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how to avoid [Content_Types].xml in .net's ZipPackage class

Tags:

.net

zip

.net-4.0

I wish to know if there is any way to avoid to have a [Content_Types].xml file inside the zip file while using .net's ZipPackage class.

like image 495
Kushal Waikar Avatar asked Sep 20 '10 05:09

Kushal Waikar


4 Answers

No.

Zip Packages are not (normal) Zip files. They must follow an imposed structure and contain that Content_Types.xml file in the root. ZipPackage = ZipArchive + Structure.

If you want to create (and especially if you want to read) normal Zip archives you will need an extra library, there is no support in the BCL.

See SharpZipLib (GPL) and DotNetZip

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Henk Holterman Avatar answered Oct 24 '22 13:10

Henk Holterman


If you don't call the .Flush() method, there will be no such file

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Yiping Avatar answered Oct 24 '22 12:10

Yiping


Yes you can create zip packages without the extra XML content file added

Inspired by this link: Using System.IO.Packaging to generate a ZIP file

Using above discovery mentioned by Yiping you can avoid the extra xml file added into the package. Save zip stream from memory stream to a physical zip file before zip archive is closed like this:

public static void AddFilesToZip(string zipFilename, List<String> filesToAdd)
{
    using (var memStream = new MemoryStream())
    {
        using (Package zip = System.IO.Packaging.Package.Open(memStream, FileMode.Create))
        {

            foreach (var fileToAdd in filesToAdd)
            {
                string destFilename = ".\\" + Path.GetFileName(fileToAdd);
                Uri uri = PackUriHelper.CreatePartUri(new Uri(destFilename, UriKind.Relative));

                //Existing parts not likely in fresh memory stream
                if (zip.PartExists(uri))
                {
                    zip.DeletePart(uri);
                }

                PackagePart part = zip.CreatePart(uri, "", CompressionOption.Normal);

                using (FileStream fileStream = new FileStream(fileToAdd, FileMode.Open, FileAccess.Read))
                {

                    using (Stream dest = part.GetStream())
                    {
                        CopyStream(fileStream, dest);
                    }
                }
            }

            //The zip Package will add an XML content type file to memeory stream when it closes
            //so before it closes we save the memorystream to physical zip file.
            using (FileStream zipfilestream = new FileStream(zipFilename, FileMode.Create, FileAccess.Write))
            {
                memStream.Position = 0;
                CopyStream(memStream, zipfilestream);
            }

            // That's it. Zip file saved to file. Things added by package after this point will be to memory stream finally disposed.
        }
    }
}
like image 27
flodis Avatar answered Oct 24 '22 13:10

flodis


Man, this helped me a lot! You rock! I had to use this because my old framework (3.5). Only to complement, see below the implementation of the CopyStream function:

    private void CopyStream(Stream source, Stream target)
    {
        const int bufSize = 0x1000;
        byte[] buf = new byte[bufSize];
        int bytesRead = 0;
        while ((bytesRead = source.Read(buf, 0, bufSize)) > 0)
            target.Write(buf, 0, bytesRead);
    }
like image 32
Rodolfo Avatar answered Oct 24 '22 13:10

Rodolfo