I am writing data into text file and using below code,
await using var file = new StreamWriter(filePath);
foreach (var packet in resultPackets)
{
file.WriteLine(JsonConvert.SerializeObject(packet));
}
And I am using below code to zip the file with password protected using `DotNetZip,
using (ZipFile zip = new ZipFile())
{
zip.Password = "password";
zip.AddFile(filePath);
zip.Save(#"C:\tmp\data4.zip");
}
Is there a way to combined both, I want to create a file on the fly as password protected.
I don't
want to create first file with data, t
then create zip file from it
and delete the base file
Is this possible? Thanks!
Okay, so since this is still unanswered, here's a small program that does the job for me:
using (var stream = new MemoryStream())
using (var streamWriter = new StreamWriter(stream))
{
// Insert your code in here, i.e.
//foreach (var packet in resultPackets)
//{
// streamWriter.WriteLine(JsonConvert.SerializeObject(packet));
//}
// ... instead I write a simple string.
streamWriter.Write("Hello World!");
// Make sure the contents from the StreamWriter are actually flushed into the stream, then seek the beginning of the stream.
streamWriter.Flush();
stream.Seek(0, SeekOrigin.Begin);
using (ZipFile zip = new ZipFile())
{
zip.Password = "password";
// Write the contents of the stream into a file that is called "test.txt"
zip.AddEntry("test.txt", stream);
// Save the archive.
zip.Save("test.zip");
}
}
Note how AddEntry does not create any form of temporary file. Instead, when the archive is saved, the contents of the stream are read and put into a compressed file within the archive. However, be aware that the whole content of the file are completely kept in memory before it the archive is written to the disk.
When I try to read the files contained in a large zip file (expands to 3.9GB) using ZipArchive, I get an InvalidDataException exception "The archive entry was compressed using an unsupported compression method" on the StreamReader line.
using (ZipArchive archive = ZipFile.OpenRead(zippedFileName))
{
foreach (ZipArchiveEntry entry in archive.Entries)
{
using (StreamReader sr = new StreamReader(entry.Open()))
{
...
}
}
}
The same code works file for a 600M (extracted) zip file.
Is there a way to get this to work?
How can I decompress (.zip) files without extracting to a new location in the .net framework? Specifically, I'm trying to read a filename.csv.zip into a DataTable.
I'm aware of .extractToDirectory (which is within ZipArchive) but I just want to extract it into an object in c# and I would like to not create a new file.
Hoping to be able to do this w/o third party libraries, but I'll take what I can get.
May be some bugs because I never tested this, but here you go:
List<byte[]> urmom = new List<byte[]>();
using (ZipArchive archive = ZipFile.OpenRead(zipPath))
foreach (ZipArchiveEntry entry in archive.Entries)
using (StreamReader r = new StreamReader(entry.Open()))
urmom.Add(r.ReadToEnd(entry));
Basically you use the ZipArchive's openread class to iterate through each entry. At this point, you can use the streamreader to read each entry. From there you can create a file from the stream and even read the filename if you want to. My code doesn't do this, a bit of laziness on my part.
Keep in mind that a compressed stream might contain multiple files. To resolve this is required to iterate through all entries of zip file in order to retrieve them and treat separately.
The sample bellow converts a sequence of bytes in a list of string where each one is the context of the files included in zipped folder:
public static IEnumerable<string> DecompressToEntriesTextContext(byte[] input)
{
var zipEntriesContext = new List<string>();
using (var compressedStream = new MemoryStream(input))
using (var zip = new ZipArchive(compressedStream, ZipArchiveMode.Read))
{
foreach(var entry in zip.Entries)
{
using (var entryStream = entry.Open())
using (var memoryEntryStream = new MemoryStream())
using (var reader = new StreamReader(memoryEntryStream))
{
entryStream.CopyTo(memoryEntryStream);
memoryEntryStream.Position = 0;
zipEntriesContext.Add(reader.ReadToEnd());
}
}
}
return zipEntriesContext;
}
I am using .NET 4.5, and the ZipFile class works great if I am trying to zip up an entire directory with "CreateFromDirectory". However, I only want to zip up one file in the directory. I tried pointing to a specific file (folder\data.txt), but that doesn't work. I considered the ZipArchive class since it has a "CreateEntryFromFile" method, but it seems this only allows you to create an entry into an existing file.
Is there no way to simply zip up one file without creating an empty zipfile (which has its issues) and then using the ZipArchiveExtension's "CreateEntryFromFile" method?
**This is also assuming I am working on a company program which cannot use third-party add-ons at the moment.
example from:http://msdn.microsoft.com/en-us/library/ms404280%28v=vs.110%29.aspx
string startPath = #"c:\example\start";
string zipPath = #"c:\example\result.zip";
string extractPath = #"c:\example\extract";
ZipFile.CreateFromDirectory(startPath, zipPath);
ZipFile.ExtractToDirectory(zipPath, extractPath);
But if startPath were to be #"c:\example\start\myFile.txt;", it would throw an error that the directory is invalid.
Use the CreateEntryFromFile off a an archive and use a file or memory stream:
Using a filestream if you are fine creating the zip file and then adding to it:
using (FileStream fs = new FileStream(#"C:\Temp\output.zip",FileMode.Create))
using (ZipArchive arch = new ZipArchive(fs, ZipArchiveMode.Create))
{
arch.CreateEntryFromFile(#"C:\Temp\data.xml", "data.xml");
}
Or if you need to do everything in memory and write the file once it is done, use a memory stream:
using (MemoryStream ms = new MemoryStream())
using (ZipArchive arch = new ZipArchive(ms, ZipArchiveMode.Create))
{
arch.CreateEntryFromFile(#"C:\Temp\data.xml", "data.xml");
}
Then you can write the MemoryStream to a file.
using (FileStream file = new FileStream("file.bin", FileMode.Create, System.IO.FileAccess.Write)) {
byte[] bytes = new byte[ms.Length];
ms.Read(bytes, 0, (int)ms.Length);
file.Write(bytes, 0, bytes.Length);
ms.Close();
}
Using file (or any) stream:
using (var zip = ZipFile.Open("file.zip", ZipArchiveMode.Create))
{
var entry = zip.CreateEntry("file.txt");
entry.LastWriteTime = DateTimeOffset.Now;
using (var stream= File.OpenRead(#"c:\path\to\file.txt"))
using (var entryStream = entry.Open())
stream.CopyTo(entryStream);
}
or briefer:
// reference System.IO.Compression
using (var zip = ZipFile.Open("file.zip", ZipArchiveMode.Create))
zip.CreateEntryFromFile("file.txt", "file.txt");
make sure you add references to System.IO.Compression
Update
Also, check out the new dotnet API documentation for ZipFile and ZipArchive too. There are a few examples there. There is also a warning about referencing System.IO.Compression.FileSystem to use ZipFile.
To use the ZipFile class, you must reference the
System.IO.Compression.FileSystem assembly in your project.
The simplest way to get this working is to use a temporary folder.
FOR ZIPPING:
Create a temp folder
Move file to folder
Zip folder
Delete folder
FOR UNZIPPING:
Unzip archive
Move file from temp folder to your location
Delete temp folder
In .NET, there are quite a few ways to tackle the problem, for a single file. If you don't want to learn everything there, you can get an abstracted library, like SharpZipLib (long standing open source library), sevenzipsharp (requires 7zip libs underneath) or DotNetZip.
just use following code for compressing a file.
public void Compressfile()
{
string fileName = "Text.txt";
string sourcePath = #"C:\SMSDBBACKUP";
DirectoryInfo di = new DirectoryInfo(sourcePath);
foreach (FileInfo fi in di.GetFiles())
{
//for specific file
if (fi.ToString() == fileName)
{
Compress(fi);
}
}
}
public static void Compress(FileInfo fi)
{
// Get the stream of the source file.
using (FileStream inFile = fi.OpenRead())
{
// Prevent compressing hidden and
// already compressed files.
if ((File.GetAttributes(fi.FullName)
& FileAttributes.Hidden)
!= FileAttributes.Hidden & fi.Extension != ".gz")
{
// Create the compressed file.
using (FileStream outFile =
File.Create(fi.FullName + ".gz"))
{
using (GZipStream Compress =
new GZipStream(outFile,
CompressionMode.Compress))
{
// Copy the source file into
// the compression stream.
inFile.CopyTo(Compress);
Console.WriteLine("Compressed {0} from {1} to {2} bytes.",
fi.Name, fi.Length.ToString(), outFile.Length.ToString());
}
}
}
}
}
}
I am trying to access a file in a .7z file. I know the name of the file in the zip folder and that it exists in the .7z file. Previously I've used the ExtractArchive(templocation) which just dumps all the files into a temporary location. Now I want to be able to grab a specific file without extracting the whole .7z file.
7Zip has a class called the SevenZipExtractor that has a method ExtractFile. I would think that is what I am looking for, but I can't find any decent documentation on it.
What I need clarification on is how to go about getting the Stream parameter passed in correctly.
I am using code like this;
//this grabs the zip file and creates a FileInfo array that hold the .7z file (assume there is only one)
DirectoryInfo directoryInfo = new DirectoryInfo(ApplicationPath);
FileInfo[] zipFile = directoryInfo.GetFiles("*.7z");
//This creates the zipextractor on the zip file I just placed in the zipFile FileInfo array
using (SevenZip.SevenZipExtractor zipExtractor = new SevenZip.SevenZipExtractor(zipFile[0].FullName))
//Here I should be able to use the ExtractFile method, however I don't understand the stream parameter, and I can't find any good documentation on the method itself. What is this method looking for?
{
zipExtractor.ExtractFile("ConfigurationStore.xml", Stream stream);
}
Setup a FileStream that SevenZip can write out to:
DirectoryInfo directoryInfo = new DirectoryInfo(ApplicationPath);
FileInfo[] zipFile = directoryInfo.GetFiles("*.7z");
using (SevenZip.SevenZipExtractor zipExtractor = new SevenZip.SevenZipExtractor(zipFile[0].FullName))
{
using (FileStream fs = new FileStream("", FileMode.Create)) //replace empty string with desired destination
{
zipExtractor.ExtractFile("ConfigurationStore.xml", fs);
}
}