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;
}
}
Related
There is a client requirement to create a zip which consist of multiple files placed inside tree-like structured folders. It contains upto 150 files at maximum. When these files exceed approximately 160MB in memory stream OutOfMemoryException is thrown.
List item
Is there a config to increase memory allocated to this operation?
Any other alternative ways to solve this?
Sample code
MemoryStream memStream = new MemoryStream();
using (var zipStream = new ZipOutputStream(memStream))
{
foreach (FileModel fileToBeAddedInZip in listOfFiles)
{
byte[] fileBytes;
fileBytes = //Read the file from DB
ZipEntry fileEntry = null;
fileEntry = new ZipEntry(fileToBeAddedInZip.fileName)
{
Size = fileBytes.Length
};
zipStream.PutNextEntry(fileEntry);
zipStream.Write(fileBytes, 0, fileBytes.Length);
}
}
zipStream.Flush();
zipStream.Close();
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());
}
}
}
}
I am attempting to download a bunch of files that I am zipping up(archiving) via ZipoutputStream.
using (var zipStream = new ZipOutputStream(outputMemStream))
{
foreach (var documentIdString in documentUniqueIdentifiers)
{
...
var blockBlob = container.GetBlockBlobReference(documentId.ToString());
var fileMemoryStream = new MemoryStream();
blockBlob.DownloadToStream(fileMemoryStream);
zipStream.SetLevel(3);
fileMemoryStream.Position = 0;
ZipEntry newEntry = new ZipEntry(document.FileName);
newEntry.DateTime = DateTime.Now;
zipStream.PutNextEntry(newEntry);
fileMemoryStream.Seek(0, SeekOrigin.Begin);
StreamUtils.Copy(fileMemoryStream, zipStream, new byte[4096]);
zipStream.IsStreamOwner = false; // False stops the Close also Closing the underlying stream.
}
outputMemStream.Seek(0, SeekOrigin.Begin);
return outputMemStream;
}
In my controller I am returning the following code that should download the Zip file i created in the previous example. The controller actions downloads the file as it should in the browser, but the Archived File is empty. I can see the content length populated returning from the method above...
file.Seek(0, SeekOrigin.Begin);
return File(file, "application/octet-stream", "Archive.zip");
Does anyone have any idea why my file that is returned by my controller is empty or corrupt?
I believe you need to close your entries and your final zip stream. You should also using and dispose all of your streams. Try this:
using (var zipStream = new ZipOutputStream(outputMemStream))
{
zipStream.IsStreamOwner = false;
// Set compression level
zipStream.SetLevel(3);
foreach (var documentIdString in documentUniqueIdentifiers)
{
...
var blockBlob = container.GetBlockBlobReference(documentId.ToString());
using (var fileMemoryStream = new MemoryStream())
{
// Populate stream with bytes
blockBlob.DownloadToStream(fileMemoryStream);
// Create zip entry and set date
ZipEntry newEntry = new ZipEntry(document.FileName);
newEntry.DateTime = DateTime.Now;
// Put entry RECORD, not actual data
zipStream.PutNextEntry(newEntry);
// Copy date to zip RECORD
StreamUtils.Copy(fileMemoryStream, zipStream, new byte[4096]);
// Mark this RECORD closed in the zip
zipStream.CloseEntry();
}
}
// Close the zip stream, parent stays open due to !IsStreamOwner
zipStream.Close();
outputMemStream.Seek(0, SeekOrigin.Begin);
return outputMemStream;
}
EDIT - you should remove:
// Reset position of stream
fileMemoryStream.Position = 0;
Pretty sure that's the problem.
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());
}
}
}
}
I am attempting to create a new FileStream object from a byte array. I'm sure that made no sense at all so I will try to explain in further detail below.
Tasks I am completing:
1) Reading the source file which was previously compressed
2) Decompressing the data using GZipStream
3) copying the decompressed data into a byte array.
What I would like to change:
1) I would like to be able to use File.ReadAllBytes to read the decompressed data.
2) I would then like to create a new filestream object usingg this byte array.
In short, I want to do this entire operating using byte arrays. One of the parameters for GZipStream is a stream of some sort, so I figured I was stuck using a filestream. But, if some method exists where I can create a new instance of a FileStream from a byte array - then I should be fine.
Here is what I have so far:
FolderBrowserDialog fbd = new FolderBrowserDialog(); // Shows a browser dialog
fbd.ShowDialog();
// Path to directory of files to compress and decompress.
string dirpath = fbd.SelectedPath;
DirectoryInfo di = new DirectoryInfo(dirpath);
foreach (FileInfo fi in di.GetFiles())
{
zip.Program.Decompress(fi);
}
// Get the stream of the source file.
using (FileStream inFile = fi.OpenRead())
{
//Create the decompressed file.
string outfile = #"C:\Decompressed.exe";
{
using (GZipStream Decompress = new GZipStream(inFile,
CompressionMode.Decompress))
{
byte[] b = new byte[blen.Length];
Decompress.Read(b,0,b.Length);
File.WriteAllBytes(outfile, b);
}
}
}
Thanks for any help!
Regards,
Evan
It sounds like you need to use a MemoryStream.
Since you don't know how many bytes you'll be reading from the GZipStream, you can't really allocate an array for it. You need to read it all into a byte array and then use a MemoryStream to decompress.
const int BufferSize = 65536;
byte[] compressedBytes = File.ReadAllBytes("compressedFilename");
// create memory stream
using (var mstrm = new MemoryStream(compressedBytes))
{
using(var inStream = new GzipStream(mstrm, CompressionMode.Decompress))
{
using (var outStream = File.Create("outputfilename"))
{
var buffer = new byte[BufferSize];
int bytesRead;
while ((bytesRead = inStream.Read(buffer, 0, BufferSize)) != 0)
{
outStream.Write(buffer, 0, bytesRead);
}
}
}
}
Here is what I ended up doing. I realize that I did not give sufficient information in my question - and I apologize for that - but I do know the size of the file I need to decompress as I am using it earlier in my program. This buffer is referred to as "blen".
string fi = #"C:\Path To Compressed File";
// Get the stream of the source file.
// using (FileStream inFile = fi.OpenRead())
using (MemoryStream infile1 = new MemoryStream(File.ReadAllBytes(fi)))
{
//Create the decompressed file.
string outfile = #"C:\Decompressed.exe";
{
using (GZipStream Decompress = new GZipStream(infile1,
CompressionMode.Decompress))
{
byte[] b = new byte[blen.Length];
Decompress.Read(b,0,b.Length);
File.WriteAllBytes(outfile, b);
}
}
}