In Java: How to zip file from byte[] array?

Question

My application is receiving email through SMTP server. There are one or more attachments in the email and email attachment return as byte[] (using sun javamail api).

I am trying to zip the attachment files on the fly without writing them to disk first.

What is/are possible way to achieve this outcome?

score 161 · Accepted Answer · answered Dec 10 '08 at 22:48

161

You can use Java's java.util.zip.ZipOutputStream to create a zip file in memory. For example:

public static byte[] zipBytes(String filename, byte[] input) throws IOException {
    ByteArrayOutputStream baos = new ByteArrayOutputStream();
    ZipOutputStream zos = new ZipOutputStream(baos);
    ZipEntry entry = new ZipEntry(filename);
    entry.setSize(input.length);
    zos.putNextEntry(entry);
    zos.write(input);
    zos.closeEntry();
    zos.close();
    return baos.toByteArray();
}

answered Dec 10 '08 at 22:48

Dave L.

43,907
11
63
62

@Dave - can ZipOutputStream be sent as jax-rs output – Varun Nov 21 '17 at 06:04
2

@Deve I have the same case but I have to zip 5 file. Is there any suggestion. – Deva Jan 24 '19 at 08:58
2

@Deva you can repeat the ZipEntry-part for as manny files you like. – kai-dj Jun 12 '19 at 10:16
1

Thank you it worked, but I am wondering why do we need ZipOutputStream, as we are working with ByteArrayOutputStream directly ( i.e. we are returning baos) – Ashish Burnwal Oct 16 '19 at 06:34
2

@AshishBurnwal The ZipOutputStream is what performs the actual compression and encoding. It takes the bytes from the array and produces a compressed stream of bytes that are fed to the ByteArrayOutputStream. Then, we access and return those compressed, encoded bytes from that underlying stream to return them. – Dave L. Oct 16 '19 at 13:46
@Dish I'm not quite sure I understand your follow up question. This question and answer illustrate how to handle input and output byte arrays. If you prefer to deal with streams of data so that it is not all in memory at the same time, you can certainly use a BufferedOutputStream for your output. – Dave L. Dec 17 '20 at 19:24

Jesús Sánchez · Answer 2 · 2020-05-21T23:55:16.037

I have the same problem but i needed a many files in a zip.

 protected byte[] listBytesToZip(Map<String, byte[]> mapReporte) throws IOException {
    String extension = ".pdf";
    ByteArrayOutputStream baos = new ByteArrayOutputStream();
    ZipOutputStream zos = new ZipOutputStream(baos);
    for (Entry<String, byte[]> reporte : mapReporte.entrySet()) {
        ZipEntry entry = new ZipEntry(reporte.getKey() + extension);
        entry.setSize(reporte.getValue().length);
        zos.putNextEntry(entry);
        zos.write(reporte.getValue());
    }
    zos.closeEntry();
    zos.close();
    return baos.toByteArray();
}

score 1 · Answer 3 · answered Mar 10 '17 at 15:03

You can create a zip file from byte array and return to ui streamedContent

public StreamedContent getXMLFile() {
        try {
            byte[] blobFromDB= null;
            ByteArrayOutputStream baos = new ByteArrayOutputStream();
            ZipOutputStream zos = new ZipOutputStream(baos);
            String fileName= "fileName";
            ZipEntry entry = new ZipEntry(fileName+".xml");
            entry.setSize(byteArray.length);
            zos.putNextEntry(entry);
            zos.write(byteArray);
            zos.closeEntry();
            zos.close();
            InputStream is = new ByteArrayInputStream(baos.toByteArray());
            StreamedContent zipedFile= new DefaultStreamedContent(is,   "application/zip", fileName+".zip", Charsets.UTF_8.name());
            return fileDownload;
        } catch (IOException e) {
            LOG.error("IOException e:{} ",e.getMessage());
        } catch (Exception ex) {
            LOG.error("Exception ex:{} ",ex.getMessage());
        }
}

score 1 · Answer 4 · answered Oct 27 '20 at 11:33

   byte[] createReport() {
    try {
     ByteArrayOutputStream byteArrayOutputStream = new ByteArrayOutputStream();
     ZipArchiveOutputStream zipOutputStream = new 
     ZipArchiveOutputStream(byteArrayOutputStream);
     
     zipOutputStream.setMethod(ZipArchiveOutputStream.STORED);
     zipOutputStream.setEncoding(ENCODING);

     String text= "text";
     byte[] textBytes = text.getBytes(StandardCharsets.UTF_8);

     ArchiveEntry zipEntryReportObject = newStoredEntry("file.txt", textBytes);
     zipOutputStream.putArchiveEntry(zipEntryReportObject);
     zipOutputStream.write(textBytes);

     zipOutputStream.closeArchiveEntry();
     zipOutputStream.close();
    
     return byteArrayOutputStream.toByteArray();
     } catch (IOException e) {
       return null;
    }

and

ArchiveEntry newStoredEntry(String name, byte[] data) {
    ZipArchiveEntry zipEntry = new ZipArchiveEntry(name);
    zipEntry.setSize(data.length);
    zipEntry.setCompressedSize(zipEntry.getSize());
    CRC32 crc32 = new CRC32();
    crc32.update(data);
    zipEntry.setCrc(crc32.getValue());
    return zipEntry;
  }

Please specify that your code requires an external library like Apache Commons Compress — Pedro Joaquín, Jul 06 '21 at 22:51

score 0 · Answer 5 · answered Dec 10 '08 at 22:36

Maybe the java.util.zip package might help you

Since you're asking about how to convert from byte array I think (not tested) you can use the ByteArrayInputStream method

int     read(byte[] b, int off, int len)
          Reads up to len bytes of data into an array of bytes from this input stream.

that you will feed to

ZipInputStream  This class implements an input stream filter for reading files in the ZIP file format.

score 0 · Answer 6 · answered Dec 10 '08 at 22:42

0

You have to use a ZipOutputStream for that.

http://java.sun.com/javase/6/docs/api/java/util/zip/ZipOutputStream.html

answered Dec 10 '08 at 22:42

OscarRyz

196,001
113
385
569

score 0 · Answer 7 · answered Jul 16 '20 at 07:35

ByteArrayInputStream bais = new ByteArrayInputStream(retByte);
                
ZipInputStream zis = new ZipInputStream(bais);
           
zis.getNextEntry();

Scanner sc = new Scanner(zis);
while (sc.hasNextLine()) {
    System.out.println("-->:" +sc.nextLine());
}

zis.closeEntry();
zis.close();

score 0 · Answer 8 · answered Jan 03 '21 at 01:56

0

public static void createZip(byte[] data) throws ZipException {
    ZipInputStream zipStream = new ZipInputStream(new ByteArrayInputStream(data));
    ZipParameters parameters = new ZipParameters();
    parameters.setFileNameInZip("bank.zip");
    new ZipFile("F:\\ssd\\bank.zip").addStream(new ByteArrayInputStream(data), parameters);
}

answered Jan 03 '21 at 01:56

sudhansu

51
1
4

use net.lingala.zip4j zip4j 2.6.4 – sudhansu Jan 03 '21 at 01:57

In Java: How to zip file from byte[] array?

8 Answers8

Linked

Related