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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

how to make file content as base64 encoded before making zip file

I have a directory where I have CSV files which I need to first encode the file content as base64 string and then make it as zip file.
I am able to make file as zip with below code, but in between on the fly how to make file content as base64 encoded? Thanks!
var csvFiles = Directory.GetFiles(#"C:\Temp", "*.csv")
.Select(f => new FileInfo(f));
foreach (var file in csvFiles)
{
using (var newFile = ZipFile.Open($#"C:\tmp\{Path.GetFileNameWithoutExtension(file.Name)}.zip",
ZipArchiveMode.Create))
{
newFile.CreateEntryFromFile($#"C:\Temp\{file.Name}",
file.Name);
}
}

Disregarding your motives or other problems (conceptual or otherwise)
Here is a fully streamed solution with minimal allocations (let's be nice to your Large Object Heap). The CryptoStream with ToBase64Transform, is just a way to stream base64 encoding
var csvFiles = Directory.GetFiles(#"D:\Temp");
using var outputStream = new FileStream(#"D:\Test.zip", FileMode.Create);
using var archive = new ZipArchive(outputStream, ZipArchiveMode.Create, true);
foreach (var file in csvFiles)
{
using var inputFile = new FileStream(file, FileMode.Open, FileAccess.Read);
using var base64Stream = new CryptoStream(inputFile, new ToBase64Transform(), CryptoStreamMode.Read);
var entry = archive.CreateEntry(Path.GetFileName(file));
using var zipStream = entry.Open();
base64Stream.CopyTo(zipStream);
}

You need to create the base64 string, convert it to a byte array, and then create the archive entry from the byte array (by creating a stream).
Something like this should do the job:
var dirInfo = new DirectoryInfo(#"C:\Temp");
var csvFiles = dirInfo.GetFiles("*.csv"); // This already returns a `FileInfo[]`.
foreach (var file in csvFiles)
{
var fileBytes = File.ReadAllBytes(file.FullName);
var base64String = Convert.ToBase64String(fileBytes);
var base64Bytes = Encoding.UTF8.GetBytes(base64String);
string newFilePath = $#"C:\tmp\{Path.GetFileNameWithoutExtension(file.Name)}.zip";
using (var newFile = ZipFile.Open(newFilePath, ZipArchiveMode.Create))
{
// You might want to change the extension
// since the file is no longer in CSV format.
var zipEntry = newFile.CreateEntry(file.Name);
using (var base64Stream = new MemoryStream(base64Bytes))
using (var zipEntryStream = zipEntry.Open())
{
base64Stream.CopyTo(zipEntryStream);
}
}
}
Alternatively, you could save the base64 string to a temporary file, create the entry from that file, and then delete it; but I don't prefer writing dummy data to the disk when the job can be done in memory.

C# create zip file using zip archive System.IO.Compression

Here is the functionality I want to achieve
Write a JSON file.
Write a PDF file.
Create an archive for these two files
I am using the System.IO.Compression ZipArchive to achieve this. From the documentation, I have not found a good use case for this. The examples in documentation assume that the zip file exists.
What I want to do
Create zipArchive stream write JSON file and pdf file as entries in the zip file.
using (var stream = new FileStream(path, FileMode.Create))
{
using (var archive = new ZipArchive(stream, ZipArchiveMode.Create, true))
{
ZipArchiveEntry manifest = archive.CreateEntry(filenameManifest);
using (StreamWriter writerManifest = new StreamWriter(manifest.Open()))
{
writerManifest.WriteLine(JSONObject_String);
}
ZipArchiveEntry pdfFile = archive.CreateEntry(filenameManifest);
using (StreamWriter writerPDF = new StreamWriter(pdfFile.Open()))
{
writerPDF.WriteLine(pdf);
}
}
}

You don't close the stream, you open with 'manifest.Open()'. Then it might not have written everything to the zip.
Wrap it in another using, like this:
using (var stream = new FileStream(path, FileMode.Create))
{
using (var archive = new ZipArchive(stream, ZipArchiveMode.Create, true))
{
ZipArchiveEntry manifest = archive.CreateEntry(filenameManifest);
using (Stream st = manifest.Open())
{
using (StreamWriter writerManifest = new StreamWriter(st))
{
writerManifest.WriteLine(JSONObject_String);
}
}
ZipArchiveEntry pdfFile = archive.CreateEntry(filenameManifest);
using (Stream st = manifest.Open())
{
using (StreamWriter writerPDF = new StreamWriter(st))
{
writerPDF.WriteLine(pdf);
}
}
}
}

C# ZipArchive "End of Central Directory record could not be found"

I created a BizTalk Custom Pipeline Component, which zips all message parts into a zip stream. After some fails I created the following test method in a separate test project.
Basically I get a XML file which contains a filename and an UUID which I use to call a stored procedure and get the Base64 encoded content of the database entry.
The base64 content seems valid, because after decoding it and saving it to the file system the files can be read by the windows explorer without problems.
After saving the archiveStream to the file system I get the following error message from 7Zip when I try to extract the file:
"Unexpected end of data". if I try to just open the file with 7Zip there is no problem. I even can open the files from inside the 7Zip explorer.
If I try to read the file from C# with the following code I get the error message:
"End of Central Directory record could not be found."
Unzip code:
private static void ReadDat() {
var path = #"...\zip\0e00128b-0a6e-4b99-944d-68e9c20a51c2.zip";
var stream = System.IO.File.OpenRead(path);
// End of Central Directory record could not be found:
var zipArchive = new ZipArchive(stream, ZipArchiveMode.Read, false);
foreach(var zipEntry in zipArchive.Entries) {
var stream = zipEntry.Open();
Console.WriteLine(stream.Length);
}
}
Zip Code:
private static void StreamUuidList() {
var path = #"...\2017-08-05T132705.xml";
var xdoc = XDocument.Load(System.IO.File.OpenRead(path));
var files = xdoc.Root.Descendants().Where(d => d.Name.LocalName.Equals("NodeName"));
using (var archiveStream = new MemoryStream())
using (var archive = new ZipArchive(archiveStream, ZipArchiveMode.Create, true)) {
foreach (var file in files) {
var fileName = file.Elements().Where(e => e.Name.LocalName.Equals("FileName")).FirstOrDefault()?.Value ?? "";
var streamUuid = file.Elements().Where(e => e.Name.LocalName.Equals("StreamUUID")).FirstOrDefault()?.Value ?? "";
// validation here...
// get base64 content and convert content
var base64Content = GetStreamContent(streamUuid);
var data = Convert.FromBase64String(base64Content);
var dataStream = new MemoryStream(data);
dataStream.Seek(0, SeekOrigin.Begin);
// debug - save to file location
using (var fileStream = new FileStream($#"...\files\{fileName}", FileMode.Create)) {
dataStream.CopyTo(fileStream);
}
dataStream.Seek(0, SeekOrigin.Begin);
// create zip entry
var zipFile = archive.CreateEntry(fileName, GetCompressionLevelFromString("Optimal"));
using (var zipFileStream = zipFile.Open()) {
// copy data from mesage part stream into zip entry stream
dataStream.Seek(0, SeekOrigin.Begin);
dataStream.CopyTo(zipFileStream);
}
Console.WriteLine(fileName + ": " + streamUuid);
}
// debug - save to file location
archiveStream.Seek(0, SeekOrigin.Begin);
using (var fileStream = new FileStream($#"...\zip\{Guid.NewGuid()}.dat", FileMode.Create)) {
archiveStream.CopyTo(fileStream);
}
// debug end
}
}

Read the content of an xml file within a zip package

I am required to read the contents of an .xml file using the Stream (Here the xml file is existing with in the zip package). Here in the below code, I need to get the file path at runtime (here I have hardcoded the path for reference). Please let me know how to read the file path at run time.
I have tried to use string s =entry.FullName.ToString(); but get the error "Could not find the Path". I have also tried to hard code the path as shown below. however get the same FileNotFound error.
string metaDataContents;
using (var zipStream = new FileStream(#"C:\OB10LinuxShare\TEST1\Temp" + "\\"+zipFileName+".zip", FileMode.Open))
using (var archive = new ZipArchive(zipStream, ZipArchiveMode.Read))
{
foreach (var entry in archive.Entries)
{
if (entry.Name.EndsWith(".xml"))
{
FileInfo metadataFileInfo = new FileInfo(entry.Name);
string metadataFileName = metadataFileInfo.Name.Replace(metadataFileInfo.Extension, String.Empty);
if (String.Compare(zipFileName, metadataFileName, true) == 0)
{
using (var stream = entry.Open())
using (var reader = new StreamReader(stream))
{
metaDataContents = reader.ReadToEnd();
clientProcessLogWriter.WriteToLog(LogWriter.LogLevel.DEBUG, "metaDataContents : " + metaDataContents);
}
}
}
}
}
I have also tried to get the contents of the .xml file using the Stream object as shown below. But here I get the error "Stream was not readable".
Stream metaDataStream = null;
string metaDataContent = string.Empty;
using (Stream stream = entry.Open())
{
metaDataStream = stream;
}
using (var reader = new StreamReader(metaDataStream))
{
metaDataContent = reader.ReadToEnd();
}
Kindly suggest, how to read the contents of the xml with in a zip file using Stream and StreamReader by specifying the file path at run time

Your section code snippet is failing because when you reach the end of the first using statement:
using (Stream stream = entry.Open())
{
metaDataStream = stream;
}
... the stream will be disposed. That's the point of a using statment. You should be fine with this sort of code, but load the XML file while the stream is open:
XDocument doc;
using (Stream stream = entry.Open())
{
doc = XDocument.Load(stream);
}
That's to load it as XML... if you really just want the text, you could use:
string text;
using (Stream stream = entry.Open())
{
using (StreamReader reader = new StreamReader(stream))
{
text = reader.ReadToEnd();
}
}
Again, note how this is reading before it hits the end of either using statement.

Here is a sample of how to read a zip file using .net 4.5
private void readZipFile(String filePath)
{
String fileContents = "";
try
{
if (System.IO.File.Exists(filePath))
{
System.IO.Compression.ZipArchive apcZipFile = System.IO.Compression.ZipFile.Open(filePath, System.IO.Compression.ZipArchiveMode.Read);
foreach (System.IO.Compression.ZipArchiveEntry entry in apcZipFile.Entries)
{
if (entry.Name.ToUpper().EndsWith(".XML"))
{
System.IO.Compression.ZipArchiveEntry zipEntry = apcZipFile.GetEntry(entry.Name);
using (System.IO.StreamReader sr = new System.IO.StreamReader(zipEntry.Open()))
{
//read the contents into a string
fileContents = sr.ReadToEnd();
}
}
}
}
}
catch (Exception)
{
throw;
}
}