Bracket.Hosting.ZipArchiveDirectory.OpenOutputFileStream C# (CSharp) Метод

        public Stream OpenOutputFileStream(string path)
        {
            string directory = VirtualFileUtils.NormalizePath(Path.GetDirectoryName(path));
            string fileName = Path.GetFileName(path);

            Stream output = new ZipArchiveOutputStream(_storage);
            _storage.AddEntry(fileName, directory, output);
            return output;
        }