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Invalid zip file reading from a Stream in C#

I have the following code:
private static byte[] ConverterStringToByte(Stream body)
{
string fileName = "data_" + DateTime.Now.ToString("yyyyMMddhhmmss") + ".zip";
// Take out the bytes from the memory stream and safely close the stream
using (var ms = new MemoryStream())
{
body.CopyTo(ms);
using (var zipArchive = new ZipArchive(ms, ZipArchiveMode.Create, false))
{
var zipEntry = zipArchive.CreateEntry(fileName, CompressionLevel.Optimal);
using (BinaryWriter writer = new BinaryWriter(zipEntry.Open()))
{
ms.Position = 0;
writer.Write(ms.ToArray());
}
}
return ms.ToArray();
}
}
I am downloading the file successfully, however I'm getting
invalid file
when trying to open

I think it should be something like this. Not sure about fileName though, because it's the name of the file being put into archive, so I don't think it should have *.zip extension. Unless you are creating a zip of zips.
static byte[] ConverterStringToByte(Stream body)
{
string fileName = #"data_" + DateTime.Now.ToString("yyyyMMddhhmmss") + ".zip";
using (var ms = new MemoryStream())
{
using (var zipArchive = new ZipArchive(ms, ZipArchiveMode.Create, false))
{
var zipEntry = zipArchive.CreateEntry(fileName, CompressionLevel.Optimal);
using (var destStream = zipEntry.Open())
{
body.CopyTo(destStream);
}
}
return ms.ToArray();
}
}

Moving file directly into zip file

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.

C# ZipArchive - How to nest internal .zip files without writing to disk

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.

Invalid zip file after creating it with System.IO.Compression

I'm trying to create a zip file that contains one or more files.
I'm using the .NET framework 4.5 and more specifically System.IO.Compression namespace.
The objective is to allow a user to download a zip file through a ASP.NET MVC application.
The zip file is being generated and sent to the client but when I try to open it by doing double click on it I get the following error:
Windows cannot open the folder.
The compressed (zipped) folder ... is invalid.
Here's my code:
[HttpGet]
public FileResult Download()
{
var fileOne = CreateFile(VegieType.POTATO);
var fileTwo = CreateFile(VegieType.ONION);
var fileThree = CreateFile(VegieType.CARROT);
IEnumerable<FileContentResult> files = new List<FileContentResult>() { fileOne, fileTwo, fileThree };
var zip = CreateZip(files);
return zip;
}
private FileContentResult CreateFile(VegieType vType)
{
string fileName = string.Empty;
string fileContent = string.Empty;
switch (vType)
{
case VegieType.BATATA:
fileName = "batata.csv";
fileContent = "THIS,IS,A,POTATO";
break;
case VegieType.CEBOLA:
fileName = "cebola.csv";
fileContent = "THIS,IS,AN,ONION";
break;
case VegieType.CENOURA:
fileName = "cenoura.csv";
fileContent = "THIS,IS,A,CARROT";
break;
default:
break;
}
var fileBytes = Encoding.GetEncoding(1252).GetBytes(fileContent);
return File(fileBytes, MediaTypeNames.Application.Octet, fileName);
}
private FileResult CreateZip(IEnumerable<FileContentResult> files)
{
byte[] retVal = null;
if (files.Any())
{
using (MemoryStream zipStream = new MemoryStream())
{
using (ZipArchive archive = new ZipArchive(zipStream, ZipArchiveMode.Create, false))
{
foreach (var f in files)
{
var entry = archive.CreateEntry(f.FileDownloadName, CompressionLevel.Fastest);
using (var entryStream = entry.Open())
{
entryStream.Write(f.FileContents, 0, f.FileContents.Length);
entryStream.Close();
}
}
zipStream.Position = 0;
retVal = zipStream.ToArray();
}
}
}
return File(retVal, MediaTypeNames.Application.Zip, "horta.zip");
}
Can anyone please shed some light on why is windows saying that my zip file is invalid when I double click on it.
A final consideration, I can open it using 7-Zip.

You need to get the MemoryStream buffer via ToArray after the ZipArchive object gets disposed. Otherwise you end up with corrupted archive.
And please note that I have changed the parameters of ZipArchive constructor to keep it open when adding entries.
There is some checksumming going on when the ZipArchive is beeing disposed so if you read the MemoryStream before, it is still incomplete.
private FileResult CreateZip(IEnumerable<FileContentResult> files)
{
byte[] retVal = null;
if (files.Any())
{
using (MemoryStream zipStream = new MemoryStream())
{
using (ZipArchive archive = new ZipArchive(zipStream, ZipArchiveMode.Create, true))
{
foreach (var f in files)
{
var entry = archive.CreateEntry(f.FileDownloadName, CompressionLevel.Fastest);
using (BinaryWriter writer = new BinaryWriter(entry.Open()))
{
writer.Write(f.FileContents, 0, f.FileContents.Length);
writer.Close();
}
}
zipStream.Position = 0;
}
retVal = zipStream.ToArray();
}
}
return File(retVal, MediaTypeNames.Application.Zip, "horta.zip");
}

Just return the stream...
private ActionResult CreateZip(IEnumerable files)
{
if (files.Any())
{
MemoryStream zipStream = new MemoryStream();
using (ZipArchive archive = new ZipArchive(zipStream, ZipArchiveMode.Create, false))
{
foreach (var f in files)
{
var entry = archive.CreateEntry(f.FileDownloadName, CompressionLevel.Fastest);
using (var entryStream = entry.Open())
{
entryStream.Write(f.FileContents, 0, f.FileContents.Length);
entryStream.Close();
}
}
}
zipStream.Position = 0;
return File(zipStream, MediaTypeNames.Application.Zip, "horta.zip");
}
return new EmptyResult();
}

Try changing
using (ZipArchive archive = new ZipArchive(zipStream, ZipArchiveMode.Create, false))
to
using (ZipArchive archive = new ZipArchive(zipStream, ZipArchiveMode.Create, true))
In this usage, the archive is forced to write to the stream when it is closed. However, if the leaveOpen argument of the constructor is set to false, it will close the underlying stream too.

When I added a wrong name for the entry as in the example
var fileToZip = "/abc.txt";
ZipArchiveEntry zipFileEntry = zipArchive.CreateEntry(fileToZip);
I got the same error. After correcting the file name, it is ok now.

I got the "The compressed (zipped) folder ... is invalid." error because my entries were named with a leading "/" in front of them. Some zip extractors had no problem with this but the Windows one does. I resolved it by removing the slash from the entry name (from "/file.txt" to "file.txt").

How to reset the stream of StreamReader to beginning?

I have referred this Return StreamReader to Beginning, but couldn't figure out this problem.
This is code to read stream of a particular file in zip file. Here there are two stream of files inside two different zip files. Now I need to compare the streams.
I am unable to set the stream of BaseFileReader stream to beginning of stream.
using (FileStream BaseZipToOpen = new FileStream(BaseArchive,FileMode.Open) , CurrentZipToOpen = new FileStream(CurrentArchive,FileMode.Open))
{
using (ZipArchive BaseZip = new ZipArchive(BaseZipToOpen, ZipArchiveMode.Read), CurrentZip = new ZipArchive(CurrentZipToOpen, ZipArchiveMode.Read))
{
ZipArchiveEntry BaseFile = BaseZip.GetEntry(requiredFile);
ZipArchiveEntry CurrentFile = CurrentZip.GetEntry(requiredFile);
using (StreamReader BaseFileReader = new StreamReader(BaseFile.Open()), CurrentFileReader = new StreamReader(CurrentFile.Open()))
{
string baseFileLine, currentFileLine;
while (!CurrentFileReader.EndOfStream)
{
currentFileLine = CurrentFileReader.ReadLine();
while (!BaseFileReader.EndOfStream)
{
baseFileLine = BaseFileReader.ReadLine();
if (!currentFileLine.Equals(baseFileLine))
{
difference = true;
}
else
{
difference = false;
break;
}
}
// how to reset BaseFileReader Stream to beginning?
BaseZipToOpen.Seek(0, SeekOrigin.Begin); //This is not working
}
}
}
}

You can use
FileStream stream = new FileStream();
stream.Position = 0;