I'm trying to compress two text files to a zip file. This is how my public method looks like:
public ActionResult Index()
{
byte[] file1 = System.IO.File.ReadAllBytes(#"C:\file1.txt");
byte[] file2 = System.IO.File.ReadAllBytes(#"C:\file2.txt");
Dictionary<string, byte[]> fileList = new Dictionary<string, byte[]>();
fileList.Add("file1.txt", file1);
fileList.Add("file2.txt", file2);
CompressToZip("zip.zip", fileList);
return View();
}
This is how my compress method looks like:
private void CompressToZip(string fileName, Dictionary<string, byte[]> fileList)
{
using (var memoryStream = new MemoryStream())
{
foreach (var file in fileList)
{
using (var archive = new ZipArchive(memoryStream, ZipArchiveMode.Create, true))
{
var demoFile = archive.CreateEntry(file.Key);
using (var entryStream = demoFile.Open())
using (var b = new BinaryWriter(entryStream))
{
b.Write(file.Value);
}
}
}
using (var fileStream = new FileStream(fileName, FileMode.Create))
{
memoryStream.Seek(0, SeekOrigin.Begin);
memoryStream.CopyTo(fileStream);
}
}
}
In this approach, the zip folder is perfectly created. However the issue is I'm getting only one file inside the zip folder (Just the second file will be created inside the zip folder).
There are no errors found.
Question: How to compress both text files into the zip folder?
Thank you in advanced!
Your code is actually saving two separate zip archives to the zip.zip file (a new ZipArchive is created for each file to be compressed). The first zip archive contains only file1.txt, the second only file2.txt. When you open zip.zip in Windows Explorer, it shows just the contents of the second zip archive.
To create a single zip archive containing both files, just move the creation of the ZipArchive outside of your fileList loop:
using (var archive = new ZipArchive(memoryStream, ZipArchiveMode.Create, true))
{
foreach (var file in fileList)
{
var demoFile = archive.CreateEntry(file.Key);
using (var entryStream = demoFile.Open())
using (var b = new BinaryWriter(entryStream))
{
b.Write(file.Value);
}
}
}
At first glance, I would suggest that your foreach statement around the using var (archive = new ZipArchive...) is the wrong way round.
This way you are creating a new ZipArchive each time you iterate the foreach loop.
Surely you want to create the ZipArchive and loop through the foreach inside of that?
Like this:
private void CompressToZip(string fileName, Dictionary<string, byte[]> fileList)
{
using (var memoryStream = new MemoryStream())
{
using (var archive = new ZipArchive(memoryStream, ZipArchiveMode.Create, true))
{
foreach (var file in fileList)
{
var demoFile = archive.CreateEntry(file.Key);
using (var entryStream = demoFile.Open())
using (var b = new BinaryWriter(entryStream))
{
b.Write(file.Value);
}
}
}
using (var fileStream = new FileStream(fileName, FileMode.Create))
{
memoryStream.Seek(0, SeekOrigin.Begin);
memoryStream.CopyTo(fileStream);
}
}
}
Hope this helps!
string startPath = #"c:\example\start";
string zipPath = #"c:\example\result.zip";
ZipFile.CreateFromDirectory(startPath, zipPath);
Related
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();
}
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").
We use ASP.NET core and have this code:
public static void Compress(string sourceDirectoryName, string destinationArchiveFileName)
{
var directoryName = Path.GetFileName(sourceDirectoryName);
var remotePath = sourceDirectoryName.Split(new[] { directoryName }, StringSplitOptions.RemoveEmptyEntries).First();
using (var zipStream = new MemoryStream())
{
using (var zip = new ZipArchive(zipStream, ZipArchiveMode.Create, true, Encoding.UTF8))
{
zip.CreateEntry(string.Concat(directoryName, directorySlash));
foreach (var path in Directory.EnumerateFileSystemEntries(sourceDirectoryName, "*", SearchOption.AllDirectories))
{
if (!Directory.Exists(path))
zip.CreateEntryFromFile(path, path.RemoveSubString(remotePath).ReplacePathSeparatorOnSlash());
else
zip.CreateEntry(string.Concat(path.RemoveSubString(remotePath).ReplacePathSeparatorOnSlash(), directorySlash));
}
}
using (var outputZip = new FileStream(destinationArchiveFileName, FileMode.Create))
{
zipStream.Seek(0, SeekOrigin.Begin);
zipStream.CopyTo(outputZip);
}
}
}
After zipping sourceDirectoryName that contains russian symbols if we open this archive with windows explorer, we see the following:
and where the name marked with green is correct, and the name marked with red has its name encoding changed.
If we use the following code:
ZipFile.CreateFromDirectory(sourceDirectoryName, destinationArchiveFileName);
We have the same problem. How to fix this problem?
I have found answer. I have to use encoding 866:
using (var zip = new ZipArchive(zipStream, ZipArchiveMode.Create, true, Encoding.GetEncoding(866)))
{
...
}
I'm trying to create a ZIP archive with multiple text files as follows:
Dictionary<string, string> Values = new Dictionary<string, string>();
using (var memoryStream = new MemoryStream())
{
string zip = #"C:\Temp\ZipFile.zip";
foreach (var item in Values)
{
using (var archive = new ZipArchive(memoryStream, ZipArchiveMode.Create, true))
{
var file = archive.CreateEntry(item.Key + ".txt");
using (var entryStream = file.Open())
using (var streamWriter = new StreamWriter(entryStream))
{
streamWriter.Write(item.Value);
}
}
}
using (var fileStream = new FileStream(zip, FileMode.Create))
{
memoryStream.Seek(0, SeekOrigin.Begin);
memoryStream.CopyTo(fileStream);
}
}
However, the ZIP is created with only the last text file, what's wrong?
You are creating ZipArchive on each iteration. Swapping foreach and using should solve it:
Dictionary<string, string> Values = new Dictionary<string, string>();
using (var memoryStream = new MemoryStream())
{
string zip = #"C:\Temp\ZipFile.zip";
using (var archive = new ZipArchive(memoryStream, ZipArchiveMode.Create, true))
{
foreach (var item in Values)
{
var file = archive.CreateEntry(item.Key + ".txt");
using (var entryStream = file.Open())
using (var streamWriter = new StreamWriter(entryStream))
{
streamWriter.Write(item.Value);
}
}
}
using (var fileStream = new FileStream(zip, FileMode.Create))
{
memoryStream.Seek(0, SeekOrigin.Begin);
memoryStream.CopyTo(fileStream);
}
}
Each time your foreach loop runs it has the ZipArchiveMode as Create. That should be the problem, so it generates new zip everytime with new content on it, such as the last text file. Create an exception for each loop run after the first one it should be solved.