[ { "Id": "656984", "ThreadId": "268802", "Html": "\r\n

Excuse me , if I want to compress stream to stream which is .zip format?

\r\n

One thing I know is GZipStream Class is .gz format.  I just want\r\n.zip format replace .gz format.Others like conclusion of\r\nhttp://dotnetzip.codeplex.com/discussions/268506   .

\r\n

public byte[] Zip(string ToCompress)
\r\n{
\r\n using (Stream input = StringToMemoryStream(ToCompress))
\r\n {
\r\n  MemoryStream raw = new MemoryStream();
\r\n  using (Stream compressor = new GZipStream(raw, CompressionMode.Compress))——？ just\r\n want .zip format replace by .gz format
\r\n  {
\r\n   byte[] buffer = new byte[2048];
\r\n   int n;
\r\n   while ((n = input.Read(buffer, 0, buffer.Length)) != 0)
\r\n   {
\r\n    compressor.Write(buffer, 0, n);
\r\n   }
\r\n   byte[] bytes = raw.ToArray(); 

\r\n

   return bytes;
\r\n  }
\r\n }
\r\n}

\r\n

 So now Which I should to choose?  ZipOutputStream and ZipInputStream?How to do it?

\r\n

 

\r\n

 

\r\n", "PostedDate": "2011-08-12T00:00:48.253-07:00", "UserRole": null, "MarkedAsAnswerDate": null }, { "Id": "657884", "ThreadId": "268802", "Html": "

ZipOutputStream.  Check the documentation for examples.

", "PostedDate": "2011-08-13T17:50:07.277-07:00", "UserRole": null, "MarkedAsAnswerDate": null }, { "Id": "658735", "ThreadId": "268802", "Html": "

1) First problem, compress file to file:

\r\n

string filesToZip = Path.GetDirectoryName(Assembly.GetExecutingAssembly().GetName().CodeBase) + \"\\\\new.xml\";       //pathname :program files/test/new.xml

\r\n

string outputFileName =  Path.GetDirectoryName(Assembly.GetExecutingAssembly().GetName().CodeBase) + \"\\\\new.zip\";   //pathname :program files/test/new.zip

\r\n

public void Zipup()
{
 using (ZipFile zip = new ZipFile())
 {
  zip.AddFile(filesToZip);
  zip.Save(outputFileName);
 }
}

\r\n

But this a problem, when I decompression this zip, there are two folders, just like program files/test/new.xml. But I want when I decompression this zip,there is just new.xml. What I can do????

\r\n

\r\n

2) Second problem, compress stream to stream, last time you say I can \"ZipOutputStream.  Check the documentation for examples\".

\r\n

           if (filesToZip.Count == 0)
            {
                return;
            }

\r\n

           using (var raw = File.Open(outputFileName, FileMode.Create, FileAccess.ReadWrite))
            {
                using (var output = new ZipOutputStream(raw))
                {
                    foreach (string inputFileName in filesToZip)
                    {
                        output.PutNextEntry(inputFileName);
                        using (var input = File.Open(inputFileName, FileMode.Open, FileAccess.Read, FileShare.Read | FileShare.Write))
                        {
                            byte[] buffer = new byte[2048];
                            int n;
                            while ((n = input.Read(buffer, 0, buffer.Length)) > 0)
                            {
                                output.Write(buffer, 0, n);
                            }
                        }
                    }
                }
            }

\r\n

filesToZip and outputFileName are all filenames, what I can do to change it to compress stream to stream????

", "PostedDate": "2011-08-16T02:15:21.61-07:00", "UserRole": null, "MarkedAsAnswerDate": null }, { "Id": "661238", "ThreadId": "268802", "Html": "

File.Open returns a stream.  in your code, replace File.Open() with the method that returns your stream.

\r\n

 

", "PostedDate": "2011-08-21T10:30:37.033-07:00", "UserRole": null, "MarkedAsAnswerDate": null } ]