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Unable to extract nested tar within zip file i.e. a .tar file inside a zip file and so on

I have gone through the link of how to extract a .tar file and several link on SOF using Java.
However, I didnt find any which can relate to my concerns which is multilevel or nested .tar/.tgz/.zip file.
my concern is with something like below
Abc.tar.gz
--DEF.tar
--sample1.txt
--sample2.txt
--FGH.tgz
--sample3.txt
-sample4.txt
This is the simple one which I can give here . As it can be in any compressed combination with the folder like .tar inside .tar and .gz and again .tgz and so on....
My problem is I am able to extract till the first level using Apache Commons Compress library. that is if Abc.tar.gz gets extracted then in the destination/output folder its only DEF.tar available . beyond that my extraction is not working.
I tried to give the output of first to the input to the second on the fly but I got stuck with FileNotFoundException. As at that point of time output file would have not been in place and the second extraction not able to get the file.
Pseudocode:
public class CommonExtraction {
TarArchiveInputStream tar = null;
if((sourcePath.trim().toLowerCase.endsWith(".tar.gz")) || sourcePath.trim().toLowerCase.endsWith(".tgz")) {
try {
tar=new TarArchiveInputStream(new GzipCompressorInputStream(new BufferedInputStream(new FileInputStream(sourcePath))));
extractTar(tar,destPath)
} catch (Exception e) {
e.printStackTrace();
}
}
}
Public static void extractTar(TarArchiveInputStream tar, String outputFolder) {
try{
TarArchiveEntry entry;
while (null!=(entry=(TarArchiveEntry)tar.getNextTarEntry())) {
if(entry.getName().trim().toLowerCase.endsWith(".tar")){
final String path = outputFolder + entry.getName()
tar=new TarArchiveInputStream(new BufferedInputStream(new FileInputStream(path))) // failing as .tar folder after decompression from .gz not available at destination path
extractTar(tar,outputFolder)
}
extractEntry(entry,tar,outputFolder)
}
tar.close();
}catch(Exception ex){
ex.printStackTrace();
}
}
Public static void extractEntry(TarArchiveEntry entry , InputStream tar, String outputFolder){
final String path = outputFolder + entry.getName()
if(entry.isDirectory()){
new File(path).mkdirs();
}else{
//create directory for the file if not exist
}
// code to read and write until last byte is encountered
}
}
Ps: please ignore the syntax and all in the code.

Try this
try (InputStream fi = file.getInputStream();
InputStream bi = new BufferedInputStream(fi);
InputStream gzi = new GzipCompressorInputStream(bi, false);
ArchiveInputStream archive = new TarArchiveInputStream(gzi)) {
withArchiveStream(archive, result::appendEntry);
}
As i see what .tar.gz and .tgz is same formats. And my method withArchiveEntry is:
private void withArchiveStream(ArchiveInputStream archInStream, BiConsumer<ArchiveInputStream, ArchiveEntry> entryConsumer) throws IOException {
ArchiveEntry entry;
while((entry = archInStream.getNextEntry()) != null) {
entryConsumer.accept(archInStream, entry);
}
}
private void appendEntry(ArchiveInputStream archive, ArchiveEntry entry) {
if (!archive.canReadEntryData(entry)) {
throw new IOException("Can`t read archive entry");
}
if (entry.isDirectory()) {
return;
}
// And for example
String content = new String(archive.readAllBytes(), StandardCharsets.UTF_8);
System.out.println(content);
}

You have a recursive problem, so you can use recursion to solve it. Here is some pseudocode to show how it can be done:
public class ArchiveExtractor
{
public void extract(File file)
{
List<File> files; // list of extracted files
if(isZip(file))
files = extractZip(file);
else if(isTGZ(file))
files = extractTGZ(file);
else if(isTar(file))
files = extractTar(file);
else if(isGZip(file))
files = extractGZip(file);
for(File f : files)
{
if(isArchive(f))
extract(f); // recursive call
}
}
private List<File> extractZip(File file)
{
// extract archive and return list of extracted files
}
private List<File> extractTGZ(File file)
{
// extract archive and return list of extracted files
}
private List<File> extractTar(File file)
{
// extract archive and return list of extracted files
}
private List<File> extractGZip(File file)
{
// extract archive and return list of extracted file
}
}
where:
isZip() tests if the file extension is zip
isTGZ() tests if the file extension is tgz
isTar() tests if the file extension is tar
isGZip() tests if the file extension is gz
isArchive() means isZip() || isTGZ() || isTar() || isGZip()
As for the directory where each archive is extracted: you are free to do as you want.
If you process test.zip for example, you may extract in the same directory as where the archive is,
or create the directory test and extract in it.

when using a zip file as a FileSystem the zip file is not being update

I want to manipulate a jar using the standard nio Files and Paths methods. So, Java has a way to do this by creating a zip FileSystem:
try {
zipFS = FileSystems.newFileSystem(zipDisk, zipFSproperties);
} catch(Exception e) {
System.out.println(e.getMessage());
}
My test program uses an existing jar file as a FileSystem and it lists the entries contained in the jar. All that works great. I then copy a new file into the jar and list the entries again. And just as you would expect, the list now contains the newly added file. The problem is after the program closes, I open up the jar file that the jar filesystem is based upon and it doesn't have the new entry added to it. So that's my question! Shouldn't the jar file itself be changed when I add a new entry. I don't know of any commands I can issue the would cause the zip FileSystem to update to the actual jar file that the zip FileSystem wraps. Am I reading more into a FileSystem; are changes in the zip filesystem suppose to cause the corresponding backend zip file to be updated.
code:
public class Main {
public static void main(String[] args) throws IOException {
ZipFileSystem zipFS = new ZipFileSystem("C:\\Temp\\mylibrary\\build\\outputs\\jar\\temp\\mylibrary-debug.zip");
Stream<Path> paths = Files.find(zipFS.zipFS.getRootDirectories().iterator().next().getRoot(),10, (path, basicFileAttributes) -> {
return !Files.isDirectory(path);
});
paths.forEach( path ->
System.out.println ("zip contains entry: " + path)
);
File file = new File("C:\\Temp\\mylibrary\\src\\main\\java\\com\\phinneyridge\\android\\myLib.java");
System.out.println("copying " + file.getPath());
Path outPath = zipFS.zipFS.getPath("myLib.java");
Files.copy (file.toPath(), outPath, StandardCopyOption.REPLACE_EXISTING);
paths = Files.find(zipFS.zipFS.getPath(""),10, (path, basicFileAttributes) -> {
return !Files.isDirectory(path);
});
paths.forEach( path ->
System.out.println ("zip contains entry: " + path)
);
}
}
I added code that shows me accessing a zip file, listing the current entries it contains, adding a new entry (via file copy), and lastly listing the contents again. All of this code works correctly. What doesn't work is that the changes to the zip filesystem don't get incorporated back into the zip file when the application ends. I was surprised that the zip file didn't get updated, but I'm now under the opinion, that it's working as it is intended to work; not doing what I wanted it to do, but that's okay. I can't find any documentation that says it would update the jar file that the FileSystem object originated from. So I'm basically asking is that the correct behavior, or is there something I'm entirely missing to cause the zip FileSystem object to update the Zip file?
Here's the code when I tried Dunc suggestion:
public static void main(String[] args) throws IOException {
ZipFileSystem zipFS = new ZipFileSystem("C:\\Temp\\mylibrary\\build\\outputs\\jar\\temp\\mylibrary-debug.zip");
try (FileSystem fs = zipFS.zipFS) {
try (Stream<Path> paths = Files.find(zipFS.zipFS.getRootDirectories().
iterator().next().getRoot(), 10, (path, basicFileAttributes) -> {
return !Files.isDirectory(path);
})) {
paths.forEach(path ->
System.out.println("zip contains entry: " + path)
);
}
File file = new File("C:\\Temp\\mylibrary\\src\\main\\java\\com\\phinneyridge\\android\\myLib.java");
System.out.println("copying " + file.getPath());
Path outPath = fs.getPath("myLib.java");
Files.copy(file.toPath(), outPath, StandardCopyOption.REPLACE_EXISTING);
try (Stream<Path> paths = Files.find(zipFS.zipFS.getRootDirectories().
iterator().next().getRoot(), 10, (path, basicFileAttributes) -> {
return !Files.isDirectory(path);
})) {
paths.forEach(path ->
System.out.println("zip contains entry: " + path)
);
}
} catch (Exception e) {
System.out.println("FileSystem Error: " + e.getClass().getName() + " - " + e.getMessage());
e.printStackTrace();
}
}
}
And by the way ZipFileSystem is a wrapper class around the FileSystem. I'll post that code too, incase that's where I 'm doing something wrong.
public class ZipFileSystem {
FileSystem zipFS;
Path zipFSPath;
/**
* Constructor for a ZipFile object
* #param zipFilePath string representing the path to the zipfile. If the path doesn't exist,
* the zip file will be automatically created. If the path exist, it must be a file (not
* a directory) and it must be a valid zip file
*/
public ZipFileSystem(String zipFilePath) {
Map<String, String> zipFSproperties = new HashMap<>();
/* set create to true if you want to create a new ZIP file */
zipFSproperties.put("create", "true");
/* specify encoding to UTF-8 */
zipFSproperties.put("encoding", "UTF-8");
/* Locate File on disk for creation */
URI zipFileUri = new File(zipFilePath).toURI();
URI zipDisk = URI.create("jar:" + zipFileUri);
zipFSPath = Paths.get(zipFileUri);
if (!Files.exists(zipFSPath)) {
try {
createEmptyZipFile(zipFSPath);
} catch (Exception e ) {
System.out.println(e.getMessage());
}
} else {
if (Files.isDirectory(zipFSPath)) {
} else {
try {
// let's open it, which will verify if it's a valid zip file
ZipFile zipFile = new ZipFile(zipFilePath);
} catch (Exception e) {
System.out.println(e.getMessage());
}
}
}
try {
zipFS = FileSystems.newFileSystem(zipDisk, zipFSproperties);
} catch(Exception e) {
System.out.println(e.getMessage());
}
try {
listFiles(zipFS.getPath("/"));
} catch (IOException e) {
e.printStackTrace();
}
}

The correct way to open a zip from a Path - and create if not exists - is:
Path zip = Path.of("/Somepath/to/xyz.zip");
Map<String, String> env = Map.of(
"create", "true"
// other args here ...
);
try (FileSystem fs = FileSystems.newFileSystem(zip, env)) {
// code to read/update here
}
You have not closed any files or streams properly so your changes are probably not flushed back to the file system and will keep hold of file handles which block some operations.
Use try with resources for every operation which will manage the modifications to zip filesystem as well as closing each Stream<Path> from Files.find, and check other places such as createEmptyZipFile for the same problem:
try (FileSystem fs = ... ) {
try (Stream<Path> paths = Files.find(...) ) {
}
Files.copy( ... );
try (Stream<Path> paths = Files.find(...) ) {
}
}
The process cannot access the file because it is being used by another process
You have unnecessary code ZipFile zipFile = new ZipFile(zipFilePath) which tests the zip is valid and you do not call close(), so it will prevent the zip changes being written back. The check can safely be deleted (as FileSystems.newFileSystem does same) or must be wrapped in try() {} so that zipFile is closed before your edits to the zip filesystem.

Using the Files.move creates a new "file" file type rather than moving the file to a directory

I am trying to make a program that extracts multiple MP4 files from there individual folders and places them in a folder that is already created (code has been changed slightly so that it doesn't mess up any more of the MP4s, rather dummy text files).
I have managed to get so far as to list all folders/files in the specified folder however am having trouble moving them to a directory.
static File dir = new File("G:\\New Folder");
static Path source;
static Path target = Paths.get("G:\\gohere");
static void showFiles(File files[]) {
for (File file : files) { // Loops through each file in the specified directory in "dir" variable.
if (file.isDirectory()) { // If the file is a directory.
File[] subDir = file.listFiles(); // Store each file in a File list.
for (File subFiles : subDir) { // Loops through the files in the sub-directory.
if (subFiles.getName().endsWith(".mp4")) { // if the file is of type MP4
source = subFiles.toPath(); // Set source to be the abs path to the file.
System.out.println(source);
try {
Files.move(source, target);
System.out.println("File Moved");
} catch (IOException e) {
e.getMessage();
}
}
}
} else {
source = file.toPath(); // abs path to file
try {
Files.move(source, target);
System.out.println("File moved - " + file.getName());
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
}
}
public static void main(String[] args) {
showFiles(dir.listFiles());
}
The problem is when I go to move the file from the source folder to the target, it removes or converts the target.

Files.move isn't like the command line. You're programming. You have to spell things out. You're literally asking Files.move to make it so that target (here, G:\GoHere) will henceforth be the location for the file you are moving. If you intended: No, the target is G:\GoHere\TheSameFileName then you have to program this.
Separately, your code is a mess. Stop using java.io.File and java.nio.Path together. Pick a side (and pick the java.nio side, it's an newer API for a good reason), and do not mix and match.
For example:
Path fromDir = Paths.get("G:\\FromHere");
Path targetDir = Paths.get(G:\\ToHere");
try (DirectoryStream ds = Files.newDirectoryStream(fromDir)) {
for (Path child : ds) {
if (Files.isRegularFile(child)) {
Path targetFile = targetDir.resolve(child.getFileName());
Files.move(child, targetFile);
}
}
}
resolve gives you a Path object that is what you need here: The actual file in the target dir.

Working directory when run in eclipse, cmd, and jar

I'm writing a Java program which will execute an external file ~/Java/exampleProject/bin/import.sh. My program is under package gqqnbig. So the directory structure is
exampleProject/
bin/
import.sh
gqqnbig/
*.class
When I debug the program in eclipse, the working directory is ~/Java/exampleProject/. I have to execute bin/import.sh.
When I run the program in cmd, the current directory is ~/Java/exampleProject/bin, my code will not find import.sh.
The program has to be portable (distribute with import.sh). With the correct directory structure, it should work in my computer as well as in your computer, so I cannot hard code the path of import.sh.
I also want to pack it into a single jar file. The desired structure is (Figure 1)
bin/
import.sh
program.jar
So how can my program find import.sh when run in eclipse, cmd and jar?
UPDATE
I ask my question in another way. Please implement getAbsolutePath function, so that no matther the code is running in eclipse, in cmd, or as a jar file in a folder which also has import.sh (See Figure 1), the output is identical.
public static void main(String[] args)
{
System.out.println("Look for "+getAbsolutePath()+"\\import.sh");
}

Here's a method pulled from one of my projects. It get's the folder that the jar file is located in as opposed to the directory if was run from if invoked on the command line.
/**
* Retrieve a File representation of the folder this application is
* located in.
*
* #return
*/
private static File getApplicationRootFolder()
{
String path = FileGetter.class.getProtectionDomain().getCodeSource()
.getLocation().getPath();
try
{
String decodedPath = URLDecoder.decode(path, "UTF-8");
File jarParentFolder = new File(decodedPath).getParentFile();
if (jarParentFolder.exists() && jarParentFolder.canRead()
{
File shellScript = new File(jarParentFolder, "import.sh")
}
catch (UnsupportedEncodingException e)
{
Main.myLog.error(TAG, "Unencoding jar path failed on:\n\t" + path);
e.printStackTrace();
return null;
}
}
You can then use that directory to make a File object for your shell script File shellScript = new File(getApplicationRootFolder(), scriptFilename);
EDIT: Follow up questions to try to help you out and a solution
So you want to be able to access one file that has three locations depending on when/where you code is run. This is how I see those cases:
Case 1: Running directly from Eclipse (unpackaged code):
shell script: X:/Java/exampleProject/bin/import.sh
class file: X:/Java/exampleProject/bin/gqqnbig/YourClass.class
Case 2: Running the packaged jar (shell script inside):
shell script: X:/Java/YourJar.jar/bin/import.sh
class file: X:/Java/YourJar.jar/bin/gqqnbig/YourClass.class
Case 3: Running the packaged jar (shell script external):
shell script: X:/Java/import.sh
class file: X:/Java/YourJar.jar/bin/gqqnbig/YourClass.class
What I think you need to do is prioritise the order you look at these locations and fall back to the next one in line if the shell script isn't found. I'd guess you want:
1. external to jar
2. inside packaged jar
3. unpackaged
So to access these you will need to write each separately and move through each until you get File.exists() == true.
Something like what follows. Note I didn't test this and there are likely errors. I'll leave you to sort them out. My code is based on the assumptions made above, again I'll leave you to modify the code based on any incorrect guesses.
So here's a class with one public method taking a filename argument and returning an InputStream. I opted for InputStream in all cases as once you package up your jar you cannot access the resources as File objects any more, only Streams.
public class FileGetter
{
private static String RESOURCE_DIRECTORY = "bin";
/**
* Retrieve an InputStream for a resource file.
*
* #param filename
* #return
*/
public InputStream getResourceFileStream(String filename)
{
// this is where you decide your preference or the priority of the locations
InputStream inputStream = null;
inputStream = getExternalFile(filename);
if (inputStream != null)
{
return inputStream;
}
inputStream = getInternalPackagedFile(filename);
if (inputStream != null)
{
return inputStream;
}
inputStream = getInternalUnpackagedFile(filename);
if (inputStream != null)
{
return inputStream;
}
// couldn't find the file anywhere so log some error or throw an exception
return null;
}
/**
* Retrieve an InputStream for a file located outside your Jar
*
* #param filename
* #return
*/
private static InputStream getExternalFile(String filename)
{
// get the jar's absolute location on disk (regardless of current 'working directory')
String appRootPath = FileGetter.class.getProtectionDomain().getCodeSource()
.getLocation().getPath();
try
{
String decodedPath = URLDecoder.decode(appRootPath, "UTF-8");
File jarfile = new File(decodedPath);
File parentDirectory = jarfile.getParentFile();
if (testExists(parentDirectory))
{
File shellScript = new File(parentDirectory, filename);
if (testExists(shellScript))
{
return new FileInputStream(shellScript);
}
}
}
catch (UnsupportedEncodingException e)
{}
catch (NullPointerException e)
{}
catch (FileNotFoundException e)
{}
// if any part fails return null
return null;
}
/**
* Retrieve an InputStream for a file located inside your Jar.
*
* #param filename
* #return
*/
private static InputStream getInternalPackagedFile(String filename)
{
// root directory is defined as the jar's root so we start with a "/".
URL resUrl = FileGetter.class.getResource(File.separator + RESOURCE_DIRECTORY
+ File.separator + filename);
String badPath = resUrl.getPath();
String goodPath = badPath.substring(badPath.indexOf("!") + 1);
InputStream input = FileGetter.class.getResourceAsStream(goodPath);
// returns null if nothing there so just
return input;
}
private static InputStream getInternalUnpackagedFile(String filename)
{
// eclipse will 'cd' to the code's directory so we use relative paths
File shellScriptFile = new File(RESOURCE_DIRECTORY + File.separator + filename);
if (testExists(shellScriptFile))
{
try
{
InputStream shellScriptStream = new FileInputStream(shellScriptFile);
if (shellScriptStream != null)
{
return shellScriptStream;
}
}
catch (FileNotFoundException e)
{}
}
return null;
}
/**
* Test that a file exists and can be read.
*
* #param file
* #return
*/
private static boolean testExists(File file)
{
return file != null && file.exists() && file.canRead();
}
}
But with all that being said a better way to sort this would be to ensure that the file exists on disk and create it if not found. Then execute the script from disk.

I would like to know a definitive answer for this myself.
As a workaround I would put 'import.sh' inside the exampleProject and change the relative path to 'import.sh'.
In theory that should work inside Eclipse, and as a packaged Jar file with program.jar and import.sh in the same directory.
It won't work on the cmd prompt unfortunately, maybe someone can suggest a better method.
-Kaz

I composed a solution. Call getExecutablePath() to get unified path.
public static File getExecutablePath()
{
String workingDirectory = System.getProperty("user.dir");
File binFile = new File(workingDirectory, "bin");
if (binFile.exists() && (new File(workingDirectory, "src")).exists())
{
return binFile;
}
else if (isRunningFromJar())
return getApplicationRootFolder();
else
return new File(workingDirectory);
}
public static boolean isRunningFromJar()
{
String className = SystemHelper.class.getName().replace('.', '/');
String classJar = SystemHelper.class.getResource("/" + className + ".class").toString();
return classJar.startsWith("jar:");
}
/**
* Retrieve a File representation of the folder this application is located in.
*
* #return
*/
private static File getApplicationRootFolder()
{
try
{
String path = SystemHelper.class.getProtectionDomain().getCodeSource().getLocation().getPath();
String decodedPath;
decodedPath = URLDecoder.decode(path, "UTF-8");
File jarfile = new File(decodedPath);
return jarfile.getParentFile();
}
catch (UnsupportedEncodingException e)
{
throw new RuntimeException(e);
}
}

How to copy file inside jar to outside the jar?

I want to copy a file from a jar. The file that I am copying is going to be copied outside the working directory. I have done some tests and all methods I try end up with 0 byte files.
EDIT: I want the copying of the file to be done via a program, not manually.

First of all I want to say that some answers posted before are entirely correct, but I want to give mine, since sometimes we can't use open source libraries under the GPL, or because we are too lazy to download the jar XD or what ever your reason is here is a standalone solution.
The function below copy the resource beside the Jar file:
/**
* Export a resource embedded into a Jar file to the local file path.
*
* #param resourceName ie.: "/SmartLibrary.dll"
* #return The path to the exported resource
* #throws Exception
*/
static public String ExportResource(String resourceName) throws Exception {
InputStream stream = null;
OutputStream resStreamOut = null;
String jarFolder;
try {
stream = ExecutingClass.class.getResourceAsStream(resourceName);//note that each / is a directory down in the "jar tree" been the jar the root of the tree
if(stream == null) {
throw new Exception("Cannot get resource \"" + resourceName + "\" from Jar file.");
}
int readBytes;
byte[] buffer = new byte[4096];
jarFolder = new File(ExecutingClass.class.getProtectionDomain().getCodeSource().getLocation().toURI().getPath()).getParentFile().getPath().replace('\\', '/');
resStreamOut = new FileOutputStream(jarFolder + resourceName);
while ((readBytes = stream.read(buffer)) > 0) {
resStreamOut.write(buffer, 0, readBytes);
}
} catch (Exception ex) {
throw ex;
} finally {
stream.close();
resStreamOut.close();
}
return jarFolder + resourceName;
}
Just change ExecutingClass to the name of your class, and call it like this:
String fullPath = ExportResource("/myresource.ext");
Edit for Java 7+ (for your convenience)
As answered by GOXR3PLUS and noted by Andy Thomas you can achieve this with:
Files.copy( InputStream in, Path target, CopyOption... options)
See GOXR3PLUS answer for more details

Given your comment about 0-byte files, I have to assume you're trying to do this programmatically, and, given your tags, that you're doing it in Java. If that's true, then just use Class.getResource() to get a URL pointing to the file in your JAR, then Apache Commons IO FileUtils.copyURLToFile() to copy it out to the file system. E.g.:
URL inputUrl = getClass().getResource("/absolute/path/of/source/in/jar/file");
File dest = new File("/path/to/destination/file");
FileUtils.copyURLToFile(inputUrl, dest);
Most likely, the problem with whatever code you have now is that you're (correctly) using a buffered output stream to write to the file but (incorrectly) failing to close it.
Oh, and you should edit your question to clarify exactly how you want to do this (programmatically, not, language, ...)

Faster way to do it with Java 7+ , plus code to get the current directory:
/**
* Copy a file from source to destination.
*
* #param source
* the source
* #param destination
* the destination
* #return True if succeeded , False if not
*/
public static boolean copy(InputStream source , String destination) {
boolean succeess = true;
System.out.println("Copying ->" + source + "\n\tto ->" + destination);
try {
Files.copy(source, Paths.get(destination), StandardCopyOption.REPLACE_EXISTING);
} catch (IOException ex) {
logger.log(Level.WARNING, "", ex);
succeess = false;
}
return succeess;
}
Testing it (icon.png is an image inside the package image of the application):
copy(getClass().getResourceAsStream("/image/icon.png"),getBasePathForClass(Main.class)+"icon.png");
About the line of code (getBasePathForClass(Main.class)): -> check the answer i have added here :) -> Getting the Current Working Directory in Java

Java 8 (actually FileSystem is there since 1.7) comes with some cool new classes/methods to deal with this. As somebody already mentioned that JAR is basically ZIP file, you could use
final URI jarFileUril = URI.create("jar:file:" + file.toURI().getPath());
final FileSystem fs = FileSystems.newFileSystem(jarFileUri, env);
(See Zip File)
Then you can use one of the convenient methods like:
fs.getPath("filename");
Then you can use Files class
try (final Stream<Path> sources = Files.walk(from)) {
sources.forEach(src -> {
final Path dest = to.resolve(from.relativize(src).toString());
try {
if (Files.isDirectory(from)) {
if (Files.notExists(to)) {
log.trace("Creating directory {}", to);
Files.createDirectories(to);
}
} else {
log.trace("Extracting file {} to {}", from, to);
Files.copy(from, to, StandardCopyOption.REPLACE_EXISTING);
}
} catch (IOException e) {
throw new RuntimeException("Failed to unzip file.", e);
}
});
}
Note: I tried that to unpack JAR files for testing

Robust solution:
public static void copyResource(String res, String dest, Class c) throws IOException {
InputStream src = c.getResourceAsStream(res);
Files.copy(src, Paths.get(dest), StandardCopyOption.REPLACE_EXISTING);
}
You can use it like this:
File tempFileGdalZip = File.createTempFile("temp_gdal", ".zip");
copyResource("/gdal.zip", tempFileGdalZip.getAbsolutePath(), this.getClass());

Use the JarInputStream class:
// assuming you already have an InputStream to the jar file..
JarInputStream jis = new JarInputStream( is );
// get the first entry
JarEntry entry = jis.getNextEntry();
// we will loop through all the entries in the jar file
while ( entry != null ) {
// test the entry.getName() against whatever you are looking for, etc
if ( matches ) {
// read from the JarInputStream until the read method returns -1
// ...
// do what ever you want with the read output
// ...
// if you only care about one file, break here
}
// get the next entry
entry = jis.getNextEntry();
}
jis.close();
See also: JarEntry

To copy a file from your jar, to the outside, you need to use the following approach:
Get a InputStream to a the file inside your jar file using getResourceAsStream()
We open our target file using a FileOutputStream
We copy bytes from the input to the output stream
We close our streams to prevent resource leaks
Example code that also contains a variable to not replace the existing values:
public File saveResource(String name) throws IOException {
return saveResource(name, true);
}
public File saveResource(String name, boolean replace) throws IOException {
return saveResource(new File("."), name, replace)
}
public File saveResource(File outputDirectory, String name) throws IOException {
return saveResource(outputDirectory, name, true);
}
public File saveResource(File outputDirectory, String name, boolean replace)
throws IOException {
File out = new File(outputDirectory, name);
if (!replace && out.exists())
return out;
// Step 1:
InputStream resource = this.getClass().getResourceAsStream(name);
if (resource == null)
throw new FileNotFoundException(name + " (resource not found)");
// Step 2 and automatic step 4
try(InputStream in = resource;
OutputStream writer = new BufferedOutputStream(
new FileOutputStream(out))) {
// Step 3
byte[] buffer = new byte[1024 * 4];
int length;
while((length = in.read(buffer)) >= 0) {
writer.write(buffer, 0, length);
}
}
return out;
}

A jar is just a zip file. Unzip it (using whatever method you're comfortable with) and copy the file normally.

${JAVA_HOME}/bin/jar -cvf /path/to.jar