creating a zip file from an object directly without disk IO

Question

I am writing a REST API which will take in a JSON request object. The request object will have to be serialized to a file in JSON format; the file has to be compressed into a zip file and the ZIP file has to be posted to another service, for which I would have to deserialize the ZIP file. All this because the service I have to call expects me to post data as ZIP file. I am trying to see if I can avoid disk IO. Is there a way to directly convert the object into a byte array representing ZIP content in-memory instead of all the above steps?

Note : I'd prefer accomplishing this using .net framework libraries (as against external libraries)

I'd be very interested if this can be accomplished without the use of IO as you asked — gpullen, Dec 15 '14 at 11:51
Hey @Aadith, did you find any of the answers useful? If so please give the author some kudos and mark one as accepted! Thanks. — Yogster, Dec 16 '14 at 11:10
Sure @Jorge .. thanks for your solution and paving the way to the required solution — Aadith Ramia, Dec 16 '14 at 12:20

score 14 · Accepted Answer · edited Jun 20 '20 at 09:12

Yes, it is possible to create a zip file completely on memory, here is an example using SharpZip Library (Update: A sample using ZipArchive added at the end):

public static void Main()
{
    var fileContent = Encoding.UTF8.GetBytes(
        @"{
            ""fruit"":""apple"",
            ""taste"":""yummy""
          }"
        );


    var zipStream = new MemoryStream();
    var zip = new ZipOutputStream(zipStream);

    AddEntry("file0.json", fileContent, zip); //first file
    AddEntry("file1.json", fileContent, zip); //second file (with same content)

    zip.Close();

    //only for testing to see if the zip file is valid!
    File.WriteAllBytes("test.zip", zipStream.ToArray());
}

private static void AddEntry(string fileName, byte[] fileContent, ZipOutputStream zip)
{
    var zipEntry = new ZipEntry(fileName) {DateTime = DateTime.Now, Size = fileContent.Length};
    zip.PutNextEntry(zipEntry);
    zip.Write(fileContent, 0, fileContent.Length);
    zip.CloseEntry();
}

You can obtain SharpZip using Nuget command PM> Install-Package SharpZipLib

Update:

Note : I'd prefer accomplishing this using .net framework libraries (as against external libraries)

Here is an example using Built-in ZipArchive from System.IO.Compression.Dll

public static void Main()
{
    var fileContent = Encoding.UTF8.GetBytes(
        @"{
            ""fruit"":""apple"",
            ""taste"":""yummy""
          }"
        );

    var zipContent = new MemoryStream();
    var archive = new ZipArchive(zipContent, ZipArchiveMode.Create);

    AddEntry("file1.json",fileContent,archive);
    AddEntry("file2.json",fileContent,archive); //second file (same content)

    archive.Dispose();

    File.WriteAllBytes("testa.zip",zipContent.ToArray());
}


private static void AddEntry(string fileName, byte[] fileContent,ZipArchive archive)
{
    var entry = archive.CreateEntry(fileName);
    using (var stream = entry.Open())
        stream.Write(fileContent, 0, fileContent.Length);

}

Couldn't you just use System.IO.Compression.ZipArchive instead of installing a NuGet? — Yogster, Dec 15 '14 at 12:34
@Jorge, yes it's possible, it was based on personal preferences, I'll update the answer. — user3473830, Dec 15 '14 at 12:35
Thanks @user3473830 your revised solution suits my requirement — Aadith Ramia, Dec 16 '14 at 12:17
How can you modify this to create a folder within the zip archive? — ElFik, Apr 19 '19 at 00:07
@user3473830 Thanks for ZipArchive version of the code. Trying to do unit test to the component which has dependency to ZipArchive. It is about 2 hours I am searching the solution, but wasn't able to find until saw your answer. — Farhad Jabiyev, Apr 28 '20 at 13:08

Yogster · Answer 2 · 2014-12-15T12:29:55.893

4

You could use the GZipStream class along with MemoryStream.

A quick example:

using System.IO;
using System.IO.Compression;

//Put JSON into a MemoryStream
var theJson = "Your JSON Here";
var jsonStream = new MemoryStream();
var jsonStreamWriter = new StreamWriter(jsonStream);
jsonStreamWriter.Write(theJson);
jsonStreamWriter.Flush();

//Reset stream so it points to the beginning of the JSON
jsonStream.Seek(0, System.IO.SeekOrigin.Begin);

//Create stream to hold your zipped JSON
var zippedStream = new MemoryStream();

//Zip JSON and put it in zippedStream via compressionStream.
var compressionStream = new GZipStream(zippedStream, CompressionLevel.Optimal);
jsonStream.CopyTo(compressionStream);

//Reset zipped stream to point at the beginning of data
zippedStream.Seek(0, SeekOrigin.Begin);

//Get ByteArray with zipped JSON
var zippedJsonBytes = zippedStream.ToArray();

edited Dec 15 '14 at 12:29

answered Dec 15 '14 at 11:34

Yogster

884
1
7
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When I try this, I get a 0-size array for zippedJsonBytes. Looks like theres a minor snag somewhere..could you pls help me out – Aadith Ramia Dec 15 '14 at 12:25
If I'm not mistaken, OP needs to create a `Zip` container, I'm afraid the proposed solution only compress data with Gzip algorithm and does not conform Zip container format. – user3473830 Dec 15 '14 at 12:26
Thanks @user3473830 ..That was going to be my other question..whats the difference between zip and gzip? – Aadith Ramia Dec 15 '14 at 12:28
@Aadith, my bad, I forgot to call jsonStreamWriter.Flush(); I have edited the example. – Yogster Dec 15 '14 at 12:30
1

@Aadith, if what you need is a whole zip archive (as opposed to Zipped data) then you can use the ZipArchive class as suggested below. There is an example in MSDN using FileStreams, but you could also use MemoryStreams as in my example. – Yogster Dec 15 '14 at 12:33
1

@Aadith, http://stackoverflow.com/questions/20762094/how-are-zlib-gzip-and-zip-related-what-are-is-common-and-how-are-they-differen – user3473830 Dec 15 '14 at 12:49

score 2 · Answer 3 · answered Dec 15 '14 at 11:36

2

You should try the ZipArchive Class streaming to a MemoryStream Class

answered Dec 15 '14 at 11:36

A. Rama

903
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score 0 · Answer 4 · answered Dec 15 '14 at 11:34

0

Yes. You can return it as a binary stream. Depending on the language, you can use special libraries. You will also need libraries on the client.

answered Dec 15 '14 at 11:34

ACV

9,964
5
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creating a zip file from an object directly without disk IO

4 Answers4