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I want to TAR a multiple files on JBOSS server in Eclipse using java

I have tried multiple examples... But non will work for me
Make tar file by Java
If I use above code, After reading the line TarOutputStream out = new TarOutputStream( new BufferedOutputStream( dest ) ); it just come out of the method(i.e.After reading that line... it won't executed the code which are after to that. The Sequence directly gone to the finally block where that method called).
For another example, I used apache common compress Jar. In that case it won't execute the java class itself it exit from the FormBean class.
The files are fetched from the database and it will be in a CSV format and all the files are placed in an Array List.
fileNames = "Security.smod."+SYSDATE;
strSavePath = D\:\\CMSREPORT\\reportZip\\
tarFunction(fileList, strSavePath, fileNames);
//Example function 1
private String tarFunction(ArrayList fileList, String strSavePath, String outFileName) {
int fileCountSize = fileList.size();
// Output file stream
FileOutputStream dest = new FileOutputStream( strSavePath );
// Create a TarOutputStream
TarOutputStream out = new TarOutputStream( new BufferedOutputStream( dest ) );
// Files to tar
File[] filesToTar=new File[3];
for(int i=0; i<fileCountSize; i++)
{
filesToTar[i]=new File((String) fileList.get(i);
}
for(File f:filesToTar){
out.putNextEntry(new TarEntry(f, f.getName()));
BufferedInputStream origin = new BufferedInputStream(new FileInputStream( f ));
int count;
byte data[] = new byte[2048];
while((count = origin.read(data)) != -1) {
out.write(data, 0, count);
}
out.flush();
origin.close();
}
}
For Example 2, I used apache.common.compress.1.2 as a jar.
In that case, It won't execute the java class, it exited from the FormBean class.
//Example function 2
private void tarFunc(){
try{
// Files to tar
File resource1 = new File((String) fileList.get(0));
File resource2 = new File((String) fileList.get(1));
File resource3 = new File((String) fileList.get(2));
// Output Stream - that will hold the physical TAR file
OutputStream tar_output = new FileOutputStream(new File(fileNames+".tar"));
// Create Archive Output Stream that attaches File Output Stream / and specifies TAR as type of compression
ArchiveOutputStream my_tar_ball = new ArchiveStreamFactory().createArchiveOutputStream(ArchiveStreamFactory.TAR, tar_output);
// Create Archieve entry - write header information
TarArchiveEntry tar_file = new TarArchiveEntry(resource1);
// length of the TAR file needs to be set using setSize method
tar_file.setSize(resource1.length());
my_tar_ball.putArchiveEntry(tar_file);
IOUtils.copy(new FileInputStream(resource1), my_tar_ball);
// Close Archieve entry, write trailer information
my_tar_ball.closeArchiveEntry();
// Repeat steps for the next file that needs to be added to the TAR
tar_file = new TarArchiveEntry(resource2);
tar_file.setSize(resource2.length());
my_tar_ball.putArchiveEntry(tar_file);
IOUtils.copy(new FileInputStream(resource2), my_tar_ball);
// Close Archieve entry, write trailer information
my_tar_ball.closeArchiveEntry();
tar_file = new TarArchiveEntry(resource3);
tar_file.setSize(resource3.length());
my_tar_ball.putArchiveEntry(tar_file);
IOUtils.copy(new FileInputStream(resource3), my_tar_ball);
// Close Archieve entry, write trailer information
my_tar_ball.closeArchiveEntry();
my_tar_ball.finish();
// Close output stream, our files are zipped
tar_output.close();
}catch(Exception e){
e.printStackTrace();
}
}
Anyone please help to solve this issue. Thank you:)

How to access a zip file into a zip file in Java

I'm trying to read .srt files that are located in zip file itself located in a zip file. I succeed to read .srt files that were in a simple zip with the extract of code below :
for (Enumeration enume = fis.entries(); enume.hasMoreElements();) {
ZipEntry entry = (ZipEntry) enume.nextElement();
fileName = entry.toString().substring(0,entry.toString().length()-4);
try {
InputStream in = fis.getInputStream(entry);
BufferedReader reader = new BufferedReader(new InputStreamReader(in));
String ext = entry.toString().substring(entry.toString().length()-4, entry.toString().length());
But now i don't know how i could get to the zip file inside the zip file.
I tried using ZipFile fis = new ZipFile(filePath) with filePath being the path of the zip file + the name of zip file inside. It didn't recognize the path so i don't know if i am clear.
Thanks.

ZipFile only works with real files, because it's intended for use as a random access mechanism which needs to be able to seek directly to specific locations in the file to read entries by name. But as VGR suggests in the comments, while you can't get random access to the zip-inside-a-zip you can use ZipInputStream, which provides strictly sequential access to the entries and works with any InputStream of zip-format data.
However, ZipInputStream has a slightly odd usage pattern compared to other streams - calling getNextEntry reads the entry metadata and positions the stream to read that entry's data, you read from the ZipInputStream until it reports EOF, then you (optionally) call closeEntry() before moving on to the next entry in the stream.
The critical point is that you must not close() the ZipInputStream until you have finished reading the final entry, so depending what you want to do with the entry data you might need to use something like the commons-io CloseShieldInputStream to guard against the stream getting closed prematurely.
try(ZipInputStream outerZip = new ZipInputStream(fis)) {
ZipEntry outerEntry = null;
while((outerEntry = outerZip.getNextEntry()) != null) {
if(outerEntry.getName().endsWith(".zip")) {
try(ZipInputStream innerZip = new ZipInputStream(
new CloseShieldInputStream(outerZip))) {
ZipEntry innerEntry = null;
while((innerEntry = innerZip.getNextEntry()) != null) {
if(innerEntry.getName().endsWith(".srt")) {
// read the data from the innerZip stream
}
}
}
}
}
}

Find the code to extract .zip files recursively:
public void extractFolder(String zipFile) throws ZipException, IOException {
System.out.println(zipFile);
int BUFFER = 2048;
File file = new File(zipFile);
ZipFile zip = new ZipFile(file);
String newPath = zipFile.substring(0, zipFile.length() - 4);
new File(newPath).mkdir();
Enumeration zipFileEntries = zip.entries();
// Process each entry
while (zipFileEntries.hasMoreElements())
{
// grab a zip file entry
ZipEntry entry = (ZipEntry) zipFileEntries.nextElement();
String currentEntry = entry.getName();
File destFile = new File(newPath, currentEntry);
//destFile = new File(newPath, destFile.getName());
File destinationParent = destFile.getParentFile();
// create the parent directory structure if needed
destinationParent.mkdirs();
if (!entry.isDirectory())
{
BufferedInputStream is = new BufferedInputStream(zip
.getInputStream(entry));
int currentByte;
// establish buffer for writing file
byte data[] = new byte[BUFFER];
// write the current file to disk
FileOutputStream fos = new FileOutputStream(destFile);
BufferedOutputStream dest = new BufferedOutputStream(fos,
BUFFER);
// read and write until last byte is encountered
while ((currentByte = is.read(data, 0, BUFFER)) != -1) {
dest.write(data, 0, currentByte);
}
dest.flush();
dest.close();
is.close();
}
if (currentEntry.endsWith(".zip"))
{
// found a zip file, try to open
extractFolder(destFile.getAbsolutePath());
}
}
}

File Path appearing in the Zipped File

I have a java program as below for zipping a folder as a whole.
public static void zipDir(String dir2zip, ZipOutputStream zos)
{
try
{
File zipDir= new File(dir2zip);
String[] dirList = zipDir.list();
byte[] readBuffer = new byte[2156];
int bytesIn = 0;
for(int i=0; i<dirList.length; i++)
{
File f = new File(zipDir, dirList[i]);
if(f.isDirectory())
{
String filePath = f.getPath();
zipDir(filePath, zos);
continue;
}
FileInputStream fis = new FileInputStream(f);
ZipEntry anEntry = new ZipEntry(f.getPath());
zos.putNextEntry(anEntry);
while((bytesIn = fis.read(readBuffer)) != -1)
{
zos.write(readBuffer, 0, bytesIn);
}
fis.close();
}
}
catch(Exception e)
{
e.printStackTrace();
}
}
public static void main(){
String date=new java.text.SimpleDateFormat("MM-dd-yyyy").format(new java.util.Date());
ZipOutputStream zos = new ZipOutputStream(new FileOutputStream("Output/" + date + "_RB" + ".zip"));
zipDir("Output/" + date + "_RB", zos);
zos.close();
}
My query here is. The target folder(+date+_RB) to be zipped is present inside the folder named Output. After successful zipping, when I extract the zipped file, I find a folder Output inside which the (+date+_RB) required folder is present. I need not want that Output folder after the extraction of the zipped file, rather it should directly extract the required folder alone. Please advise on the same.
UPDATE:
I tried Isaac's answer. While extracting the resultant zip file, no folders are getting extracted. Only the files inside all the folders are getting extracted. I just dont need the folder "Output" alone in the resultant zip file. But what the program does is, it doesnt extracts all other folders inside the Output folder, rather it just extracts the files inside those folders. Kindly advise on how to proceed...

It happens because of this:
ZipEntry anEntry = new ZipEntry(f.getPath());
f.getPath() will return Output/ at the beginning of the string. This is due to the flow of your program and how it (mis)uses File objects.
I suggest you construct a File object called, say, tmp:
File tmp = new File(dirList[i]);
The change the construction of f:
File f = new File(zipDir, tmp.getPath());
Then, change this:
ZipEntry anEntry = new ZipEntry(f.getPath());
To this:
ZipEntry anEntry = new ZipEntry(tmp.getPath());
I didn't have time to actually test it, but in a nutshell, your problem is due to how the File object is constructed.

How to use java.util.zip to archive/deflate string in java for use in Google Earth?

Use Case
I need to package up our kml which is in a String into a kmz response for a network link in Google Earth. I would like to also wrap up icons and such while I'm at it.
Problem
Using the implementation below I receive errors from both WinZip and Google Earth that the archive is corrupted or that the file cannot be opened respectively. The part that deviates from other examples I'd built this from are the lines where the string is added:
ZipEntry kmlZipEntry = new ZipEntry("doc.kml");
out.putNextEntry(kmlZipEntry);
out.write(kml.getBytes("UTF-8"));
Please point me in the right direction to correctly write the string so that it is in doc.xml in the resulting kmz file. I know how to write the string to a temporary file, but I would very much like to keep the operation in memory for understandability and efficiency.
private static final int BUFFER = 2048;
private static void kmz(OutputStream os, String kml)
{
try{
BufferedInputStream origin = null;
ZipOutputStream out = new ZipOutputStream(os);
out.setMethod(ZipOutputStream.DEFLATED);
byte data[] = new byte[BUFFER];
File f = new File("./icons"); //folder containing icons and such
String files[] = f.list();
if(files != null)
{
for (String file: files) {
LOGGER.info("Adding to KMZ: "+ file);
FileInputStream fi = new FileInputStream(file);
origin = new BufferedInputStream(fi, BUFFER);
ZipEntry entry = new ZipEntry(file);
out.putNextEntry(entry);
int count;
while((count = origin.read(data, 0, BUFFER)) != -1) {
out.write(data, 0, count);
}
origin.close();
}
}
ZipEntry kmlZipEntry = new ZipEntry("doc.kml");
out.putNextEntry(kmlZipEntry);
out.write(kml.getBytes("UTF-8"));
}
catch(Exception e)
{
LOGGER.error("Problem creating kmz file", e);
}
}
Bonus points for showing me how to put the supplementary files from the icons folder into a similar folder within the archive as opposed to at the same layer as the doc.kml.
Update Even when saving the string to a temp file the errors occur. Ugh.
Use Case Note The use case is for use in a web app, but the code to get the list of files won't work there. For details see how-to-access-local-files-on-server-in-jboss-application

You forgot to call close() on ZipOutputStream. Best place to call it is the finally block of the try block where it's been created.
Update: To create a folder, just prepend its name in the entry name.
ZipEntry entry = new ZipEntry("icons/" + file);

How do I create a ZIP file in Java?

What is the Java equivalent to this jar command:
C:\>jar cvf myjar.jar directory
I'd like to create this jar file programmatically as I can't be assured that the jar command will be on the system path where I could just run the external process.
Edit: All I want is to archive (and compress) a directory. Doesn't have to follow any java standard. Ie: a standard zip is fine.

// These are the files to include in the ZIP file
String[] source = new String[]{"source1", "source2"};
// Create a buffer for reading the files
byte[] buf = new byte[1024];
try {
// Create the ZIP file
String target = "target.zip";
ZipOutputStream out = new ZipOutputStream(new FileOutputStream(target));
// Compress the files
for (int i=0; i<source.length; i++) {
FileInputStream in = new FileInputStream(source[i]);
// Add ZIP entry to output stream.
out.putNextEntry(new ZipEntry(source[i]));
// Transfer bytes from the file to the ZIP file
int len;
while ((len = in.read(buf)) > 0) {
out.write(buf, 0, len);
}
// Complete the entry
out.closeEntry();
in.close();
}
// Complete the ZIP file
out.close();
} catch (IOException e) {
}
You can also use the answer from this post How to use JarOutputStream to create a JAR file?

Everything you'll want is in the java.util.jar package:
http://java.sun.com/javase/6/docs/api/java/util/jar/package-summary.html