ZipArchive creates invalid ZIP file

Question

I am trying to create a new ZIP package from code with one entry and save the ZIP package to a file. I am trying to achive this with the System.IO.Compression.ZipArchive class. I am creating the ZIP package with the following code:

using (MemoryStream zipStream = new MemoryStream())
{
    using (ZipArchive zip = new ZipArchive(zipStream, ZipArchiveMode.Create))
    {
        var entry = zip.CreateEntry("test.txt");
        using (StreamWriter sw = new StreamWriter(entry.Open()))
        {
            sw.WriteLine(
                "Etiam eros nunc, hendrerit nec malesuada vitae, pretium at ligula.");
        }

Then I save the ZIP to a file either in WinRT:

        var file = await Windows.Storage.ApplicationData.Current.LocalFolder.CreateFileAsync("test.zip", CreationCollisionOption.ReplaceExisting);
        zipStream.Position = 0;
        using (Stream s = await file.OpenStreamForWriteAsync())
        {
            zipStream.CopyTo(s);
        }

Or in normal .NET 4.5:

        using (FileStream fs = new FileStream(@"C:\Temp\test.zip", FileMode.Create))
        {
            zipStream.Position = 0;
            zipStream.CopyTo(fs);
        }

However, I can't open the produced files neither in Windows Explorer, WinRAR, etc. (I checked that the size of the produced file matches the Length of the zipStream, so the stream itself was saved to the file correctly.)
Am I doing something wrong or is there a problem with the ZipArchive class?

Here's a good example if you are adding binary data instead of strings: https://stackoverflow.com/questions/48927574/create-zip-file-in-memory-from-bytes-text-with-arbitrary-encoding — John Gilmer, Apr 26 '20 at 18:05

Mark Vincze · Accepted Answer · 2021-03-04T11:07:31.847

126

I found the—in hindsight, obvious—error in my code. The ZipArchive has to be disposed to make it write its content to its underlying stream. So I had to save the stream to a file after the end of the using block of the ZipArchive.
And it was important to set the leaveOpen argument of its constructor to true, to make it not close the underlying stream. So here is the complete working solution:

using (MemoryStream zipStream = new MemoryStream())
{
    using (ZipArchive zip = new ZipArchive(zipStream, ZipArchiveMode.Create, true))
    {
        var entry = zip.CreateEntry("test.txt");
        using (StreamWriter sw = new StreamWriter(entry.Open()))
        {
            sw.WriteLine(
                "Etiam eros nunc, hendrerit nec malesuada vitae, pretium at ligula.");
        }
    }

    var file = await Windows.Storage.ApplicationData.Current.LocalFolder.CreateFileAsync(
        "test.zip",
        CreationCollisionOption.ReplaceExisting);

    zipStream.Position = 0;
    using (Stream s = await file.OpenStreamForWriteAsync())
    {
        zipStream.CopyTo(s);
    }
}

edited Mar 04 '21 at 11:07

answered Sep 10 '12 at 10:48

Mark Vincze

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I get the error, "The name 'Windows' does not exist in the current context". How can I rectify this? – Michael Murphy Apr 25 '15 at 10:17
1

Michael, this is a build error, right? Are you developing a Windows Store app, or a normal .NET app? In a normal .NET app you should use the `FileStream` class, and not the storage classes in the `Windows.Storage` namespace. You can see an example in the question. – Mark Vincze Apr 25 '15 at 20:47
Thanks @MarkVincze that was not entirely obvious at all to me. Saved me tearing my hair out. – The Senator Nov 23 '22 at 11:15
I would argue it wasn't obvious at all! It has no flush or close indicating that it does this. Relying on disposal to implement functionality is disgusting design. But thanks for the post, saved my bacon. – Adam Hardy Aug 30 '23 at 03:57

Andrew Dragan · Answer 2 · 2018-03-16T22:00:28.973

5

// Create file "archive.zip" in current directory use it as destination for ZIP archive
using (var zipArchive = new ZipArchive(File.OpenWrite("archive.zip"), 
ZipArchiveMode.Create))
{
    // Create entry inside ZIP archive with name "test.txt"
    using (var entry = zipArchive.CreateEntry("test.txt").Open())
    {
        // Copy content from current directory file "test.txt" into created ZIP entry 
        using (var file = File.OpenRead("test.txt"))
        {
            file.CopyTo(entry);
        }
    }
}

In result you will get archive "archive.zip" with single entry file "test.txt". You need this cascade of using to avoid any interaction with already disposed resources.

edited Mar 16 '18 at 22:00

answered May 31 '17 at 09:04

Andrew Dragan

71
1
2

Please add an explanation for your code, and what it does to solve the question. – Douglas Gaskell Mar 15 '18 at 20:45
@DouglasGaskell done. Probably code above don't answer question directly but show how to ZIP some file using ZipArchive – Andrew Dragan Mar 16 '18 at 22:02

score 4 · Answer 3 · answered Sep 10 '12 at 08:24

4

On all of your Stream Object you must rewind the streams from the beggining in order from them to be read correctly by other applications using the .Seek method.

Example:

zipStream.Seek(0, SeekOrigin.Begin);

answered Sep 10 '12 at 08:24

Freeman

5,691
3
29
41

1

This did not help. This is why I set the Position property to 0, I guess that achieves the same effect. – Mark Vincze Sep 10 '12 at 08:28
the zip library has to have its own SaveAs method, try looking into that, also on the project peage there are various documentation with different save scenarios. – Freeman Sep 10 '12 at 08:29
1

Umm, the ZipArchive class (http://msdn.microsoft.com/en-us/library/system.io.compression.ziparchive.aspx) has no SaveAs method. What do you mean by "project page"? – Mark Vincze Sep 10 '12 at 08:43
sorry, i thought it was a 3rd party library. – Freeman Sep 10 '12 at 08:45
@Freeman this did the trick in my case. Thanks! – Leniel Maccaferri Jan 26 '21 at 13:03

score 1 · Answer 4 · answered Feb 28 '13 at 21:43

You can follow the same idea, only in reverse order, using the file stream like source. I did the form below and the file was opened normally:

string fileFormat = ".zip"; // any format
string filename = "teste" + fileformat;

StorageFile zipFile = await Windows.Storage.ApplicationData.Current.LocalFolder.CreateFileAsync(filename,CreationCollisionOption.ReplaceExisting);

using (Stream zipStream = await zipFile.OpenStreamForWriteAsync()){

   using (ZipArchive archive = new ZipArchive(zipStream, ZipArchiveMode.Create)){
        //Include your content here
   }
}

score 1 · Answer 5 · answered Jul 10 '19 at 05:17

In My case the Issue was that I was not disposing "zipArchive" when the zip file Task Completes. Even though I was flushing and closing all the streams.

So, Either use using as suggested in answers above

using (var zipArchive = new ZipArchive(File.OpenWrite("archive.zip"), 
ZipArchiveMode.Create))

Or Dispose the variable at the end of the Task...

zipArchive.Dispose();

score 0 · Answer 6 · answered Feb 28 '13 at 21:48

The complete code looks like this:

var file = await Windows.Storage.ApplicationData.Current.LocalFolder.CreateFileAsync("test.zip",CreationCollisionOption.ReplaceExisting);
using (Stream zipStream = await zipFile.OpenStreamForWriteAsync())
{
    using (ZipArchive zip = new ZipArchive(zipStream, ZipArchiveMode.Create, true))
    {
        var entry = zip.CreateEntry("test.txt");
        using (StreamWriter sw = new StreamWriter(entry.Open()))
        {
            sw.WriteLine("Etiam eros nunc, hendrerit nec malesuada vitae, pretium at ligula.");
        }
    }   
}

ZipArchive creates invalid ZIP file

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