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Moving from one directory to another java [duplicate]

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;
}

Java: How to copy a folder from one jar to another?

Solved, see my post below
I want to copy files from one .jar to another using java. But not all files of the jar shall be copied. I only want one specific folder to be copied which also can contain subfolders. I already have the code to copy files from one jar to another:
public static void copyFiles(final File file) throws IOException {
final JarFile targetJar = new JarFile(file);
final JarOutputStream output = new JarOutputStream(new FileOutputStream(file));
final JarFile localJar = new JarFile(new File(JarEditor.class.getProtectionDomain().getCodeSource().getLocation().getFile()));
// WRAPPER_CONTENT is a String[] which contains all names of the files i want to copy
for (final String entryName : StringResource.WRAPPER_CONTENT) {
final JarEntry entryToCopy = localJar.getJarEntry(entryName);
final JarEntry targetEntry = new JarEntry(entryName);
output.putNextEntry(targetEntry);
final InputStream input = localJar.getInputStream(entryToCopy);
final byte[] buffer = new byte[10240];
int readByte = 0;
while ((readByte = input.read(buffer)) >= 0) {
output.write(buffer, 0, readByte);
}
}
output.flush();
output.close();
localJar.close();
targetJar.close();
}
Currently im copying the files by running through a hardcoded list of their names. But id prefer a more flexible way so i could add and remove resources from my jar and it still will work. They all have the same parent-folder in the jar. So how would i go through that one folder and only copy those files?
Also maybe worth mentioning, all files which shall be copied are located in the jarfile my runtime is coming from as you probably got from looking at my code.
Thanks for help
Baschdi

Bad luck i guess. I've been searching for the solution for a while now and just when i create this thread i manage to find the solution. I simply check if the name of the jarentry starts with the name of the folder i want to copy. If so, i use the same way as above to copy the jarentry to the other jar. Solved

Getting a FileNotFoundException if not running from Eclipse [duplicate]

I'm trying to access an XML file within a jar file, from a separate jar that's running as a desktop application. I can get the URL to the file I need, but when I pass that to a FileReader (as a String) I get a FileNotFoundException saying "The file name, directory name, or volume label syntax is incorrect."
As a point of reference, I have no trouble reading image resources from the same jar, passing the URL to an ImageIcon constructor. This seems to indicate that the method I'm using to get the URL is correct.
URL url = getClass().getResource("/xxx/xxx/xxx/services.xml");
ServicesLoader jsl = new ServicesLoader( url.toString() );
Inside the ServicesLoader class I have
XMLReader xr = XMLReaderFactory.createXMLReader();
xr.setContentHandler( this );
xr.setErrorHandler( this );
xr.parse( new InputSource( new FileReader( filename )));
What's wrong with using this technique to read the XML file?

Looks like you want to use java.lang.Class.getResourceAsStream(String), see
https://docs.oracle.com/javase/8/docs/api/java/lang/Class.html#getResourceAsStream-java.lang.String-

You don't say if this is a desktop or web app. I would use the getResourceAsStream() method from an appropriate ClassLoader if it's a desktop or the Context if it's a web app.

It looks as if you are using the URL.toString result as the argument to the FileReader constructor. URL.toString is a bit broken, and instead you should generally use url.toURI().toString(). In any case, the string is not a file path.
Instead, you should either:
Pass the URL to ServicesLoader and let it call openStream or similar.
Use Class.getResourceAsStream and just pass the stream over, possibly inside an InputSource. (Remember to check for nulls as the API is a bit messy.)

The problem was that I was going a step too far in calling the parse method of XMLReader. The parse method accepts an InputSource, so there was no reason to even use a FileReader. Changing the last line of the code above to
xr.parse( new InputSource( filename ));
works just fine.

I'd like to point out that one issues is what if the same resources are in multiple jar files.
Let's say you want to read /org/node/foo.txt, but not from one file, but from each and every jar file.
I have run into this same issue several times before.
I was hoping in JDK 7 that someone would write a classpath filesystem, but alas not yet.
Spring has the Resource class which allows you to load classpath resources quite nicely.
I wrote a little prototype to solve this very problem of reading resources form multiple jar files. The prototype does not handle every edge case, but it does handle looking for resources in directories that are in the jar files.
I have used Stack Overflow for quite sometime. This is the second answer that I remember answering a question so forgive me if I go too long (it is my nature).
This is a prototype resource reader. The prototype is devoid of robust error checking.
I have two prototype jar files that I have setup.
<pre>
<dependency>
<groupId>invoke</groupId>
<artifactId>invoke</artifactId>
<version>1.0-SNAPSHOT</version>
</dependency>
<dependency>
<groupId>node</groupId>
<artifactId>node</artifactId>
<version>1.0-SNAPSHOT</version>
</dependency>
The jar files each have a file under /org/node/ called resource.txt.
This is just a prototype of what a handler would look like with classpath://
I also have a resource.foo.txt in my local resources for this project.
It picks them all up and prints them out.
package com.foo;
import java.io.File;
import java.io.FileReader;
import java.io.InputStreamReader;
import java.io.Reader;
import java.net.URI;
import java.net.URL;
import java.util.Enumeration;
import java.util.zip.ZipEntry;
import java.util.zip.ZipFile;
/**
* Prototype resource reader.
* This prototype is devoid of error checking.
*
*
* I have two prototype jar files that I have setup.
* <pre>
* <dependency>
* <groupId>invoke</groupId>
* <artifactId>invoke</artifactId>
* <version>1.0-SNAPSHOT</version>
* </dependency>
*
* <dependency>
* <groupId>node</groupId>
* <artifactId>node</artifactId>
* <version>1.0-SNAPSHOT</version>
* </dependency>
* </pre>
* The jar files each have a file under /org/node/ called resource.txt.
* <br />
* This is just a prototype of what a handler would look like with classpath://
* I also have a resource.foo.txt in my local resources for this project.
* <br />
*/
public class ClasspathReader {
public static void main(String[] args) throws Exception {
/* This project includes two jar files that each have a resource located
in /org/node/ called resource.txt.
*/
/*
Name space is just a device I am using to see if a file in a dir
starts with a name space. Think of namespace like a file extension
but it is the start of the file not the end.
*/
String namespace = "resource";
//someResource is classpath.
String someResource = args.length > 0 ? args[0] :
//"classpath:///org/node/resource.txt"; It works with files
"classpath:///org/node/"; //It also works with directories
URI someResourceURI = URI.create(someResource);
System.out.println("URI of resource = " + someResourceURI);
someResource = someResourceURI.getPath();
System.out.println("PATH of resource =" + someResource);
boolean isDir = !someResource.endsWith(".txt");
/** Classpath resource can never really start with a starting slash.
* Logically they do, but in reality you have to strip it.
* This is a known behavior of classpath resources.
* It works with a slash unless the resource is in a jar file.
* Bottom line, by stripping it, it always works.
*/
if (someResource.startsWith("/")) {
someResource = someResource.substring(1);
}
/* Use the ClassLoader to lookup all resources that have this name.
Look for all resources that match the location we are looking for. */
Enumeration resources = null;
/* Check the context classloader first. Always use this if available. */
try {
resources =
Thread.currentThread().getContextClassLoader().getResources(someResource);
} catch (Exception ex) {
ex.printStackTrace();
}
if (resources == null || !resources.hasMoreElements()) {
resources = ClasspathReader.class.getClassLoader().getResources(someResource);
}
//Now iterate over the URLs of the resources from the classpath
while (resources.hasMoreElements()) {
URL resource = resources.nextElement();
/* if the resource is a file, it just means that we can use normal mechanism
to scan the directory.
*/
if (resource.getProtocol().equals("file")) {
//if it is a file then we can handle it the normal way.
handleFile(resource, namespace);
continue;
}
System.out.println("Resource " + resource);
/*
Split up the string that looks like this:
jar:file:/Users/rick/.m2/repository/invoke/invoke/1.0-SNAPSHOT/invoke-1.0-SNAPSHOT.jar!/org/node/
into
this /Users/rick/.m2/repository/invoke/invoke/1.0-SNAPSHOT/invoke-1.0-SNAPSHOT.jar
and this
/org/node/
*/
String[] split = resource.toString().split(":");
String[] split2 = split[2].split("!");
String zipFileName = split2[0];
String sresource = split2[1];
System.out.printf("After split zip file name = %s," +
" \nresource in zip %s \n", zipFileName, sresource);
/* Open up the zip file. */
ZipFile zipFile = new ZipFile(zipFileName);
/* Iterate through the entries. */
Enumeration entries = zipFile.entries();
while (entries.hasMoreElements()) {
ZipEntry entry = entries.nextElement();
/* If it is a directory, then skip it. */
if (entry.isDirectory()) {
continue;
}
String entryName = entry.getName();
System.out.printf("zip entry name %s \n", entryName);
/* If it does not start with our someResource String
then it is not our resource so continue.
*/
if (!entryName.startsWith(someResource)) {
continue;
}
/* the fileName part from the entry name.
* where /foo/bar/foo/bee/bar.txt, bar.txt is the file
*/
String fileName = entryName.substring(entryName.lastIndexOf("/") + 1);
System.out.printf("fileName %s \n", fileName);
/* See if the file starts with our namespace and ends with our extension.
*/
if (fileName.startsWith(namespace) && fileName.endsWith(".txt")) {
/* If you found the file, print out
the contents fo the file to System.out.*/
try (Reader reader = new InputStreamReader(zipFile.getInputStream(entry))) {
StringBuilder builder = new StringBuilder();
int ch = 0;
while ((ch = reader.read()) != -1) {
builder.append((char) ch);
}
System.out.printf("zip fileName = %s\n\n####\n contents of file %s\n###\n", entryName, builder);
} catch (Exception ex) {
ex.printStackTrace();
}
}
//use the entry to see if it's the file '1.txt'
//Read from the byte using file.getInputStream(entry)
}
}
}
/**
* The file was on the file system not a zip file,
* this is here for completeness for this example.
* otherwise.
*
* #param resource
* #param namespace
* #throws Exception
*/
private static void handleFile(URL resource, String namespace) throws Exception {
System.out.println("Handle this resource as a file " + resource);
URI uri = resource.toURI();
File file = new File(uri.getPath());
if (file.isDirectory()) {
for (File childFile : file.listFiles()) {
if (childFile.isDirectory()) {
continue;
}
String fileName = childFile.getName();
if (fileName.startsWith(namespace) && fileName.endsWith("txt")) {
try (FileReader reader = new FileReader(childFile)) {
StringBuilder builder = new StringBuilder();
int ch = 0;
while ((ch = reader.read()) != -1) {
builder.append((char) ch);
}
System.out.printf("fileName = %s\n\n####\n contents of file %s\n###\n", childFile, builder);
} catch (Exception ex) {
ex.printStackTrace();
}
}
}
} else {
String fileName = file.getName();
if (fileName.startsWith(namespace) && fileName.endsWith("txt")) {
try (FileReader reader = new FileReader(file)) {
StringBuilder builder = new StringBuilder();
int ch = 0;
while ((ch = reader.read()) != -1) {
builder.append((char) ch);
}
System.out.printf("fileName = %s\n\n####\n contents of file %s\n###\n", fileName, builder);
} catch (Exception ex) {
ex.printStackTrace();
}
}
}
}
}
You can see a fuller example here with the sample output.

Here's a sample code on how to read a file properly inside a jar file (in this case, the current executing jar file)
Just change executable with the path of your jar file if it is not the current running one.
Then change the filePath to the path of the file you want to use inside the jar file. I.E. if your file is in
someJar.jar\img\test.gif
. Set the filePath to "img\test.gif"
File executable = new File(BrowserViewControl.class.getProtectionDomain().getCodeSource().getLocation().toURI());
JarFile jar = new JarFile(executable);
InputStream fileInputStreamReader = jar.getInputStream(jar.getJarEntry(filePath));
byte[] bytes = new byte[fileInputStreamReader.available()];
int sizeOrig = fileInputStreamReader.available();
int size = fileInputStreamReader.available();
int offset = 0;
while (size != 0){
fileInputStreamReader.read(bytes, offset, size);
offset = sizeOrig - fileInputStreamReader.available();
size = fileInputStreamReader.available();
}

Outside of your technique, why not use the standard Java JarFile class to get the references you want? From there most of your problems should go away.

If you use resources extensively, you might consider using
Commons VFS.
Also supports:
* Local Files
* FTP, SFTP
* HTTP and HTTPS
* Temporary Files "normal FS backed)
* Zip, Jar and Tar (uncompressed, tgz or tbz2)
* gzip and bzip2
* resources
* ram - "ramdrive"
* mime
There's also JBoss VFS - but it's not much documented.

I have 2 CSV files that I use to read data. The java program is exported as a runnable jar file. When you export it, you will figure out it doesn't export your resources with it.
I added a folder under project called data in eclipse. In that folder i stored my csv files.
When I need to reference those files I do it like this...
private static final String ZIP_FILE_LOCATION_PRIMARY = "free-zipcode-database-Primary.csv";
private static final String ZIP_FILE_LOCATION = "free-zipcode-database.csv";
private static String getFileLocation(){
String loc = new File("").getAbsolutePath() + File.separatorChar +
"data" + File.separatorChar;
if (usePrimaryZipCodesOnly()){
loc = loc.concat(ZIP_FILE_LOCATION_PRIMARY);
} else {
loc = loc.concat(ZIP_FILE_LOCATION);
}
return loc;
}
Then when you put the jar in a location so it can be ran via commandline, make sure that you add the data folder with the resources into the same location as the jar file.

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;
}