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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

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;

ZipArchive Created with System.IO.Compression is Damaged

I am having a hard time creating a ZipArchive successfully on Asp.net core MVC. I have an excel file generated with data that works and I need to put in an archive. This is what I've done so far
public FileResult ExportGoodsReceiptData()
{
var records = _salesService.GetAllReceipts();
var lineRecords = _salesService.GetAllReceiptLines();
var result = _salesService.ExportGoodsReceiptData(records);
var lineResult = _salesService.ExportGoodsReceiptLineData(lineRecords);
byte[] resultArr = StreamToByteArray(result);
byte[] lineResultArr = StreamToByteArray(lineResult);
using(MemoryStream stream = new MemoryStream())
{
using (var archive = new ZipArchive(stream, ZipArchiveMode.Create, true))
{
var zipArchiveEntry = archive.CreateEntry("GoodsReceipts.csv", CompressionLevel.Fastest);
using (var zipStream = zipArchiveEntry.Open())
using (var resultCom = new MemoryStream(resultArr))
{
resultCom.CopyTo(zipStream);
}
}
return new FileStreamResult(stream, "application/zip") { FileDownloadName = "GoodsReceiptsArchive.zip" };
}
}
When I run it, I get the zipfile, but can't open it. It throws error stating that it may have been damaged. I debugged the code to notice that one of the properties (length property) throws an invalidOperation exception. My approach looks identical to most samples I found online. Don't know how else to solve this. Please help.

Your problem is that you're disposing of your memory stream before you return it. Remove this using:
using(MemoryStream stream = new MemoryStream())
Replace it with:
var stream = new MemoryStream();
Asp.Net MVC will automatically dispose of the stream for you.

Save a zip file to memory and unzip file from stream and get content

I am currently working on integrating Amazon Prime on our system and being stuck at getting the label back as ZPL format.
Basically, Amazon returns a base64 string, we will need to convert that string to a byte array, then save that array as a *.gzip file. From that gzip file, we can extract the content and get the zpl label content.
My question is, how we can do all of above without storing any temp files to system. I have researched some solutions but none is working for me.
My current code as below:
var str = "base64string";
var label = Convert.FromBase64String(str);
using (var memoryStream = new MemoryStream())
{
using (var archive = new ZipArchive(memoryStream, ZipArchiveMode.Create, true))
{
var demoFile = archive.CreateEntry("label.zip");
var entryStream = demoFile.Open();
using (var bw = new BinaryWriter(entryStream))
{
bw.Write(label);
}
var data = new MemoryStream();
using (var zip = ZipFile.Read(entryStream))
{
zip["label"].Extract(data);
}
data.Seek(0, SeekOrigin.Begin);
entryStream.Close();
}
using (var fileStream = new FileStream(#"D:\test.zip", FileMode.Create))
{
memoryStream.Seek(0, SeekOrigin.Begin);
memoryStream.CopyTo(fileStream);
}
}
If I save the file as test.zip, I can successfully get the label back. But if I try to extract it directly to another stream, I get an error
A stream from ZipArchiveEntry has been disposed

I've done something similar, taking PNG label data from a zipped web response. This is how I went about that
using (WebClient webClient = new WebClient())
{
// Download. Expect this to be a zip file
byte[] data = webClient.DownloadData(urlString);
MemoryStream memoryStream = new MemoryStream(data);
ZipArchive zipArchive = new ZipArchive(memoryStream);
foreach (var zipEntry in zipArchive.Entries)
{
// Can check file name here and ignore anything in zip we're not expecting
if (!zipEntry.Name.EndsWith(".png")) continue;
// Open zip entry as stream
Stream extractedFile = zipEntry.Open();
// Convert stream to memory stream
MemoryStream extractedMemoryStream = new MemoryStream();
extractedFile.CopyTo(extractedMemoryStream);
// At this point the extractedMemoryStream is a sequence of bytes containing image data.
// In this test project I'm pushing that into a bitmap image, just to see something on screen, but could as easily be written to a file or passed for storage to sql or whatever.
BitmapDecoder decoder = PngBitmapDecoder.Create(extractedMemoryStream, BitmapCreateOptions.None, BitmapCacheOption.OnLoad);
BitmapFrame frame = decoder.Frames.First();
frame.Freeze();
this.LabelImage.Source = frame;
}
}

I was overthinking it. I finally found a simple way to do it. We just need to convert that base64 string to bytes array and use GzipStream to directly decompress it. I leave the solution here in case someone needs it. Thanks!
var label = Convert.FromBase64String(str);
using (var compressedStream = new MemoryStream(label))
using (var zipStream = new GZipStream(compressedStream, CompressionMode.Decompress))
using (var resultStream = new MemoryStream())
{
zipStream.CopyTo(resultStream);
return resultStream.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.