I have a memory stream that contains a zip file in byte[]
format. Is there any way I can unzip this memory stream, without any need of writing the file to disk?
In general I am using ICSharpCode.SharpZipLib.Zip.FastZip
to unzip a file, but is there any way to unzip a memory stream, maybe by storing the files in another MemoryStream
or in byte[]
format according to the files/folders present in the zip?
Any way I can use the Memorymapped files feature in this scenario ?
Yes, .Net 4.5 now supports more Zip functionality.
Here is a code example based on your description.
In your project, right click on the References folder and add a reference to System.IO.Compression
using System.IO.Compression; Stream data = new MemoryStream(); // The original data Stream unzippedEntryStream; // Unzipped data from a file in the archive ZipArchive archive = new ZipArchive(data); foreach (ZipArchiveEntry entry in archive.Entries) { if(entry.FullName.EndsWith(".txt", StringComparison.OrdinalIgnoreCase)) { unzippedEntryStream = entry.Open(); // .Open will return a stream // Process entry data here } }
Hope this helps.
We use DotNetZip, and I can unzip the contents of a zip file from a Stream
into memory. Here's the sample code for extracting a specifically named file from a stream (LocalCatalogZip
) and returning a stream to read that file, but it'd be easy to expand on it.
private static MemoryStream UnZipCatalog() { MemoryStream data = new MemoryStream(); using (ZipFile zip = ZipFile.Read(LocalCatalogZip)) { zip["ListingExport.txt"].Extract(data); } data.Seek(0, SeekOrigin.Begin); return data; }
It's not the library you're using now, but if you can change, you can get that functionality.
Here's a variation which would return a Dictionary<string,MemoryStream>
of for the contents of every file of a zip file.
private static Dictionary<string,MemoryStream> UnZipToMemory() { var result = new Dictionary<string,MemoryStream>(); using (ZipFile zip = ZipFile.Read(LocalCatalogZip)) { foreach (ZipEntry e in zip) { MemoryStream data = new MemoryStream(); e.Extract(data); result.Add(e.FileName, data); } } return result; }
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