Using System.IO.Compression, create a ZIP archive in memory.

Problem

I’m trying to use a MemoryStream to construct a ZIP archive with a basic example text file like follows:

using (var memoryStream = new MemoryStream())
using (var archive = new ZipArchive(memoryStream , ZipArchiveMode.Create))
{
    var demoFile = archive.CreateEntry("foo.txt");

    using (var entryStream = demoFile.Open())
    using (var streamWriter = new StreamWriter(entryStream))
    {
        streamWriter.Write("Bar!");
    }

    using (var fileStream = new FileStream(@"C:\Temp\test.zip", FileMode.Create))
    {
        stream.CopyTo(fileStream);
    }
}

The archive file is formed when I run this code, but foo.txt is not.

When I directly replace the MemoryStream with the file stream, however, the archive is formed correctly:

using (var fileStream = new FileStream(@"C:\Temp\test.zip", FileMode.Create))
using (var archive = new ZipArchive(fileStream, FileMode.Create))
{
    // ...
}

Is it possible to construct the ZIP archive without the FileStream by using a MemoryStream?

Asked by Marius Schulz

Solution #1

I got the following as a result of ZipArchive creating an incorrect ZIP file:

using (var memoryStream = new MemoryStream())
{
   using (var archive = new ZipArchive(memoryStream, ZipArchiveMode.Create, true))
   {
      var demoFile = archive.CreateEntry("foo.txt");

      using (var entryStream = demoFile.Open())
      using (var streamWriter = new StreamWriter(entryStream))
      {
         streamWriter.Write("Bar!");
      }
   }

   using (var fileStream = new FileStream(@"C:\Temp\test.zip", FileMode.Create))
   {
      memoryStream.Seek(0, SeekOrigin.Begin);
      memoryStream.CopyTo(fileStream);
   }
}

That meant we had to execute Dispose on ZipArchive before we could use it, which, as Amir notes, is likely because it adds last bytes to the archive, such as a checksum, to make it complete. However, you must supply true as the third option to ZipArchive in order to keep the stream open and reuse it afterwards.

Answered by Michael

Solution #2

It’s just another variation of zipping that doesn’t create any files.

string fileName = "export_" + DateTime.Now.ToString("yyyyMMddhhmmss") + ".xlsx";
byte[] fileBytes = here is your file in bytes
byte[] compressedBytes;
string fileNameZip = "Export_" + DateTime.Now.ToString("yyyyMMddhhmmss") + ".zip";

using (var outStream = new MemoryStream())
{
    using (var archive = new ZipArchive(outStream, ZipArchiveMode.Create, true))
    {
        var fileInArchive = archive.CreateEntry(fileName, CompressionLevel.Optimal);
        using (var entryStream = fileInArchive.Open())
        using (var fileToCompressStream = new MemoryStream(fileBytes))
        {
            fileToCompressStream.CopyTo(entryStream);
        }
    }
    compressedBytes = outStream.ToArray();
}

Answered by luci79rom

Solution #3

Before transferring the stream to the zip stream, set its position to 0.

using (var memoryStream = new MemoryStream())
{
 using (var archive = new ZipArchive(memoryStream, ZipArchiveMode.Create, true))
 {
  var demoFile = archive.CreateEntry("foo.txt");

  using (var entryStream = demoFile.Open())
  using (var streamWriter = new StreamWriter(entryStream))
  {
     streamWriter.Write("Bar!");
  }
 }

 using (var fileStream = new FileStream(@"C:\Temp\test.zip", FileMode.Create))
   {
     memoryStream.Position=0;
     memoryStream.WriteTo(fileStream);
   }
 }

Answered by Mahesh

Solution #4

MVC solution that actually works

    public ActionResult Index()
    {
        string fileName = "test.pdf";
        string fileName1 = "test.vsix";
        string fileNameZip = "Export_" + DateTime.Now.ToString("yyyyMMddhhmmss") + ".zip";

        byte[] fileBytes = System.IO.File.ReadAllBytes(@"C:\test\test.pdf");
        byte[] fileBytes1 = System.IO.File.ReadAllBytes(@"C:\test\test.vsix");
        byte[] compressedBytes;
        using (var outStream = new MemoryStream())
        {
            using (var archive = new ZipArchive(outStream, ZipArchiveMode.Create, true))
            {
                var fileInArchive = archive.CreateEntry(fileName, CompressionLevel.Optimal);
                using (var entryStream = fileInArchive.Open())
                using (var fileToCompressStream = new MemoryStream(fileBytes))
                {
                    fileToCompressStream.CopyTo(entryStream);
                }

                var fileInArchive1 = archive.CreateEntry(fileName1, CompressionLevel.Optimal);
                using (var entryStream = fileInArchive1.Open())
                using (var fileToCompressStream = new MemoryStream(fileBytes1))
                {
                    fileToCompressStream.CopyTo(entryStream);
                }


            }
            compressedBytes = outStream.ToArray();
        }
        return File(compressedBytes, "application/zip", fileNameZip);
    }

Answered by Debajit Mukhopadhyay

Solution #5

You must finish writing the memory stream before returning to the buffer.

        using (var memoryStream = new MemoryStream())
        {
            using (var archive = new ZipArchive(memoryStream, ZipArchiveMode.Create))
            {
                var demoFile = archive.CreateEntry("foo.txt");

                using (var entryStream = demoFile.Open())
                using (var streamWriter = new StreamWriter(entryStream))
                {
                    streamWriter.Write("Bar!");
                }
            }

            using (var fileStream = new FileStream(@"C:\Temp\test.zip", FileMode.Create))
            {
                var bytes = memoryStream.GetBuffer();
                fileStream.Write(bytes,0,bytes.Length );
            }
        }

Answered by Rattle

Post is based on https://stackoverflow.com/questions/17232414/creating-a-zip-archive-in-memory-using-system-io-compression