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System.IO.Compression.ZipArchive corrupt when downloading [duplicate]

I'm trying to create a zip stream on the fly with some byte array data and make it download via my MVC action.
But the downloaded file always gives the following corrupted error when opened in windows.
And this error when I try to xtract from 7z
But note that the files extracted from the 7z is not corrupted.
I'm using ZipArchive and the below is my code.
private byte[] GetZippedPods(IEnumerable<POD> pods, long consignmentID)
{
using (var zipStream = new MemoryStream())
{
//Create an archive and store the stream in memory.
using (var zipArchive = new ZipArchive(zipStream, ZipArchiveMode.Create, true))
{
int index = 1;
foreach (var pod in pods)
{
var zipEntry = zipArchive.CreateEntry($"POD{consignmentID}{index++}.png", CompressionLevel.NoCompression);
using (var originalFileStream = new MemoryStream(pod.ByteData))
{
using (var zipEntryStream = zipEntry.Open())
{
originalFileStream.CopyTo(zipEntryStream);
}
}
}
return zipStream.ToArray();
}
}
}
public ActionResult DownloadPOD(long consignmentID)
{
var pods = _consignmentService.GetPODs(consignmentID);
var fileBytes = GetZippedPods(pods, consignmentID);
return File(fileBytes, MediaTypeNames.Application.Octet, $"PODS{consignmentID}.zip");
}
What am I doing wrong here.
Any help would be highly appreciated as I'm struggling with this for a whole day.
Thanks in advance

Move zipStream.ToArray() outside of the zipArchive using.
The reason for your problem is that the stream is buffered. There's a few ways to deal wtih it:
You can set the stream's AutoFlush property to true.
You can manually call .Flush() on the stream.
Or, since it's MemoryStream and you're using .ToArray(), you can simply allow the stream to be Closed/Disposed first (which we've done by moving it outside the using).

I Dispose ZipArchive And error solved
public static byte[] GetZipFile(Dictionary<string, List<FileInformation>> allFileInformations)
{
MemoryStream compressedFileStream = new MemoryStream();
//Create an archive and store the stream in memory.
using (var zipArchive = new ZipArchive(compressedFileStream, ZipArchiveMode.Create, true))
{
foreach (var fInformation in allFileInformations)
{
var files = allFileInformations.Where(x => x.Key == fInformation.Key).SelectMany(x => x.Value).ToList();
for (var i = 0; i < files.Count; i++)
{
ZipArchiveEntry zipEntry = zipArchive.CreateEntry(fInformation.Key + "/" + files[i].FileName);
var caseAttachmentModel = Encoding.UTF8.GetBytes(files[i].Content);
//Get the stream of the attachment
using (var originalFileStream = new MemoryStream(caseAttachmentModel))
using (var zipEntryStream = zipEntry.Open())
{
//Copy the attachment stream to the zip entry stream
originalFileStream.CopyTo(zipEntryStream);
}
}
}
//i added this line
zipArchive.Dispose();
return compressedFileStream.ToArray();
}
}
public void SaveZipFile(){
var zipFileArray = Global.GetZipFile(allFileInformations);
var zipFile = new MemoryStream(zipFileArray);
FileStream fs = new FileStream(path + "\\111.zip",
FileMode.Create,FileAccess.Write);
zipFile.CopyTo(fs);
zipFile.Flush();
fs.Close();
zipFile.Close();
}

I was also having problems with this and I found my issue was not the generation of the archive itself but rather how I was handing my GET request in AngularJS.
This post helped me: how to download a zip file using angular
The key was adding responseType: 'arraybuffer' to my $http call.
factory.serverConfigExportZIP = function () {
return $http({
url: dataServiceBase + 'serverConfigExport',
method: "GET",
responseType: 'arraybuffer'
})
};

you can remove "using" and use Dispose and Close methods
it's work for me
...
zip.Dispose();
zipStream.Close();
return zipStream.ToArray();

I know this is a C# question but for managed C++, delete the ZipArchive^ after you're done with it to fix the error.
ZipArchive^ zar = ZipFile::Open(starget, ZipArchiveMode::Create);
ZipFileExtensions::CreateEntryFromFile(zar, sfile1, "file.txt");
ZipFileExtensions::CreateEntryFromFile(zar, sfile2, "file2.txt");
delete zar;

when i wanted to create zip file directly from MemoryStream which i used for ZipArchive i was getting error ( "unexpected end of data" or zero length file )
there are three points to get ride of this error
set the last parameter of ZipArchive constructor to true ( it leaves to leave stream open after ZipArchive disposed )
call dispose() on ZipArchive and dispose it manually.
create another MemoryStream based on which you set in ZipArchive constructor, by calling ToArray() method.
here is sample code :
using (var memoryStream = new MemoryStream())
{
using (var archive = new ZipArchive(memoryStream, ZipArchiveMode.Create,))
{
foreach (var s3Object in objectList.S3Objects)
{
var entry = archive.CreateEntry(s3Object.Key, CompressionLevel.NoCompression);
using (var entryStream = entry.Open())
{
var request = new GetObjectRequest { BucketName = command.BucketName, Key = s3Object.Key };
using (var getObjectResponse = await client.GetObjectAsync(request))
{
await getObjectResponse.ResponseStream.CopyToAsync(entryStream);
}
}
}
archive.Dispose();
using (var fileStream = new FileStream(outputFileName, FileMode.Create, FileAccess.Write))
{
var zipFileMemoryStream = new MemoryStream(memoryStream.ToArray());
zipFileMemoryStream.CopyTo(fileStream);
zipFileMemoryStream.Flush();
fileStream.Close();
zipFileMemoryStream.Close();
}
}
}

I had the same problem... In this case I just needed to move the ToArray() (byte[]) from MemoryStream outside the using (var zipArchive = new ZipArchive...
I think it is necessary for using related to ZipArchive to completely close and dispose of the file before converting it into a byte array

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());
}
}

Download zip file from the server and parsing it

I am trying to download a zipped file from the server and trying to show the content of each files in zipped folder to the view.
I wrote a separate code where the file is on my laptop and I ran across each file and dislpayed the content such as
static void Main(string[] args)
{
string filePath = "C:\\ACL Data\\New folder\\files.zip";
var zip= new ZipInputStream(File.OpenRead(filePath));
var filestream=new FileStream(filePath,FileMode.Open,FileAccess.Read);
ZipFile zipfile = new ZipFile(filestream);
ZipEntry item;
while ((item = zip.GetNextEntry()) != null)
{
Console.WriteLine(item.Name);
using (StreamReader s = new StreamReader(zipfile.GetInputStream(item)))
{
Console.WriteLine(s.ReadToEnd());
}
}
Console.Read();
}
I am using sharplibzip library to implement this
This is the case when the zip file is located locally in the system. My next task scenario is what if the zipped file is located on the server. I am figuring out the way to implement it, below is the code what I assume should work
static void Main(string[] args)
{
string url = "https://test/code/304fd9c6-7e53-42a2-845a-624608bfd2ce.zip";
WebRequest webRequest = WebRequest.Create(url);
webRequest.Method = "GET";
WebResponse webResponse = webRequest.GetResponse();
var zip = new ZipInputStream(webResponse.GetResponseStream());
ZipEntry item1;
//var zip= new ZipInputStream(File.OpenRead(filePath));
var filestream = new FileStream(filepath, FileMode.Open, FileAccess.Read);
ZipFile zipfile = new ZipFile(filestream);
ZipEntry item;
while ((item = zip.GetNextEntry()) != null)
{
Console.WriteLine(item.Name);
using (StreamReader s = new StreamReader(zipfile.GetInputStream(item)))
{
Console.WriteLine(s.ReadToEnd());
}
}
Console.Read();
}
I am stuck at this part: var filestream = new FileStream(filepath, FileMode.Open, FileAccess.Read);
This expect the first parameter to be path of the zip file. Since in the new scenario zip file is located remotely on the server. What should be the parameter in this case?

Your original code opens the stream twice on the following rows, which I think is causing some confusion:
var zip= new ZipInputStream(File.OpenRead(filePath));
var filestream=new FileStream(filePath,FileMode.Open,FileAccess.Read);
There is an overload to the ZipFile constructor that takes "any" Stream rather than specifically a FileStream, which you - unsurprisingly - can only create for files.
However, you cannot use the stream returned by GetResponseStream directly, because it's CanSeek property is false. This is because it's a NetworkStream, which can only be read once from beginning to end. SharpZipLib needs random access to read the file contents.
Depending on the size of the ZIP file, loading it in memory may be an option. If you expect large files, writing it to a temporary file may be better.
This should do the trick, without using both ZipInputStream and ZipFile, by enumerating through ZipFile instead:
string url = "https://test/code/304fd9c6-7e53-42a2-845a-624608bfd2ce.zip";
WebRequest webRequest = WebRequest.Create(url);
webRequest.Method = "GET";
WebResponse webResponse = webRequest.GetResponse();
using (var responseStream = webResponse.GetResponseStream())
using (var ms = new MemoryStream())
{
// Copy entire file into memory. Use a file if you expect a lot of data
responseStream.CopyTo(ms);
var zipFile = new ZipFile(ms);
foreach (ZipEntry item in zipFile)
{
Console.WriteLine(item.Name);
using (var s = new StreamReader(zipFile.GetInputStream(item)))
{
Console.WriteLine(s.ReadToEnd());
}
}
}
Console.Read();
PS: starting .NET 4.5, there is support for ZIP files built in. See the ZipArchive class.

Adding multiple files to an in memory zip archive in c#

I'm trying to use the Zip Archive library in .NET 4.5 to create a .zip file in memory of a bunch of byte[] attachments:
using (var memoryStream = new MemoryStream())
{
using (var archive = new ZipArchive(memoryStream, ZipArchiveMode.Create, true))
{
string zipFilename = string.Format(#"c:\temp\{0} - {1}.zip",
"test",
System.DateTime.Now.ToString("yyyyMMddHHmm"));
using (var fileStream = new FileStream(zipFilename, FileMode.Create))
{
foreach (var attachment in attachments)
{
ZipArchiveEntry entry = archive.CreateEntry(attachment.FileName);
using (Stream ZipFile = entry.Open())
{
byte[] data = attachment.Data;
ZipFile.Write(data, 0, data.Length);
}
}
}
}
}
PdfAttachment is a class with a byte[] Data and string Filename.
My problem is twofold. Once, the zip archive is empty. 2, rather than save it to a file, I'd like to use the response outputs to download the file in the users browser, which I have tried with:
Response.Clear();
Response.ClearContent();
Response.ClearHeaders();
Response.ContentType = "application/pdf";
Response.AddHeader("Content-Disposition", string.Format("attachment;filename={0}.zip; size={1}", "test.zip", memoryStream.Length));
Response.BinaryWrite(memoryStream);
Response.Flush();
Response.End();
I haven't been able to find many examples online, hence the vagueness.

The fileStream is never written to because it is not associated with the archive.
So the archive is being written to the memoryStream.
BinaryWrite only accepts a byte[] so use memoryStream.ToArray().
Also, Response.ContentType value is wrong.