I'm working on a project where I want to obtain the file from a compressed archive which only holds a single file. The files I'm interested in opening are of the file extension .als (Ableton Live Set) and from 7Zip I've found they use the gzip compression type.
I've found some code which let's me easily obtain the file from the archive:
using (ZipArchive archive = new ZipArchive(zipToOpen, ZipArchiveMode.Read))
{
var entryName = Path.GetFileNameWithoutExtension(txt_file_select.Text);
ZipArchiveEntry entry = archive.GetEntry(entryName);
using (StreamReader reader = new StreamReader(entry.Open()))
{
String fileContent = reader.ReadToEnd();
Debug.WriteLine(fileContent);
}
}
This code opens the file from the stream zipToOpen which has the same filepath as txt_file_select.Text. The single entry that I'm interested in seems to always have the same name as the zipped file itself so this should always give me an entry for any compressed archive that uses zip type. But this code does not accept gzip types in the current state as I get:
Exception thrown: 'System.IO.InvalidDataException' in System.IO.Compression.dll
An exception of type 'System.IO.InvalidDataException' occurred in System.IO.Compression.dll but was not handled in user code
End of Central Directory record could not be found.
I've found that you can open gzip type archives using the GZipStream class. so I added this line:
using (GZipStream zipToOpen = new GZipStream(new FileStream(txt_file_select.Text, FileMode.Open), CompressionMode.Decompress))
which still results in the same exception being thrown. I'm assuming the problem is in converting from GZipStream to ZipArchive.
Here is a simple reproducible example:
using System.Diagnostics;
using System.IO.Compression;
String filepath = #"path\to\als\file.als";
using (GZipStream zipToOpen = new GZipStream(new FileStream(filepath, FileMode.Open), CompressionMode.Decompress))
{
using (ZipArchive archive = new ZipArchive(zipToOpen, ZipArchiveMode.Read))
{
var entryName = Path.GetFileNameWithoutExtension(filepath);
ZipArchiveEntry entry = archive.GetEntry(entryName);
using (StreamReader reader = new StreamReader(entry.Open()))
{
String fileContent = reader.ReadToEnd();
Debug.WriteLine(fileContent);
}
}
}
How do I convert the GZipStream into a ZipArchive?
Related
Can someone tell me what's wrong with my code? I want to zip multiple xml into one file yet the result file is always empty.
using (MemoryStream zipStream = new MemoryStream())
{
using (ZipArchive zip = new ZipArchive(zipStream, ZipArchiveMode.Create, true))
{
string[] xmls = Directory.GetFiles(#"c:\temp\test", "*.xml");
foreach (string xml in xmls)
{
var file = zip.CreateEntry(xml);
using (var entryStream = file.Open())
using (var streamWriter = new StreamWriter(entryStream))
{
streamWriter.Write(xml);
}
}
}
using (FileStream fs = new FileStream(#"C:\Temp\test.zip", FileMode.Create))
{
zipStream.Position = 0;
zipStream.CopyTo(fs);
}
}
See the remarks in the documentation (emphasis mine):
The entryName string should reflect the relative path of the entry you want to create within the zip archive. There is no restriction on the string you provide. However, if it is not formatted as a relative path, the entry is created, but you may get an exception when you extract the contents of the zip archive. If an entry with the specified path and name already exists in the archive, a second entry is created with the same path and name.
You are using an absolute path here:
var file = zip.CreateEntry(xml);
My guess is that when you try to open the archive, it is failing silently to show the entries.
Change your code to use the names of the files without their path:
var file = zip.CreateEntry(Path.GetFileName(xml));
As a separate issue, notice that you're just writing the name of the file to the ZIP entry, rather than the actual file. I imagine you want something like this instead:
var zipEntry = zip.CreateEntry(Path.GetFileName(xml));
using (var entryStream = file.Open())
{
using var fileStream = File.OpenRead(xml);
fileStream.CopyTo(entryStream);
}
I am making a GET request using HttpClient to download a zip file from the internet.
I want to extract all the files contained in the zip file without saving the zip file to disk.
Currently, I am able to download and save the zip file to disk, extract its contents and then delete the zip file from disk. This perfectly fine. However, I want to optimize the process.
I found a way to extract the contents directly from the downloaded zip stream but I have to specify the filenames and extensions.
I am not sure how to extract the contents while preserving their original filenames and extensions without me specifying them.
Current Approach:
string requestUri = "https://www.nuget.org/api/v2/package/" + PackageName + "/" + PackageVersion;
HttpResponseMessage response = await client.GetAsync(requestUri);
response.EnsureSuccessStatusCode();
using Stream PackageStream = await response.Content.ReadAsStreamAsync();
SaveStream($"{DownloadPath}.zip", PackageStream);
ZipFile.ExtractToDirectory($"{DownloadPath}.zip", ExtractPath);
File.Delete($"{DownloadPath}.zip");
// Directly extract Zip contents without saving file and without losing filename and extension
using (ZipArchive archive = new ZipArchive(await response.Content.ReadAsStreamAsync()))
{
foreach (ZipArchiveEntry entry in archive.Entries)
{
using (Stream stream = entry.Open())
{
using (FileStream file = new FileStream("file.txt", FileMode.Create, FileAccess.Write))
{
stream.CopyTo(file);
}
}
}
}
.NET 4.8
.NET Core 3.1
C# 8.0
Any help in this regards would be appreciated.
Please feel free to comment on alternative approaches or suggestions.
Thank you in advance.
ZipArchiveEntry has a Name and FullName property that can be used to get the names of the files within the archive while preserving their original filenames and extensions
The FullName property contains the relative path, including the subdirectory hierarchy, of an entry in a zip archive. (In contrast, the Name property contains only the name of the entry and does not include the subdirectory hierarchy.)
For example
using (ZipArchive archive = new ZipArchive(await response.Content.ReadAsStreamAsync())) {
foreach (ZipArchiveEntry entry in archive.Entries) {
using (Stream stream = entry.Open()) {
string destination = Path.GetFullPath(Path.Combine(downloadPath, entry.FullName));
var directory = Path.GetDirectoryName(destination);
if (!Directory.Exists(directory))
Directory.CreateDirectory(directory);
using (FileStream file = new FileStream(destination, FileMode.Create, FileAccess.Write)) {
await stream.CopyToAsync(file);
}
}
}
}
will extract the files in the same subdirectory hierarchy as they were stored in the archive while if entry.Name was used, all the files would be extracted to the same location.
Using MailKit in .NET CORE an attachement can be loaded using:
bodyBuilder.Attachments.Add(FILE);
I'm trying to attach a file from inside a ZIP file using:
using System.IO.Compression;
string zipPath = #"./html-files.ZIP";
using (ZipArchive archive = ZipFile.OpenRead(zipPath))
{
// bodyBuilder.Attachments.Add("msg.html");
bodyBuilder.Attachments.Add(archive.GetEntry("msg.html"));
}
But it did not work, and gave me APP\"msg.html" not found, which means it is trying to load a file with the same name from the root directory instead of the zipped one.
bodyBuilder.Attachments.Add() doesn't have an overload that takes a ZipArchiveEntry, so using archive.GetEntry("msg.html") has no chance of working.
Most likely what is happening is that the compiler is casting the ZipArchiveEntry to a string which happens to be APP\"msg.html" which is why you get that error.
What you'll need to do is extract the content from the zip archive and then add that to the list of attachments.
using System.IO;
using System.IO.Compression;
string zipPath = #"./html-files.ZIP";
using (ZipArchive archive = ZipFile.OpenRead (zipPath)) {
ZipArchiveEntry entry = archive.GetEntry ("msg.html");
var stream = new MemoryStream ();
// extract the content from the zip archive entry
using (var content = entry.Open ())
content.CopyTo (stream);
// rewind the stream
stream.Position = 0;
bodyBuilder.Attachments.Add ("msg.html", stream);
}
I have a file which I upload to a temp folder after having done some image resizing as follows:
/uploads/temp/myfile.jpg
How can I read this file into a binary reader as follows if all I have is the file path as indicated above i.e. In the example below I get error:
Error in saving fileSystem.IO.DirectoryNotFoundException: Could not find a part of the path 'C:\\uploads\\temp\\myfile.jpg'
var fs = new FileStream('/uploads/temp/myfile.jpg', FileMode.Open);
using (var reader = new BinaryReader(fs))
{
image.ImageContent = reader.ReadBytes((int)fs.Length);
}
Managed to fix by doing the following:
var fs = new FileStream(HostingEnvironment.MapPath("~/") + "/uploads/temp/myfile.jpg", FileMode.Open);
I have converted a .zip file into a byte[], and now I am trying to convert the byte[] back to the original .zip file. I am running out of the options that I have tried. Anyone give me a pointer how can I achieve this?
You want the System.IO.Compression.ZipArchive class:
using (ZipArchive zip = ZipFile.Open("test.zip", ZipArchiveMode.Create))
{
var entry = zip.CreateEntry("File Name.txt");
using (StreamWriter sw = new StreamWriter(entry.Open()))
{
sw.Write("Some Text");
}
}
using (ZipArchive zip = ZipFile.Open("test.zip", ZipArchiveMode.Read))
{
foreach (ZipArchiveEntry entry in zip.Entries)
{
using (StreamReader sr = new StreamReader(entry.Open()))
{
var result = sr.ReadToEnd();
}
}
}
You probably don't want to read in the raw zip file into a byte array first and then try to decompress it. Instead, access it through this helper method.
Note the use of ZipArchive.Entries to access the sub-files stored in the single zip archive; this tripped me up when first learning to use zip files.