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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.

Create directory in Java but don't throw error if it already exists [duplicate]

The condition is if the directory exists it has to create files in that specific directory without creating a new directory.
The below code only creates a file with the new directory but not for the existing directory . For example the directory name would be like "GETDIRECTION":
String PATH = "/remote/dir/server/";
String fileName = PATH.append(id).concat(getTimeStamp()).append(".txt");
String directoryName = PATH.append(this.getClassName());
File file = new File(String.valueOf(fileName));
File directory = new File(String.valueOf(directoryName));
if (!directory.exists()) {
directory.mkdir();
if (!file.exists() && !checkEnoughDiskSpace()) {
file.getParentFile().mkdir();
file.createNewFile();
}
}
FileWriter fw = new FileWriter(file.getAbsoluteFile());
BufferedWriter bw = new BufferedWriter(fw);
bw.write(value);
bw.close();

Java 8+ version:
Files.createDirectories(Paths.get("/Your/Path/Here"));
The Files.createDirectories() creates a new directory and parent directories that do not exist. This method does not throw an exception if the directory already exists.

This code checks for the existence of the directory first and creates it if not, and creates the file afterwards. Please note that I couldn't verify some of your method calls as I don't have your complete code, so I'm assuming the calls to things like getTimeStamp() and getClassName() will work. You should also do something with the possible IOException that can be thrown when using any of the java.io.* classes - either your function that writes the files should throw this exception (and it be handled elsewhere), or you should do it in the method directly. Also, I assumed that id is of type String - I don't know as your code doesn't explicitly define it. If it is something else like an int, you should probably cast it to a String before using it in the fileName as I have done here.
Also, I replaced your append calls with concat or + as I saw appropriate.
public void writeFile(String value){
String PATH = "/remote/dir/server/";
String directoryName = PATH.concat(this.getClassName());
String fileName = id + getTimeStamp() + ".txt";
File directory = new File(directoryName);
if (! directory.exists()){
directory.mkdir();
// If you require it to make the entire directory path including parents,
// use directory.mkdirs(); here instead.
}
File file = new File(directoryName + "/" + fileName);
try{
FileWriter fw = new FileWriter(file.getAbsoluteFile());
BufferedWriter bw = new BufferedWriter(fw);
bw.write(value);
bw.close();
}
catch (IOException e){
e.printStackTrace();
System.exit(-1);
}
}
You should probably not use bare path names like this if you want to run the code on Microsoft Windows - I'm not sure what it will do with the / in the filenames. For full portability, you should probably use something like File.separator to construct your paths.
Edit: According to a comment by JosefScript below, it's not necessary to test for directory existence. The directory.mkdir() call will return true if it created a directory, and false if it didn't, including the case when the directory already existed.

Trying to make this as short and simple as possible. Creates directory if it doesn't exist, and then returns the desired file:
/** Creates parent directories if necessary. Then returns file */
private static File fileWithDirectoryAssurance(String directory, String filename) {
File dir = new File(directory);
if (!dir.exists()) dir.mkdirs();
return new File(directory + "/" + filename);
}

I would suggest the following for Java8+.
/**
* Creates a File if the file does not exist, or returns a
* reference to the File if it already exists.
*/
public File createOrRetrieve(final String target) throws IOException {
final File answer;
Path path = Paths.get(target);
Path parent = path.getParent();
if(parent != null && Files.notExists(parent)) {
Files.createDirectories(path);
}
if(Files.notExists(path)) {
LOG.info("Target file \"" + target + "\" will be created.");
answer = Files.createFile(path).toFile();
} else {
LOG.info("Target file \"" + target + "\" will be retrieved.");
answer = path.toFile();
}
return answer;
}
Edit: Updated to fix bug as indicated by #Cataclysm and #Marcono1234. Thx guys:)

code:
// Create Directory if not exist then Copy a file.
public static void copyFile_Directory(String origin, String destDir, String destination) throws IOException {
Path FROM = Paths.get(origin);
Path TO = Paths.get(destination);
File directory = new File(String.valueOf(destDir));
if (!directory.exists()) {
directory.mkdir();
}
//overwrite the destination file if it exists, and copy
// the file attributes, including the rwx permissions
CopyOption[] options = new CopyOption[]{
StandardCopyOption.REPLACE_EXISTING,
StandardCopyOption.COPY_ATTRIBUTES
};
Files.copy(FROM, TO, options);
}

Simple Solution using using java.nio.Path
public static Path createFileWithDir(String directory, String filename) {
File dir = new File(directory);
if (!dir.exists()) dir.mkdirs();
return Paths.get(directory + File.separatorChar + filename);
}

If you create a web based application, the better solution is to check the directory exists or not then create the file if not exist. If exists, recreate again.
private File createFile(String path, String fileName) throws IOException {
ClassLoader classLoader = getClass().getClassLoader();
File file = new File(classLoader.getResource(".").getFile() + path + fileName);
// Lets create the directory
try {
file.getParentFile().mkdir();
} catch (Exception err){
System.out.println("ERROR (Directory Create)" + err.getMessage());
}
// Lets create the file if we have credential
try {
file.createNewFile();
} catch (Exception err){
System.out.println("ERROR (File Create)" + err.getMessage());
}
return file;
}

A simple solution using Java 8
public void init(String multipartLocation) throws IOException {
File storageDirectory = new File(multipartLocation);
if (!storageDirectory.exists()) {
if (!storageDirectory.mkdir()) {
throw new IOException("Error creating directory.");
}
}
}

If you're using Java 8 or above, then Files.createDirectories() method works the best.

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

How do I move a file from one location to another in Java?

How do you move a file from one location to another? When I run my program any file created in that location automatically moves to the specified location. How do I know which file is moved?

myFile.renameTo(new File("/the/new/place/newName.file"));
File#renameTo does that (it can not only rename, but also move between directories, at least on the same file system).
Renames the file denoted by this abstract pathname.
Many aspects of the behavior of this method are inherently platform-dependent: The rename operation might not be able to move a file from one filesystem to another, it might not be atomic, and it might not succeed if a file with the destination abstract pathname already exists. The return value should always be checked to make sure that the rename operation was successful.
If you need a more comprehensive solution (such as wanting to move the file between disks), look at Apache Commons FileUtils#moveFile

With Java 7 or newer you can use Files.move(from, to, CopyOption... options).
E.g.
Files.move(Paths.get("/foo.txt"), Paths.get("bar.txt"), StandardCopyOption.REPLACE_EXISTING);
See the Files documentation for more details

Java 6
public boolean moveFile(String sourcePath, String targetPath) {
File fileToMove = new File(sourcePath);
return fileToMove.renameTo(new File(targetPath));
}
Java 7 (Using NIO)
public boolean moveFile(String sourcePath, String targetPath) {
boolean fileMoved = true;
try {
Files.move(Paths.get(sourcePath), Paths.get(targetPath), StandardCopyOption.REPLACE_EXISTING);
} catch (Exception e) {
fileMoved = false;
e.printStackTrace();
}
return fileMoved;
}

File.renameTo from Java IO can be used to move a file in Java. Also see this SO question.

To move a file you could also use Jakarta Commons IOs FileUtils.moveFile
On error it throws an IOException, so when no exception is thrown you know that that the file was moved.

Just add the source and destination folder paths.
It will move all the files and folder from source folder to
destination folder.
File destinationFolder = new File("");
File sourceFolder = new File("");
if (!destinationFolder.exists())
{
destinationFolder.mkdirs();
}
// Check weather source exists and it is folder.
if (sourceFolder.exists() && sourceFolder.isDirectory())
{
// Get list of the files and iterate over them
File[] listOfFiles = sourceFolder.listFiles();
if (listOfFiles != null)
{
for (File child : listOfFiles )
{
// Move files to destination folder
child.renameTo(new File(destinationFolder + "\\" + child.getName()));
}
// Add if you want to delete the source folder
sourceFolder.delete();
}
}
else
{
System.out.println(sourceFolder + " Folder does not exists");
}

Files.move(source, target, REPLACE_EXISTING);
You can use the Files object
Read more about Files

You could execute an external tool for that task (like copy in windows environments) but, to keep the code portable, the general approach is to:
read the source file into memory
write the content to a file at the new location
delete the source file
File#renameTo will work as long as source and target location are on the same volume. Personally I'd avoid using it to move files to different folders.

Try this :-
boolean success = file.renameTo(new File(Destdir, file.getName()));

Wrote this method to do this very thing on my own project only with the replace file if existing logic in it.
// we use the older file i/o operations for this rather than the newer jdk7+ Files.move() operation
private boolean moveFileToDirectory(File sourceFile, String targetPath) {
File tDir = new File(targetPath);
if (tDir.exists()) {
String newFilePath = targetPath+File.separator+sourceFile.getName();
File movedFile = new File(newFilePath);
if (movedFile.exists())
movedFile.delete();
return sourceFile.renameTo(new File(newFilePath));
} else {
LOG.warn("unable to move file "+sourceFile.getName()+" to directory "+targetPath+" -> target directory does not exist");
return false;
}
}

Please try this.
private boolean filemovetoanotherfolder(String sourcefolder, String destinationfolder, String filename) {
boolean ismove = false;
InputStream inStream = null;
OutputStream outStream = null;
try {
File afile = new File(sourcefolder + filename);
File bfile = new File(destinationfolder + filename);
inStream = new FileInputStream(afile);
outStream = new FileOutputStream(bfile);
byte[] buffer = new byte[1024 * 4];
int length;
// copy the file content in bytes
while ((length = inStream.read(buffer)) > 0) {
outStream.write(buffer, 0, length);
}
// delete the original file
afile.delete();
ismove = true;
System.out.println("File is copied successful!");
} catch (IOException e) {
e.printStackTrace();
}finally{
inStream.close();
outStream.close();
}
return ismove;
}

Rename a file using Java

Can we rename a file say test.txt to test1.txt ?
If test1.txt exists will it rename ?
How do I rename it to the already existing test1.txt file so the new contents of test.txt are added to it for later use?

Copied from http://exampledepot.8waytrips.com/egs/java.io/RenameFile.html
// File (or directory) with old name
File file = new File("oldname");
// File (or directory) with new name
File file2 = new File("newname");
if (file2.exists())
throw new java.io.IOException("file exists");
// Rename file (or directory)
boolean success = file.renameTo(file2);
if (!success) {
// File was not successfully renamed
}
To append to the new file:
java.io.FileWriter out= new java.io.FileWriter(file2, true /*append=yes*/);

In short:
Files.move(source, source.resolveSibling("newname"));
More detail:
import java.nio.file.Files;
import java.nio.file.Path;
import java.nio.file.Paths;
import java.nio.file.StandardCopyOption;
The following is copied directly from http://docs.oracle.com/javase/7/docs/api/index.html:
Suppose we want to rename a file to "newname", keeping the file in the same directory:
Path source = Paths.get("path/here");
Files.move(source, source.resolveSibling("newname"));
Alternatively, suppose we want to move a file to new directory, keeping the same file name, and replacing any existing file of that name in the directory:
Path source = Paths.get("from/path");
Path newdir = Paths.get("to/path");
Files.move(source, newdir.resolve(source.getFileName()), StandardCopyOption.REPLACE_EXISTING);

You want to utilize the renameTo method on a File object.
First, create a File object to represent the destination. Check to see if that file exists. If it doesn't exist, create a new File object for the file to be moved. call the renameTo method on the file to be moved, and check the returned value from renameTo to see if the call was successful.
If you want to append the contents of one file to another, there are a number of writers available. Based on the extension, it sounds like it's plain text, so I would look at the FileWriter.

For Java 1.6 and lower, I believe the safest and cleanest API for this is Guava's Files.move.
Example:
File newFile = new File(oldFile.getParent(), "new-file-name.txt");
Files.move(oldFile.toPath(), newFile.toPath());
The first line makes sure that the location of the new file is the same directory, i.e. the parent directory of the old file.
EDIT:
I wrote this before I started using Java 7, which introduced a very similar approach. So if you're using Java 7+, you should see and upvote kr37's answer.

Renaming the file by moving it to a new name. (FileUtils is from Apache Commons IO lib)
String newFilePath = oldFile.getAbsolutePath().replace(oldFile.getName(), "") + newName;
File newFile = new File(newFilePath);
try {
FileUtils.moveFile(oldFile, newFile);
} catch (IOException e) {
e.printStackTrace();
}

This is an easy way to rename a file:
File oldfile =new File("test.txt");
File newfile =new File("test1.txt");
if(oldfile.renameTo(newfile)){
System.out.println("File renamed");
}else{
System.out.println("Sorry! the file can't be renamed");
}

import java.nio.file.Files;
import java.nio.file.Path;
import java.nio.file.Paths;
import static java.nio.file.StandardCopyOption.*;
Path yourFile = Paths.get("path_to_your_file\text.txt");
Files.move(yourFile, yourFile.resolveSibling("text1.txt"));
To replace an existing file with the name "text1.txt":
Files.move(yourFile, yourFile.resolveSibling("text1.txt"),REPLACE_EXISTING);

Try This
File file=new File("Your File");
boolean renameResult = file.renameTo(new File("New Name"));
// todo: check renameResult
Note :
We should always check the renameTo return value to make sure rename file is successful because it’s platform dependent(different Operating system, different file system) and it doesn’t throw IO exception if rename fails.

Yes, you can use File.renameTo(). But remember to have the correct path while renaming it to a new file.
import java.util.Arrays;
import java.util.List;
public class FileRenameUtility {
public static void main(String[] a) {
System.out.println("FileRenameUtility");
FileRenameUtility renameUtility = new FileRenameUtility();
renameUtility.fileRename("c:/Temp");
}
private void fileRename(String folder){
File file = new File(folder);
System.out.println("Reading this "+file.toString());
if(file.isDirectory()){
File[] files = file.listFiles();
List<File> filelist = Arrays.asList(files);
filelist.forEach(f->{
if(!f.isDirectory() && f.getName().startsWith("Old")){
System.out.println(f.getAbsolutePath());
String newName = f.getAbsolutePath().replace("Old","New");
boolean isRenamed = f.renameTo(new File(newName));
if(isRenamed)
System.out.println(String.format("Renamed this file %s to %s",f.getName(),newName));
else
System.out.println(String.format("%s file is not renamed to %s",f.getName(),newName));
}
});
}
}
}

If it's just renaming the file, you can use File.renameTo().
In the case where you want to append the contents of the second file to the first, take a look at FileOutputStream with the append constructor option or The same thing for FileWriter. You'll need to read the contents of the file to append and write them out using the output stream/writer.

As far as I know, renaming a file will not append its contents to that of an existing file with the target name.
About renaming a file in Java, see the documentation for the renameTo() method in class File.

Files.move(file.toPath(), fileNew.toPath());
works, but only when you close (or autoclose) ALL used resources (InputStream, FileOutputStream etc.) I think the same situation with file.renameTo or FileUtils.moveFile.

Here is my code to rename multiple files in a folder successfully:
public static void renameAllFilesInFolder(String folderPath, String newName, String extension) {
if(newName == null || newName.equals("")) {
System.out.println("New name cannot be null or empty");
return;
}
if(extension == null || extension.equals("")) {
System.out.println("Extension cannot be null or empty");
return;
}
File dir = new File(folderPath);
int i = 1;
if (dir.isDirectory()) { // make sure it's a directory
for (final File f : dir.listFiles()) {
try {
File newfile = new File(folderPath + "\\" + newName + "_" + i + "." + extension);
if(f.renameTo(newfile)){
System.out.println("Rename succesful: " + newName + "_" + i + "." + extension);
} else {
System.out.println("Rename failed");
}
i++;
} catch (Exception e) {
e.printStackTrace();
}
}
}
}
and run it for an example:
renameAllFilesInFolder("E:\\Downloads\\Foldername", "my_avatar", "gif");

I do not like java.io.File.renameTo(...) because sometimes it does not renames the file and you do not know why! It just returns true of false. It does not thrown an exception if it fails.
On the other hand, java.nio.file.Files.move(...) is more useful as it throws an exception when it fails.

Running code is here.
private static void renameFile(File fileName) {
FileOutputStream fileOutputStream =null;
BufferedReader br = null;
FileReader fr = null;
String newFileName = "yourNewFileName"
try {
fileOutputStream = new FileOutputStream(newFileName);
fr = new FileReader(fileName);
br = new BufferedReader(fr);
String sCurrentLine;
while ((sCurrentLine = br.readLine()) != null) {
fileOutputStream.write(("\n"+sCurrentLine).getBytes());
}
fileOutputStream.flush();
} catch (IOException e) {
e.printStackTrace();
} finally {
try {
fileOutputStream.close();
if (br != null)
br.close();
if (fr != null)
fr.close();
} catch (IOException ex) {
ex.printStackTrace();
}
}
}