ZipArchive Created with System.IO.Compression is Damaged - c#

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.

Related

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

How to unzip a file in memory to an Azure CloudBlob [duplicate]

I'm probably doing something obviously stupid here. Please point it out!
I have some C# code that is pulling down a bunch of .gz files from SFTP (using the SSH.NET Nuget package - works great!). Each gz contains only a single .CSV file inside of them. I want to keep these files in memory without hitting disk (yes, I know, server memory management concerns exist - that's fine as these files are fairly small), decompress them in memory to extract the CSV file inside, and then return a collection of CSV files in a custom DTO (FtpFile).
My problem is that while my MemoryStream from the SFTP connection has data in it, either it doesn't ever seem to be populated in my GZipStream or the copy from the GZipStream to my output MemoryStream is failing. I have tried with the more traditional looping over Read with my own buffer but it had the same results as this code.
Aside from connection details (it connects successfully, so no worries there), here's all of my code:
Logic:
public static List<FtpFile> Foo()
{
var connectionInfo = new ConnectionInfo("example.com",
"username",
new PasswordAuthenticationMethod("username", "password"));
using (var client = new SftpClient(connectionInfo))
{
client.Connect();
var searchResults = client.ListDirectory("/testdir")
.Where(obj => obj.IsRegularFile
&& obj.Name.ToLowerInvariant().StartsWith("test_")
&& obj.Name.ToLowerInvariant().EndsWith(".gz"))
.Take(2)
.ToList();
var fileResults = new List<FtpFile>();
foreach (var file in searchResults)
{
var ftpFile = new FtpFile { FileName = file.Name, FileSize = file.Length };
using (var fileStream = new MemoryStream())
{
client.DownloadFile(file.FullName, fileStream); // Success! All is good here, so far. :)
using (var gzStream = new GZipStream(fileStream, CompressionMode.Decompress))
{
using (var outputStream = new MemoryStream())
{
gzStream.CopyTo(outputStream);
byte[] outputBytes = outputStream.ToArray(); // No data. Sad panda. :'(
ftpFile.FileContents = Encoding.ASCII.GetString(outputBytes);
fileResults.Add(ftpFile);
}
}
}
}
return fileResults;
}
}
FtpFile (just a simple DTO I'm populating):
public class FtpFile
{
public string FileName { get; set; }
public long FileSize { get; set; }
public string FileContents { get; set; }
}
PSA If anybody comes and copies this code, be aware that this is NOT good code in that you could have some serious memory management problems with this code! It's best practice to instead stream it to disk, which is not being done in this code! My needs are very specific in that I have to have these files simultaneously in memory for what I'm building with them.
If you are inserting data into the stream, make sure to seek back to its origin before un-gzipping it.
The following should fix your troubles:
using (var fileStream = new MemoryStream())
{
client.DownloadFile(file.FullName, fileStream); // Success! All is good here, so far. :)
fileStream.Seek(0, SeekOrigin.Begin);
using (var gzStream = new GZipStream(fileStream, CompressionMode.Decompress))
{
using (var outputStream = new MemoryStream())
{
gzStream.CopyTo(outputStream);
byte[] outputBytes = outputStream.ToArray(); // No data. Sad panda. :'(
ftpFile.FileContents = Encoding.ASCII.GetString(outputBytes);
fileResults.Add(ftpFile);
}
}
}

How to go from byte[], to MemoryStream, Unzip, then write to FileStream

I am unsure what I am doing wrong. The files that I create after grabbing a byte[] (which is emailAttachment.Body) and passing it to the method ExtractZipFile, converting it to MemoryStream and then unzipping it, returning it as a KeyValuePair and then Writing to a file using FileStream.
However when I go to open the new created files there is an error in opening them. They are not able to be opened.
The below are in the same class
using Ionic.Zip;
var extractedFiles = ExtractZipFile(emailAttachment.Body);
foreach (KeyValuePair<string, MemoryStream> extractedFile in extractedFiles)
{
string FileName = extractedFile.Key;
using (FileStream file = new FileStream(CurrentFileSystem +
FileName.FileFullPath(),FileMode.Create, System.IO.FileAccess.Write))
{
byte[] bytes = new byte[extractedFile.Value.Length];
extractedFile.Value.Read(bytes, 0, (int) xtractedFile.Value.Length);
file.Write(bytes,0,bytes.Length);
extractedFile.Value.Close();
}
}
private Dictionary<string, MemoryStream> ExtractZipFile(byte[] messagePart)
{
Dictionary<string, MemoryStream> result = new Dictionary<string,MemoryStream>();
MemoryStream data = new MemoryStream(messagePart);
using (ZipFile zip = ZipFile.Read(data))
{
foreach (ZipEntry ent in zip)
{
MemoryStream memoryStream = new MemoryStream();
ent.Extract(memoryStream);
result.Add(ent.FileName,memoryStream);
}
}
return result;
}
Is there something I am missing? I do not want to save the original zip file just the extracted Files from MemoryStream.
What am I doing wrong?
After writing to your MemoryStream, you're not setting the position back to 0:
MemoryStream memoryStream = new MemoryStream();
ent.Extract(memoryStream);
result.Add(ent.FileName,memoryStream);
Because of this, the stream position will be at the end when you try to read from it, and you'll read nothing. Make sure to rewind it:
memoryStream.Position = 0;
Also, you don't have to handle the copy manually. Just let the CopyTo method take care of it:
extractedFile.Value.CopyTo(file);
I'd suggest that you clean up your use of MemoryStream in your code.
I agree that calling memoryStream.Position = 0; will allow this code to work correctly, but it's an easy thing to miss when reading and writing memory streams.
It's better to write code that avoids the bug.
Try this:
private IEnumerable<(string Path, byte[] Content)> ExtractZipFile(byte[] messagePart)
{
using (var data = new MemoryStream(messagePart))
{
using (var zipFile = ZipFile.Read(data))
{
foreach (var zipEntry in zipFile)
{
using (var memoryStream = new MemoryStream())
{
zipEntry.Extract(memoryStream);
yield return (Path: zipEntry.FileName, Content: memoryStream.ToArray());
}
}
}
}
}
Then your calling code would look something like this:
foreach (var extractedFile in ExtractZipFile(emailAttachment.Body))
{
File.WriteAllBytes(Path.Combine(CurrentFileSystem, extractedFile.Path.FileFullPath()), extractedFile.Content);
}
It's just a lot less code and a much better chance of avoiding bugs. The number one predictor of bugs in code is the number of lines of code you write.
Since I find it all a lot of code for a simple operation, here's my two cents.
using Ionic.Zip;
using (var s = new MemoryStream(emailAttachment.Body))
using (ZipFile zip = ZipFile.Read(s))
{
foreach (ZipEntry ent in zip)
{
string path = Path.Combine(CurrentFileSystem, ent.FileName.FileFullPath())
using (FileStream file = new FileStream(path, FileAccess.Write))
{
ent.Extract(file);
}
}
}

Create Zip File with multiple files C#

I'm creating a Zip file from files I'm downloading from the internet in bytes[],
but I have an Issue, I don't know what I'm doing wrong... I generate the zip file but it's corrupted, the file size is correct(is not 0).
Could you help me please?
Maybe I didn't understand it well.
public <ActionResult> SomeFunction()
{
var invoices = GetInvoices();
WebClient client = new WebClient();
byte[] zipBytes = null;
using (var compressedFileStream = new MemoryStream())
{
using (ZipArchive zipArchive = new ZipArchive(compressedFileStream, ZipArchiveMode.Create, leaveOpen: true))
{
foreach (var invoice in invoices)
{
// This has correct values.
byte[] fileBytes = client.DownloadData(invoice.XmlUri);
// Create the instance of the file.
var zipEntry = zipArchive.CreateEntry(invoice.XmlFileName);
// Get the stream of the file.
using (var entryStream = new MemoryStream(fileBytes))
// Get the Stream of the zipEntry
using (var zipEntryStream = zipEntry.Open())
{
// Adding the file to the zip file.
entryStream.CopyTo(zipEntryStream);
}
}
}
zipBytes = compressedFileStream.ToArray();
}
return File(zipBytes , System.Net.Mime.MediaTypeNames.Application.Octet, "test.zip");
}
Move
zipBytes = compressedFileStream.ToArray();
To after the archive has been disposed so that all data is flushed to the underlying stream.
public <ActionResult> SomeFunction() {
var invoices = GetInvoices();
WebClient client = new WebClient();
byte[] zipBytes = null;
using (var compressedFileStream = new MemoryStream()) {
using (ZipArchive zipArchive = new ZipArchive(compressedFileStream, ZipArchiveMode.Create, leaveOpen: true)) {
foreach (var invoice in invoices) {
// This has correct values.
byte[] fileBytes = client.DownloadData(invoice.XmlUri);
// Create the instance of the file.
var zipEntry = zipArchive.CreateEntry(invoice.XmlFileName);
// Get the stream of the file.
using (var entryStream = new MemoryStream(fileBytes))
// Get the Stream of the zipEntry
using (var zipEntryStream = zipEntry.Open()) {
// Adding the file to the zip file.
entryStream.CopyTo(zipEntryStream);
}
}
}
zipBytes = compressedFileStream.ToArray();
}
return File(zipBytes , System.Net.Mime.MediaTypeNames.Application.Octet, "test.zip");
}
Reference ZipArchive.Dispose()
This method finishes writing the archive and releases all resources used by the ZipArchive object.
Unless you construct the object by using the ZipArchive(Stream, ZipArchiveMode, Boolean) constructor overload and set its leaveOpen parameter to true, all underlying streams are closed and no longer available for subsequent write operations.
When you are finished using this instance of ZipArchive, call Dispose() to release all resources used by this instance. You should eliminate further references to this ZipArchive instance so that the garbage collector can reclaim the memory of the instance instead of keeping it alive for finalization.

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();
}

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