In Java: How to Zip File from Byte[] Array

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

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();
}

How can I convert byte array to ZIP file

To get the contents from the bytes you can use

ZipInputStream zipStream = new ZipInputStream(new ByteArrayInputStream(bytes));
ZipEntry entry = null;
while ((entry = zipStream.getNextEntry()) != null) {

    String entryName = entry.getName();

    FileOutputStream out = new FileOutputStream(entryName);

    byte[] byteBuff = new byte[4096];
    int bytesRead = 0;
    while ((bytesRead = zipStream.read(byteBuff)) != -1)
    {
        out.write(byteBuff, 0, bytesRead);
    }

    out.close();
    zipStream.closeEntry();
}
zipStream.close(); 

How to convert zip file into bytes in Java

You can read a file from disk into a byte[] using

 byte[] ba = java.nio.file.Files.readAllBytes(filePath);

This is available from Java 7.

How to convert a byte array to ZIP file and download it using Java?

I am presuming here that you want to write a byte array to a ZIP file. As the data sent is also a ZIP file and to be saved is also ZIP file, shouldn't be a problem.

Two steps are needed: save it on disk and return the file.

1) Save on disk part:

File file = new File(/path/to/directory/save.zip);
    if (file.exists() && file.isDirectory()) {
        try {
            OutputStream outputStream = new FileOutputStream(new File(/path/to/directory/save.zip));
            outputStream.write(bytes);
            outputStream.close();
        } catch (IOException ignored) {

        }
    } else {
        // create directory and call same code
    }
}

2) Now to get it back and download it, you need a controller :

@RequestMapping(value = "/download/attachment/", method = RequestMethod.GET)
public void getAttachmentFromDatabase(HttpServletResponse response) {
    response.setContentType("application/octet-stream");
    response.setHeader("Content-Disposition", "attachment; filename=\"" + file.getFileName() + "\"");
    response.setContentLength(file.length);

    FileCopyUtils.copy(file as byte-array, response.getOutputStream());
    response.flushBuffer();
}

I have edited the code I have, so you will have to make some changes before it suits you 100%. Let me know if this is what you were looking for. If not, I will delete my answer. Enjoy.

Writing a file from byte array into a zip file

Don't use a BufferedWriter to write binary content! A Writer is made to write text content.

Use that instead:

final Path zip = file.toPath();

final Map<String, ?> env = Collections.emptyMap();
final URI uri = URI.create("jar:" + zip.toUri());

try (
    final FileSystem zipfs = FileSystems.newFileSystem(uri, env);
)  {
    Files.write(zipfs.getPath("into/zip"), buf,
        StandardOpenOption.CREATE, StandardOpenOption.APPEND);
}

(note: APPEND is a guess here; it looks from your question that you want to append if the file already exists; by default the contents will be overwritten)

Java Zip File Created With ByteArrayOutputStream is smaller than with FileOutputStream

The following statement is wrong:

baos.writeTo(new FileOutputStream("TEST_AS_BYTESTREAM_1.zip"));

The javadoc of writeTo(OutputStream out) does not say anything about closing the OutputStream, which means it doesn't, so the last of the data is still sitting buffered in the un-closed, un-flushed FileOutputStream.

The correct way is:

try (OutputStream out = new FileOutputStream("TEST_AS_BYTESTREAM_1.zip")) {
    baos.writeTo(out);
}

---

Also, a ByteArrayOutputStream does not need to be closed or flushed. As the javadoc of close() says it:

Closing a ByteArrayOutputStream has no effect.

You will however need to call finish() on an ZipOutputStream to complete the content. Since closing the zip stream automatically finishes it for you, the correct way to do this is:

ByteArrayOutputStream baos = new ByteArrayOutputStream();
try (ZipOutputStream zipOS = new ZipOutputStream(baos)) {
    // create the Zip file content here
}
try (OutputStream out = new FileOutputStream("TEST_AS_BYTESTREAM_1.zip")) {
    baos.writeTo(out);
}

When writing directly to a file:

try (ZipOutputStream zipOS = new ZipOutputStream(new FileOutputStream("TEST_AS_FILE.zip"))) {
    // create the Zip file content here
}

Consider adding a BufferedOutputStream for better performance.

How to extract and read entries of zip file which is sent as a byte array

Seeing as the source is a byte[], you'll need to use a ByteArrayInputStream to read the file.

public List<ZipEntry> extractZipEntries(byte[] content) throws IOException {
    List<ZipEntry> entries = new ArrayList<>();

    ZipInputStream zi = null;
    try {
        zi = new ZipInputStream(new ByteArrayInputStream(content));

        ZipEntry zipEntry = null;
        while ((zipEntry = zi.getNextEntry()) != null) {
            entries.add(zipEntry);
        }
    } finally {
        if (zi != null) {
            zi.close();
        }
    }
    return entries;
}

How can I write A map of byte Array into ZIP file

Change your code to:

ZipEntry entry = new ZipEntry(pair.getKey());
entry.setSize(pair.getValue().length);
zos.putNextEntry(entry);
zos.write(pair.getValue());
zos.flush();

You were writing zero bytes (int size = 0;) out of an empty byte[] (byte[] tmp = new byte[4 * 1024];) rather than the bytes from the Map.

How to convert byte array to ZipOutputStream in spring MVC?

You have to copy the byte data (content) of the zip file to the output as well...

This should work (untested):

while ((ent = zipStream.getNextEntry()) != null) {
    zipOutputStream.putNextEntry(ent);
    // copy byte stream
    org.apache.commons.io.IOUtils.copy(zis.getInputStream(ent), zipOutputStream);
}

BTW: why you do not just simply forward the original zip byte content?

try (InputStream is = new ByteArrayInputStream(fileData));) {
    IOUtils.copy(is, response.getOutputStream());
}

or even better (thanks to comment by @M. Deinum)

IOUtils.copy(fileData, response.getOutputStream());

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