = new MemoryStream();
StreamWriter writer = new StreamWriter(stream);
writer.Write(s);
writer.Flush();
stream.Position = 0;
return stream;
}
zip.AddEntry("manifest.json", manifestStream);
zip.AddFile(Path.Combine(assetsFolder, "XXXXXXX"), "/");
zip.Save(outputStream);
I must using the .JSON file type to zip it, Can any one told me where have a mistake?
To create a Gzipped Json you need to use GZipStream. Try method below.
https://www.dotnetperls.com/gzipstream
GZipStream compresses data. It saves data efficiently—such as in
compressed log files. We develop a utility method in the C# language
that uses the System.IO.Compression namespace. It creates GZIP files.
It writes them to the disk.
public static void CompressStringToFile(string fileName, string value)
{
// A.
// Write string to temporary file.
string temp = Path.GetTempFileName();
File.WriteAllText(temp, value);
// B.
// Read file into byte array buffer.
byte[] b;
using (FileStream f = new FileStream(temp, FileMode.Open))
{
b = new byte[f.Length];
f.Read(b, 0, (int)f.Length);
}
// C.
// Use GZipStream to write compressed bytes to target file.
using (FileStream f2 = new FileStream(fileName, FileMode.Create))
using (GZipStream gz = new GZipStream(f2, CompressionMode.Compress, false))
{
gz.Write(b, 0, b.Length);
}
}

How to save downloaded files in MemoryStream when using SSH.NET

I am using SSH.NET library to download files. I want to save the downloaded file as a file in memory, rather than a file on disk but it is not happening.
This is my code which works fine:
using (var sftp = new SftpClient(sFTPServer, sFTPPassword, sFTPPassword))
{
sftp.Connect();
sftp.DownloadFile("AFile.txt", System.IO.File.Create("AFile.txt"));
sftp.Disconnect();
}
and this is the code which doesn't work fine as it gives 0 bytes stream.
using (var sftp = new SftpClient(sFTPServer, sFTPPassword, sFTPPassword))
{
sftp.Connect();
System.IO.MemoryStream mem = new System.IO.MemoryStream();
System.IO.TextReader textReader = new System.IO.StreamReader(mem);
sftp.DownloadFile("file.txt", mem);
System.IO.TextReader textReader = new System.IO.StreamReader(mem);
string s = textReader.ReadToEnd(); // it is empty
sftp.Disconnect();
}
You can try the following code, which opens the file on the server and reads it back into a stream:
using (var sftp = new SftpClient(sFTPServer, sFTPUsername, sFTPPassword))
{
sftp.Connect();
// Load remote file into a stream
using (var remoteFileStream = sftp.OpenRead("file.txt"))
{
var textReader = new System.IO.StreamReader(remoteFileStream);
string s = textReader.ReadToEnd();
}
}
For simple text files, it's even easier:
var contents = sftp.ReadAllText(fileSpec);
I had a similar issue with the ScpClient, I needed to reset the stream position to the beginning after downloading the file.
using (var sftp = new SftpClient(sFTPServer, sFTPPassword, sFTPPassword))
{
sftp.Connect();
System.IO.MemoryStream mem = new System.IO.MemoryStream();
System.IO.TextReader textReader = new System.IO.StreamReader(mem);
sftp.DownloadFile("file.txt", mem);
// Reset stream to the beginning
mem.Seek(0, SeekOrigin.Begin);
System.IO.TextReader textReader = new System.IO.StreamReader(mem);
string s = textReader.ReadToEnd();
sftp.Disconnect();
}

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