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
Related
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 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);
}
}
}
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.
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.
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;