Develop Reference

C# - How can I download a zip file from url, unzip it, and read the extracted files, all in memory? [duplicate]

I have files (from 3rd parties) that are being FTP'd to a directory on our server. I download them and process them even 'x' minutes. Works great.
Now, some of the files are .zip files. Which means I can't process them. I need to unzip them first.
FTP has no concept of zip/unzipping - so I'll need to grab the zip file, unzip it, then process it.
Looking at the MSDN zip api, there seems to be no way i can unzip to a memory stream?
So is the only way to do this...
Unzip to a file (what directory? need some -very- temp location ...)
Read the file contents
Delete file.
NOTE: The contents of the file are small - say 4k <-> 1000k.

Zip compression support is built in:
using System.IO;
using System.IO.Compression;
// ^^^ requires a reference to System.IO.Compression.dll
static class Program
{
const string path = ...
static void Main()
{
using(var file = File.OpenRead(path))
using(var zip = new ZipArchive(file, ZipArchiveMode.Read))
{
foreach(var entry in zip.Entries)
{
using(var stream = entry.Open())
{
// do whatever we want with stream
// ...
}
}
}
}
}
Normally you should avoid copying it into another stream - just use it "as is", however, if you absolutely need it in a MemoryStream, you could do:
using(var ms = new MemoryStream())
{
stream.CopyTo(ms);
ms.Position = 0; // rewind
// do something with ms
}

You can use ZipArchiveEntry.Open to get a stream.
This code assumes the zip archive has one text file.
using (FileStream fs = new FileStream(path, FileMode.Open))
using (ZipArchive zip = new ZipArchive(fs) )
{
var entry = zip.Entries.First();
using (StreamReader sr = new StreamReader(entry.Open()))
{
Console.WriteLine(sr.ReadToEnd());
}
}

using (ZipArchive archive = new ZipArchive(webResponse.GetResponseStream()))
{
foreach (ZipArchiveEntry entry in archive.Entries)
{
Stream s = entry.Open();
var sr = new StreamReader(s);
var myStr = sr.ReadToEnd();
}
}

Looks like here is what you need:
using (var za = ZipFile.OpenRead(path))
{
foreach (var entry in za.Entries)
{
using (var r = new StreamReader(entry.Open()))
{
//your code here
}
}
}

You can use SharpZipLib among a variety of other libraries to achieve this.
You can use the following code example to unzip to a MemoryStream, as shown on their wiki:
using ICSharpCode.SharpZipLib.Zip;
// Compresses the supplied memory stream, naming it as zipEntryName, into a zip,
// which is returned as a memory stream or a byte array.
//
public MemoryStream CreateToMemoryStream(MemoryStream memStreamIn, string zipEntryName) {
MemoryStream outputMemStream = new MemoryStream();
ZipOutputStream zipStream = new ZipOutputStream(outputMemStream);
zipStream.SetLevel(3); //0-9, 9 being the highest level of compression
ZipEntry newEntry = new ZipEntry(zipEntryName);
newEntry.DateTime = DateTime.Now;
zipStream.PutNextEntry(newEntry);
StreamUtils.Copy(memStreamIn, zipStream, new byte[4096]);
zipStream.CloseEntry();
zipStream.IsStreamOwner = false; // False stops the Close also Closing the underlying stream.
zipStream.Close(); // Must finish the ZipOutputStream before using outputMemStream.
outputMemStream.Position = 0;
return outputMemStream;
// Alternative outputs:
// ToArray is the cleaner and easiest to use correctly with the penalty of duplicating allocated memory.
byte[] byteArrayOut = outputMemStream.ToArray();
// GetBuffer returns a raw buffer raw and so you need to account for the true length yourself.
byte[] byteArrayOut = outputMemStream.GetBuffer();
long len = outputMemStream.Length;
}

Ok so combining all of the above, suppose you want to in a very simple way take a zip file called
"file.zip" and extract it to "C:\temp" folder. (Note: This example was only tested for compress text files) You may need to do some modifications for binary files.
using System.IO;
using System.IO.Compression;
static void Main(string[] args)
{
//Call it like this:
Unzip("file.zip",#"C:\temp");
}
static void Unzip(string sourceZip, string targetPath)
{
using (var z = ZipFile.OpenRead(sourceZip))
{
foreach (var entry in z.Entries)
{
using (var r = new StreamReader(entry.Open()))
{
string uncompressedFile = Path.Combine(targetPath, entry.Name);
File.WriteAllText(uncompressedFile,r.ReadToEnd());
}
}
}
}

ZipArchive gives Unexpected end of data corrupted error

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

Converting DotNetZip memory stream to string

I am trying to read a file within a zip to check if that file has a certain string in it. But I can seem to get the "file" (memory stream) into a string in order to search it.
When I use the following code "stringOfStream" is always blank, what am I doing wrong? The reader always has a length and read byte returns different numbers.
using (ZipFile zip = ZipFile.Read(currentFile.FullName))
{
ZipEntry e = zip[this.searchFile.Text];
using (MemoryStream reader = new MemoryStream())
{
e.Extract(reader);
var stringReader = new StreamReader(reader);
var stringOfStream = stringReader.ReadToEnd();
}
}
Thanks

I think when you call Extract the position of the stream goes to the end of the file, so you need to reposition again to get the data.
Can you try this please :
using (ZipFile zip = ZipFile.Read(currentFile.FullName))
{
ZipEntry e = zip[this.searchFile.Text];
using (MemoryStream reader = new MemoryStream())
{
e.Extract(reader);
reader.Position = 0;
var stringReader = new StreamReader(reader);
var stringOfStream = stringReader.ReadToEnd();
}
}
Check if it works or not.

Merging Multiple Text Files - StreamWriter Not Writing One File?

I simply want to merge all text files in a given directory, similar to the following command prompt command:
cd $directory
copy * result.txt
I've written the following code, which almost accomplishes what I want, but it's doing something strange. When the StreamWriter writes the first file (or when i = 0), it doesn't actually write any content - the file size remains 0 bytes, despite the first file being ~300 KB. However, the other file writes execute successfully.
If I compare the output from the command prompt to the output from the C# code in diff, you can see that a large block of text is missing. Additionally, the command prompt result is 1,044 KB where the C# result is 700 KB.
string[] txtFiles = Directory.GetFiles(filepath);
using (StreamWriter writer = new StreamWriter(filepath + "result.txt"))
{
for (int i = 0; i < txtFiles.Length; i++)
{
using (StreamReader reader = File.OpenText(txtFiles[i]))
{
writer.Write(reader.ReadToEnd());
}
}
}
Am I using the StreamWriter / StreamReader incorrectly?

Minimalistic implementation, reading the bytes and writing them instead of using a stream for reading - please note, that you should handle the IOException correctly to avoid misbehavior:
var newline = Encoding.ASCII.GetBytes(Environment.NewLine);
var files = Directory.GetFiles(filepath);
try
{
using (var writer = File.Open(Path.Combine(filepath, "result.txt"), FileMode.Create))
foreach (var text in files.Select(File.ReadAllBytes))
{
writer.Write(text, 0, text.Length);
writer.Write(newline, 0, newline.Length);
}
}
catch (IOException)
{
// File might be used by different process or you have insufficient permissions
}

Here, hope it helps you. Note: By copying from a stream to another you save some ram and greatly improve performance.
class Program
{
static void Main(string[] args)
{
string filePath = #"C:\Users\FunkyName\Desktop";
string[] txtFiles = Directory.GetFiles(filePath, "*.txt");
using (Stream stream = File.Open(Path.Combine(filePath, "result.txt"), FileMode.OpenOrCreate))
{
for (int i = 0; i < txtFiles.Length; i++)
{
string fileName = txtFiles[i];
try
{
using (Stream fileStream = File.Open(fileName, FileMode.Open, FileAccess.Read))
{
fileStream.CopyTo(stream);
}
}
catch (IOException e)
{
// Handle file open exception
}
}
}
}
}

I wrote your code , it works properly! only change the line :
using (StreamWriter writer = new StreamWriter(filepath + "result.txt"))
to:
using (StreamWriter writer = new StreamWriter(filepath + "/result.txt"))
i guess you can't see the file because it is saved in another folder .

Zip files and attach them to MailMessage without saving a file

I'm working on a little C# ASP.NET web app that pulls 3 files from my server, creates a zip of those files, and sends the zip file to an e-mail recipient.
The problem I'm having is finding a way to combine those 3 files without creating a zip file on the hard drive of the server. I think I need to use some sort of memorystream or filestream, but I'm in a little beyond my understanding when it comes to merging them into 1 zip file. I've tried SharpZipLib and DotNetZip, but I haven't been able to figure it out.
The reason I don't want the zip saved locally is that there might be a number of users on this app at once, and I don't want to clog up my server machine with those zips. I'm looking for 2 answers, how to zip files without saving the zip as a file, and how to attach that zip to a MailMessage.

Check this example for SharpZipLib:
https://github.com/icsharpcode/SharpZipLib/wiki/Zip-Samples#wiki-anchorMemory
using ICSharpCode.SharpZipLib.Zip;
// Compresses the supplied memory stream, naming it as zipEntryName, into a zip,
// which is returned as a memory stream or a byte array.
//
public MemoryStream CreateToMemoryStream(MemoryStream memStreamIn, string zipEntryName) {
MemoryStream outputMemStream = new MemoryStream();
ZipOutputStream zipStream = new ZipOutputStream(outputMemStream);
zipStream.SetLevel(3); //0-9, 9 being the highest level of compression
ZipEntry newEntry = new ZipEntry(zipEntryName);
newEntry.DateTime = DateTime.Now;
zipStream.PutNextEntry(newEntry);
StreamUtils.Copy(memStreamIn, zipStream, new byte[4096]);
zipStream.CloseEntry();
zipStream.IsStreamOwner = false; // False stops the Close also Closing the underlying stream.
zipStream.Close(); // Must finish the ZipOutputStream before using outputMemStream.
outputMemStream.Position = 0;
return outputMemStream;
// Alternative outputs:
// ToArray is the cleaner and easiest to use correctly with the penalty of duplicating allocated memory.
byte[] byteArrayOut = outputMemStream.ToArray();
// GetBuffer returns a raw buffer raw and so you need to account for the true length yourself.
byte[] byteArrayOut = outputMemStream.GetBuffer();
long len = outputMemStream.Length;
}

Try this:
public static Attachment CreateAttachment(string fileNameAndPath, bool zipIfTooLarge = true, int bytes = 1 << 20)
{
if (!zipIfTooLarge)
{
return new Attachment(fileNameAndPath);
}
var fileInfo = new FileInfo(fileNameAndPath);
// Less than 1Mb just attach as is.
if (fileInfo.Length < bytes)
{
var attachment = new Attachment(fileNameAndPath);
return attachment;
}
byte[] fileBytes = File.ReadAllBytes(fileNameAndPath);
using (var memoryStream = new MemoryStream())
{
string fileName = Path.GetFileName(fileNameAndPath);
using (var zipArchive = new ZipArchive(memoryStream, ZipArchiveMode.Create))
{
ZipArchiveEntry zipArchiveEntry = zipArchive.CreateEntry(fileName, CompressionLevel.Optimal);
using (var streamWriter = new StreamWriter(zipArchiveEntry.Open()))
{
streamWriter.Write(Encoding.Default.GetString(fileBytes));
}
}
var attachmentStream = new MemoryStream(memoryStream.ToArray());
string zipname = $"{Path.GetFileNameWithoutExtension(fileName)}.zip";
var attachment = new Attachment(attachmentStream, zipname, MediaTypeNames.Application.Zip);
return attachment;
}
}