I have a bunch of Jpg images in byte array form. I want to add these to a zip file, turn the zip file into a byte array, and pass it to somewhere else. In a method, I have this code:
var response = //some response object that will hold a byte array
using (var ms = new MemoryStream())
{
using (var zipArchive = new ZipArchive(ms, ZipArchiveMode.Create, true))
{
var i = 1;
foreach (var image in images) // some collection that holds byte arrays.
{
var entry = zipArchive.CreateEntry(i + ".jpg");
using (var entryStream = entry.Open())
using (var compressStream = new MemoryStream(photo.ImageOriginal))
{
compressStream.CopyTo(entryStream);
}
i++;
}
response.ZipFile = ms.ToArray();
}
using (var fs = new FileStream(#"C:\Users\MyName\Desktop\image.zip", FileMode.Create))
{
ms.Position = 0;
ms.CopyTo(fs);
}
}
return response;
Now, I've added a filestream near the bottom to write it to a zipfile right away for testing purposes. This works, I get a zip file with 1 or more images in it on my desktop. However: response.ZipFile can not be made into a valid zipfile in the same way. I have tried this:
using (var ms2 = new MemoryStream(response.ZipFile))
using (var fs = new FileStream(#"C:\Users\Bara\Desktop\image.zip", FileMode.Create))
{
ms2.Position = 0;
ms2.CopyTo(fs);
}
But that creates a zipfile that can not be opened.
What I'm trying to do: turn response.ZipFileinto an array that can be turned into a working zipfile again. What am I doing wrong in this code?
How do you know that ZipArchive's Dispose doesn't write more to the underlying stream?
You should move this line to be after disposing the ZipArchive:
response.ZipFile = ms.ToArray();
Full code:
var response = //some response object that will hold a byte array
using (var ms = new MemoryStream())
{
using (var zipArchive = new ZipArchive(ms, ZipArchiveMode.Create, true))
{
var i = 1;
foreach (var image in images) // some collection that holds byte arrays.
{
var entry = zipArchive.CreateEntry(i + ".jpg");
using (var entryStream = entry.Open())
using (var compressStream = new MemoryStream(photo.ImageOriginal))
{
compressStream.CopyTo(entryStream);
}
i++;
}
}
response.ZipFile = ms.ToArray();
}
return response;
Related
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
I am creating a zip file that, appears, valid but is always invalid after I have put it to a Amazon S3 bucket. I am using System.IO.Compression for the task and AmazonS3Client for uploading:
private byte[] GenerateZipFile(string tenant)
{
byte[] zipData;
var results = QueryAggregateTable(tenant);
using (var memoryStream = new MemoryStream())
using (var archive = new ZipArchive(memoryStream, ZipArchiveMode.Create, true))
{
foreach (var item in results)
{
var archiveEntry = archive.CreateEntry($"{item.RowKey:D3}.json", CompressionLevel.Fastest);
using (var entryStream = archiveEntry.Open())
{
var entryBytes = Encoding.UTF8.GetBytes(item.Data);
entryStream.Write(entryBytes, 0, item.Data.Length);
}
}
zipData = memoryStream.ToArray();
}
return zipData;
}
and
private async Task UploadToAmazon(byte[] zipData, string tenant)
{
var bucketName = _config["egestionBucketName"];
var configCreds = _config["egestionAwsCredentials"].Split(":");
var awsCreds = new BasicAWSCredentials(configCreds[0], configCreds[1]);
var awsRegion = Amazon.RegionEndpoint.GetBySystemName(_config["egestionRegionEndpointSystemName"]);
var s3Client = new AmazonS3Client(awsCreds, awsRegion);
using (var stream = new MemoryStream(zipData))
{
var putRequest = new PutObjectRequest
{
BucketName = bucketName,
Key = $"{tenant}-{DateTime.UtcNow.ToString("s")}.zip",
InputStream = stream,
CannedACL = S3CannedACL.BucketOwnerFullControl
};
await s3Client.PutObjectAsync(putRequest);
}
}
The byte array looks good after returning from generation and the upload method does, in fact, load a file with the correct name to the bucket. When I attempt to download the file to check it I cannot open it with a message stating it is invalid.
I have had some problems with async/await and suspect it may be something related but there is no non async option for PutObject that I can find. Any help appreciated.
This is not an async-await issue.
The bytes from the memory stream are being collected before the archive has had a chance to write all the data to the stream. The uploaded archive is incomplete and therefore invalid when downloaded.
Move
zipData = memoryStream.ToArray();
to outside of the archive using block so that any buffered data is flushed to the backing stream when the archive is disposed.
//...
using (var memoryStream = new MemoryStream()) {
using (var archive = new ZipArchive(memoryStream, ZipArchiveMode.Create, true)) {
foreach (var item in results) {
var archiveEntry = archive.CreateEntry($"{item.RowKey:D3}.json", CompressionLevel.Fastest);
using (var entryStream = archiveEntry.Open()) {
var entryBytes = Encoding.UTF8.GetBytes(item.Data);
entryStream.Write(entryBytes, 0, entryBytes.Length);
}
}
}//Archive disposed and pushed any remaining buffered data to the stream.
zipData = memoryStream.ToArray();
}
//...
I'm trying move Source_File.mp4 into Destination_Zip.zip directly. I'm currently trying it by creating a new entry Transfer.mp4 and copying the bytes over. The code runs to the end but the file is never added to the ZIP file. I'm not sure if I'm missing something or if this isn't possible.
string sourceFile = basePath + "Source_File.mp4";
string destinationZip = basePath + "Desintation_Zip.zip";
using (var file = File.OpenRead(destinationZip))
{
MemoryStream memoryStream = new MemoryStream();
memoryStream.SetLength(file.Length);
file.Read(memoryStream.GetBuffer(), 0, (int)file.Length);
using (var zip = new ZipArchive(memoryStream, ZipArchiveMode.Update))
{
var entry = zip.CreateEntry("Transfer.mp4");
using (var destinationStream = entry.Open())
using (var sourceStream = File.OpenRead(sourceFile))
{
sourceStream.CopyTo(destinationStream);
}
}
}
My guess is that even though you've read the file and altered it, you didn't write anything back, i.e. changes persisted to MemoryStream, and didn't go anywhere after that.
You could try this (assumes you're using System.IO.Compression.ZipFile):
using (var zip = ZipFile.Open(destinationZip, ZipArchiveMode.Update))
{
var entry = zip.CreateEntry("Transfer.mp4");
using (var destinationStream = entry.Open())
using (var sourceStream = File.OpenRead(sourceFile))
{
sourceStream.CopyTo(destinationStream);
}
}
Or, if you're using ICSharpCode.SharpZipLib.Zip.ZipFile, do this:
using (var fileStream = new FileStream(destinationZip, FileMode.Open))
using (var zip = new ZipArchive(fileStream, ZipArchiveMode.Update))
{
var entry = zip.CreateEntry("Transfer.mp4");
using (var destinationStream = entry.Open())
using (var sourceStream = File.OpenRead(sourceFile))
{
sourceStream.CopyTo(destinationStream);
}
}
I'm still not sure why my initial code did not work, but I got it working by doing this instead:
string sourceFile = basePath + "Source_File.mp4";
string destinationZip = basePath + "Desintation_Zip.zip";
using (var destinationZipFileStream = new FileStream(destinationZip, FileMode.Open))
using (var destinationZipArchive = new ZipArchive(destinationZipFileStream, ZipArchiveMode.Update))
{
ZipArchiveEntry destinationWriter = destinationZipArchive.CreateEntry("Transfer.mp4");
using (var writer = new StreamWriter(destinationWriter.Open()))
using (var sourceStream = File.OpenRead(sourceFile))
{
sourceStream.CopyTo(writer.BaseStream);
}
}
Note: you will need to double check to make sure the file doesn't exist already or else you'll make duplicate files this way.
I need to create a zip file in memory, then send the zip file to the client. However, there are cases where the created zip file will need to contain other zip files that were also generated in memory. For instance, the file structure might look like this:
SendToClient.zip
InnerZip1.zip
File1.xml
File2.xml
InnerZip2.zip
File3.xml
File4.xml
I've been attempting to use the System.IO.Compression.ZipArchive library. I cannot use the System.IO.Compression.ZipFile library because my project's version of .NET is not compatible with it.
Here's an example of what I've tried.
public Stream GetMemoryStream() {
var memoryStream = new MemoryStream();
string fileContents = "Lorem ipsum dolor sit amet";
string entryName = "Lorem.txt";
string innerZipName = "InnerZip.zip";
using (var archive = new ZipArchive(memoryStream, ZipArchiveMode.Create, true)) {
ZipArchiveEntry entry = archive.CreateEntry(Path.Combine(innerZipName, entryName), CompressionLevel.Optimal);
using (var writer = new StreamWriter(entry.Open())) {
writer.Write(fileContents);
}
}
return memoryStream
}
However, this just puts Lorem.txt in a folder called "Inner.zip" (instead of in an actual zip file).
I can create an empty inner zip file if I create an entry called "Inner.zip" without writing to it. I can't add anything to it, though, and writing to an entry called "Inner.zip\Lorem.txt" afterward just creates a folder again (alongside the identically named empty .zip file).
I've also tried creating a separate archive, serializing it with a memory stream, then writing that to the original archive as a .zip.
public Stream CreateOuterZip() {
var memoryStream = new MemoryStream();
using (var archive = new ZipArchive(memoryStream, ZipArchiveMode.Create, true)) {
ZipArchiveEntry entry = archive.CreateEntry("Outer.zip", CompressionLevel.NoCompression);
using (var writer = new BinaryWriter(entry.Open())) {
writer.Write(GetMemoryStream().ToArray());
}
}
return memoryStream;
}
This just creates an invalid .zip file that windows doesn't know how to open, though.
Thanks in advance!
So I created a FileStream instead of a MemoryStream so the code can be tested easier
public static Stream CreateOuterZip()
{
string fileContents = "Lorem ipsum dolor sit amet";
// Final zip file
var fs = new FileStream(
Path.Combine(AppDomain.CurrentDomain.BaseDirectory, "SendToClient.zip"), FileMode.OpenOrCreate);
// Create inner zip 1
var innerZip1 = new MemoryStream();
using (var archive = new ZipArchive(innerZip1, ZipArchiveMode.Create, true))
{
var file1 = archive.CreateEntry("File1.xml");
using (var writer = new BinaryWriter(file1.Open()))
{
writer.Write(fileContents); // Change fileContents to real XML content
}
var file2 = archive.CreateEntry("File2.xml");
using (var writer = new BinaryWriter(file2.Open()))
{
writer.Write(fileContents); // Change fileContents to real XML content
}
}
// Create inner zip 2
var innerZip2 = new MemoryStream();
using (var archive = new ZipArchive(innerZip2, ZipArchiveMode.Create, true))
{
var file3 = archive.CreateEntry("File3.xml");
using (var writer = new BinaryWriter(file3.Open()))
{
writer.Write(fileContents); // Change fileContents to real XML content
}
var file4 = archive.CreateEntry("File4.xml");
using (var writer = new BinaryWriter(file4.Open()))
{
writer.Write(fileContents); // Change fileContents to real XML content
}
}
using (var archive = new ZipArchive(fs, ZipArchiveMode.Create, true))
{
// Create inner zip 1
var innerZipEntry = archive.CreateEntry("InnerZip1.zip");
innerZip1.Position = 0;
using (var s = innerZipEntry.Open())
{
innerZip1.WriteTo(s);
}
// Create inner zip 2
var innerZipEntry2 = archive.CreateEntry("InnerZip2.zip");
innerZip2.Position = 0;
using (var s = innerZipEntry2.Open())
{
innerZip2.WriteTo(s);
}
}
fs.Close();
return fs; // The file is written, can probably just close this
}
You can obviously modify this method to return a MemoryStream, or change the method to Void to just have the zip file written out to disk
You should create ZipArchive for internal zip file also. Write it to stream (memorystream). And after write this stream as general stream into main zip.
static Stream Inner() {
var memoryStream = new MemoryStream();
using (var archive = new ZipArchive(memoryStream, ZipArchiveMode.Create, true)) {
var demoFile = archive.CreateEntry("foo2.txt");
using (var entryStream = demoFile.Open())
using (var streamWriter = new StreamWriter(entryStream)) {
streamWriter.Write("Bar2!");
}
}
return memoryStream;
}
static void Main(string[] args) {
using (var memoryStream = new MemoryStream()) {
using (var archive = new ZipArchive(memoryStream, ZipArchiveMode.Create, true)) {
var demoFile = archive.CreateEntry("foo.txt");
using (var entryStream = demoFile.Open())
using (var streamWriter = new StreamWriter(entryStream)) {
streamWriter.Write("Bar!");
}
var zip = archive.CreateEntry("inner.zip");
using (var entryStream = zip.Open()) {
var inner = Inner();
inner.Seek(0, SeekOrigin.Begin);
inner.CopyTo(entryStream);
}
}
using (var fileStream = new FileStream(#"d:\test.zip", FileMode.Create)) {
memoryStream.Seek(0, SeekOrigin.Begin);
memoryStream.CopyTo(fileStream);
}
}
Thanks to this answer.
Given a zip file, I need to re-create it with a specified compression level (eg, no compression).
I'm nearly there, but get the error:
Failed: Number of entries expected in End Of Central Directory does not correspond to number of entries in Central Directory.
If I save the recreated zip file to windows, it looks like it's correct (correct file size, entries all exist with correct file sizes) but none of the files are extractable.
public static byte[] ReCompress(byte[] originalArchive, CompressionLevel newCompressionLevel)
{
var entries = new Dictionary<string, byte[]>();
///////////////////////////
// STEP 1: EXTRACT ALL FILES
///////////////////////////
using (var ms = new MemoryStream(originalArchive))
using (var originalZip = new ZipArchive(ms, ZipArchiveMode.Read))
{
foreach (var entry in originalZip.Entries)
{
var isFolder = entry.FullName.EndsWith("/");
if (!isFolder)
{
using (var stream = entry.Open())
using (var entryMS = new MemoryStream())
{
stream.CopyTo(entryMS);
entries.Add(entry.FullName, entryMS.ToArray());
}
}
else
{
entries.Add(entry.FullName, new byte[0]);
}
}
}
///////////////////////////
// STEP 2: BUILD ZIP FILE
///////////////////////////
using (var ms = new MemoryStream())
using (var newArchive = new ZipArchive(ms, ZipArchiveMode.Create, true))
{
foreach (var uncompressedEntry in entries)
{
var newEntry = newArchive.CreateEntry(uncompressedEntry.Key, newCompressionLevel);
using (var entryStream = newEntry.Open())
using (var writer = new BinaryWriter(entryStream, Encoding.UTF8))
{
writer.Write(uncompressedEntry.Value);
}
}
return ms.ToArray();
}
}
At the end of the function if I do:
File.WriteAllBytes(#"D:\test.zip", ms.ToArray());
It creates a correctly structure archive sized 90mb but no files are extractable.
If I end with return ms.ToArray() it returns a ~130kb byte array.
Zip archive is broken because you read its content from MemoryStream before it is finished. In order to finish archive creation you need to call newArchive.Dispose() before calling ms.ToArray().
In this particular case you can do it like this:
using (var ms = new MemoryStream())
{
using (var newArchive = new ZipArchive(ms, ZipArchiveMode.Create, true))
{
foreach (var uncompressedEntry in entries)
{
var newEntry = newArchive.CreateEntry(uncompressedEntry.Key, newCompressionLevel);
using (var entryStream = newEntry.Open())
using (var writer = new BinaryWriter(entryStream, Encoding.UTF8))
{
writer.Write(uncompressedEntry.Value);
}
}
}
return ms.ToArray();
}