Create Java-Zip-Archive from existing OutputStream

Question

Is it possible to create a Zip-Archive in Java if I do not want to write the resulting archive to disk but send it somewhere else?

The idea is that it might be a waste to create a file on disk when you want to send the Zip-Archive to a user via HTTP (e.g. from a Database-Blob or any other Data-Store).

I would like to create a

java.util.zip.ZipOutputStream

or a

apache.commons.ZipArchiveOutputStream

where the Feeder would be a ByteArrayOutputStream coming from my Subversion Repository

score 9 · Accepted Answer · answered Oct 25 '10 at 16:45

9

Yes this is absolutely possible!

Create your Zip entry using the putNextEntry method on the ZipOutputStream then put the bytes into the file in the zip by calling write on the ZipOutputStream. For the parameter for that method, the byte[], just extract them from the ByteArrayOutputStream with its toByteArray method.

And the ZipOutputStream can be sent anywhere, as its constructor just takes an OutputStream so could be e.g. your HTTP response.

answered Oct 25 '10 at 16:45

Adrian Smith

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Alright, that worked! But how do I create Directories that actually contain the files themselves then? Right now I get all files, fine, but the directories are unpacked as files with zero-byte length in the root node, so no hierarchy at all :-( – trajectory Oct 25 '10 at 20:47
1

ZIP files just contain a flat list of files. The file names, however, may contain path information. Just add a `ZipEntry` with a name such as `dir/file.txt`: that is `file.txt` within a directory `dir`. – Adrian Smith Oct 25 '10 at 22:03
Yes, by adding the absolute URL everything works a expected. Thank you very much! – trajectory Oct 26 '10 at 09:52

score 5 · Answer 2 · answered Oct 25 '10 at 16:47

5

Something like that would work:

ZipOutputStream zs = new ZipOutputStream(outputStream) ;
ZipEntry e = new ZipEntry(fileName);
zs.putNextEntry(e);
zs.write(...);
zs.close();

answered Oct 25 '10 at 16:47

Eugene Kuleshov

31,461
5
66
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score 1 · Answer 3 · answered May 08 '15 at 08:08

Input: D:/in.xml

Output: D:/final.zip (having 2 files 001zip.txt,002zip.txt)

Code:

package com.stackoverflow.filezip;

import java.io.FileInputStream;
import java.io.FileOutputStream;
import java.io.InputStream;
import java.io.OutputStream;
import java.util.zip.ZipEntry;
import java.util.zip.ZipOutputStream;

public class fileZip {

public static void main(String[] args) {

    try {
InputStream in = new FileInputStream("D:/in.xml");
OutputStream out=   new FileOutputStream("D:/final.zip");
ZipOutputStream zs = new ZipOutputStream(out);

            ZipEntry e1 = new ZipEntry("001zip.txt");
            ZipEntry e2 = new ZipEntry("002zip.txt");
            zs.putNextEntry(e1);
            zs.write("test content in file1".getBytes());
            zs.putNextEntry(e2);
            zs.write("test content in file2".getBytes());

            zs.close();
       }
       catch (Exception e) {
            e.printStackTrace();
                           }
                                       }
                       }

Create Java-Zip-Archive from existing OutputStream

3 Answers3

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