ZipArchive.CreateEntry for a Stream

Question

User1000638852 posted

'Sup Nerds:

I want to know if there is anyway that I can create a ZipArchive with a Stream; then return that ZipArchive as a byte[]:

pseudocode:

   Stream myStream;

   myZipArchive.CreateEntry(myStream)

   byte[] myByteArray = myZipArchive.ToByteArray()

Is this possible?

Answer 1

User-271186128 posted

Hi jjmonty,<!--?xml:namespace prefix = "o" ns = "urn:schemas-microsoft-com:office:office" /--><o:p></o:p>

From my point of view, I don’t think you can do that. The parameters of the ZipArchive.CreateEntry method should be string type.

For more information, please refer to this article.<o:p></o:p>

ZipArchive.CreateEntry Method: https://msdn.microsoft.com/en-us/library/system.io.compression.ziparchive.createentry(v=vs.110).aspx <o:p></o:p>

Beside, from your description, I suppose perhaps you want to upload the Zip file. If that is the case, I suggest you could try to use the FileUpload control and use the following code.

                Stream fs = FileUpload1.PostedFile.InputStream;
                BinaryReader br = new BinaryReader(fs);
                Byte[] bytes = br.ReadBytes((Int32)fs.Length);

More information, please see: http://www.aspsnippets.com/Articles/Save-Files-to-SQL-Server-Database-using-FileUpload-Control.aspx

Best Regards,
Dillion

Answer 2

User1000638852 posted

I found my solution by adapting this code:

http://stackoverflow.com/questions/17217077/create-zip-file-from-byte

My version takes a stream and compresses it into a zipped stream then returns it as a byte array:

                using (var compressedFileStream = new MemoryStream())
                {
                    //Create an archive and store the stream in memory.
                    using (ZipArchive zipArchive = new ZipArchive(compressedFileStream, ZipArchiveMode.Update, false))
                    {
                        foreach (string strFileName in arrayFileNames)
                        {
                            strErrorFileName = strFileName;

                            using (Stream msFileBody = new MemoryStream())
                            {

                               // this is custom code that copies a file location's data to a Stream (msFileBody); this is the Stream needing compression that prompted my initial question
                               

                               objFormSystem.OpenFile(strFileName, "myFileName").CopyTo(msFileBody); 

                               // this is custom code that copies a file location's data to a Stream (msFileBody); this is the Stream needing compression that prompted my initial question

                                //Create a zip entry for each FileName in the FileName array
                                zipArchiveEntry = zipArchive.CreateEntry("myFileName");

                                using (Stream originalFileStream = msFileBody)
                                {
                                    using (Stream zipEntryStream = zipArchiveEntry.Open())
                                    {
                                        //Copy the attachment stream to the zip entry stream
                                        originalFileStream.CopyTo(zipEntryStream);
                                    }
                                }
                            }
                        }
                    }

                    arrayByteReturn = compressedFileStream.ToArray();
                }