Develop Reference

How to copy folder's out of resources from jar both in runtime and dev-env?

I am currently making a game which have levels which are pre-made and i currently store them in resources. I want a solution to how i can extract a folder out of a jar in production and development environment.
I have tried copying the folder by using the given method below and pass the src as File defaultWorld = new File(GameData.class.getClassLoader().getResource("worlds/").getFile());
and destination as private static File worldsDir = new File("run/worlds");
public static void copyFolder(File src, File dest) {
try {
if (src.isDirectory()) {
if (!dest.exists()) {
dest.mkdir();
}
String[] files = src.list();
for (String file : files) {
copyFolder(new File(src, file), new File(dest, file));
}
} else {
try (InputStream in = new FileInputStream(src)) {
try (OutputStream out = new FileOutputStream(dest)) {
byte[] buffer = new byte[1024];
int length;
//copy the file content in bytes
while ((length = in.read(buffer)) > 0) {
out.write(buffer, 0, length);
}
}
}
}
} catch (IOException e) {
e.printStackTrace();
}
}
I expected the above method to work on both dev and production env but it throws FileNotFoundException when opening file output stream.

You cannot list resources in a jar.
Any workarounds you’re thinking of, are unreliable.
Never call the getFile() method of URL. It does not return a valid file name; it just returns the path and query portions of the URL, with any percent-escapes intact. Furthermore, jar entries are not file: URLs, so a resource path can never be a valid file name when it refers to a jar entry.
The only way to list things in a jar file is by iterating through all jar entries, but you aren’t even guaranteed to have access to your jar, because ClassLoaders are not guaranteed to be URLClassLoaders, and in general are not guaranteed to use jar: URLs.
You can’t even rely on MyApplication.class.getProtectionDomain().getCodeSource(), because getCodeSource() can return null.
If you want to copy multiple files from your jar, here are some reliable ways to do it:
Hard code the list of resources you plan to copy. It’s your application, so you know what files you’re putting in the jar.
Keep a single text file in your jar which contains a list of resource paths to copy.
Store your resources in a single zip archive which is embedded in your jar, and extract it yourself with a ZipInputStream which wraps MyApplication.class.getResourceAsStream.

Java nio move folder with content throws exception [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;
}

Convert JarEntry to File

I'm using a library which wants a File() as an argument.
The file I want to pass it is one I want to package with my app, as part of the .jar
Is there any way to convert the JarEntry that I get from within my .jar to a File object I can pass?
If not and I have to copy the resource to disk temporarily, where's the best place to put the temporary file?
Thanks.

You cannot get a path to a file within a JARFile, only a stream, so you should extract it to the temporary directory and then pass that extracted file.
Here's a function I wrote to do this when I provided a db with a jar previously.
/**
* This method is responsible for extracting resource files from within the .jar to the temporary directory.
* #param filePath The filepath relative to the 'Resources/' directory within the .jar from which to extract the file.
* #return A file object to the extracted file
**/
public File extract(String filePath)
{
try
{
File f = File.createTempFile(filePath, null);
FileOutputStream resourceOS = new FileOutputStream(f);
byte[] byteArray = new byte[1024];
int i;
InputStream classIS = getClass().getClassLoader().getResourceAsStream("Resources/"+filePath);
//While the input stream has bytes
while ((i = classIS.read(byteArray)) > 0)
{
//Write the bytes to the output stream
resourceOS.write(byteArray, 0, i);
}
//Close streams to prevent errors
classIS.close();
resourceOS.close();
return f;
}
catch (Exception e)
{
System.out.println("An error has occurred while extracting the database. This may mean the program is unable to have any database interaction, please contact the developer.\nError Description:\n"+e.getMessage());
return null;
}
}

A File represents a real entry in the filesystem; a JarEntry doesn't exist on the file system. The mapping won't be there unless you extract the JAR entry to an actual file.
You can create a temp file using File.createTempFile. More details are available at this SO answer.

Copy a file to the Relative path of myClass

This is my app directory
----dist
+-- lib //library folder
+-- backup //folder with my database backup filename = database_file.sqlite
|__ app.jar
|__ database_file.sqlite
When I perform the action to a button I want to close database connection and after that to copy (overwrite) the database_file.sqlite from backup folder to the root folder (dist) where the database_file.sqlite it's loaded into my app. Shortly, I want to reload a backup/original database when a button is clicked.
Note: I use a default package
For now I have this code (see bellow), but I can't figure out how to make it works.
private void cmd_backupActionPerformed(java.awt.event.ActionEvent evt) {
int p = JOptionPane.showConfirmDialog(null, "Do you really want to reset your data / Backup database?", "Backup", JOptionPane.YES_NO_OPTION);
if (p == 0) {
InputStream inStream = null;
OutputStream outStream = null;
try {
rs.close();
pst.close();
File afile = new File("C:\\Users\\Tzontonel\\Documents\\NetBeansProjects\\RDSS\\dist\\backup\\database_file.sqlite");
File bfile = new File("C:\\Users\\Tzontonel\\Documents\\NetBeansProjects\\RDSS\\dist\\database_file.sqlite");
System.out.println(afile.getCanonicalPath());
inStream = new FileInputStream(afile);
outStream = new FileOutputStream(bfile);
byte[] buffer = new byte[1024];
int length;
//copy the file content in bytes
while ((length = inStream.read(buffer)) > 0) {
outStream.write(buffer, 0, length);
}
inStream.close();
outStream.close();
// delete the original file
// afile.delete();
System.out.println("File is copied successful!");
} catch (Exception e) {
JOptionPane.showMessageDialog(null, e);
}
}
}

The class loader doesn't access the file system. It accesses resources that are in the classpath. You don't want to use the class loader here, but file IO. So the call to getResourceAsStream() doesn't make sense (and you replace the input stream it returns by another one anyway).
Put your class in a package, as all classes should be.
Now, when you create a File like this:
new File("database_file.sqlite")
you open a file that is in the current directory. And the current directory is the directory from which the java command is executed. So, unless you are always in the dist directory when executing your program, you won't be able to open the file by using such a relative path.
Pass the path of the dist directory as an argument or system property to your application, and use this directory as a base directory to open your files.
The next problem is that you open an input stream and an output stream to the same file, at the same time. You should open an InputStream to the backup file, and an output stream to the database file.

How to get a path to a resource in a Java JAR file

I am trying to get a path to a Resource but I have had no luck.
This works (both in IDE and with the JAR) but this way I can't get a path to a file, only the file contents:
ClassLoader classLoader = getClass().getClassLoader();
PrintInputStream(classLoader.getResourceAsStream("config/netclient.p"));
If I do this:
ClassLoader classLoader = getClass().getClassLoader();
File file = new File(classLoader.getResource("config/netclient.p").getFile());
The result is:
java.io.FileNotFoundException: file:/path/to/jarfile/bot.jar!/config/netclient.p (No such file or directory)
Is there a way to get a path to a resource file?

This is deliberate. The contents of the "file" may not be available as a file. Remember you are dealing with classes and resources that may be part of a JAR file or other kind of resource. The classloader does not have to provide a file handle to the resource, for example the jar file may not have been expanded into individual files in the file system.
Anything you can do by getting a java.io.File could be done by copying the stream out into a temporary file and doing the same, if a java.io.File is absolutely necessary.

When loading a resource make sure you notice the difference between:
getClass().getClassLoader().getResource("com/myorg/foo.jpg") //relative path
and
getClass().getResource("/com/myorg/foo.jpg")); //note the slash at the beginning
I guess, this confusion is causing most of problems when loading a resource.
Also, when you're loading an image it's easier to use getResourceAsStream():
BufferedImage image = ImageIO.read(getClass().getResourceAsStream("/com/myorg/foo.jpg"));
When you really have to load a (non-image) file from a JAR archive, you might try this:
File file = null;
String resource = "/com/myorg/foo.xml";
URL res = getClass().getResource(resource);
if (res.getProtocol().equals("jar")) {
try {
InputStream input = getClass().getResourceAsStream(resource);
file = File.createTempFile("tempfile", ".tmp");
OutputStream out = new FileOutputStream(file);
int read;
byte[] bytes = new byte[1024];
while ((read = input.read(bytes)) != -1) {
out.write(bytes, 0, read);
}
out.close();
file.deleteOnExit();
} catch (IOException ex) {
Exceptions.printStackTrace(ex);
}
} else {
//this will probably work in your IDE, but not from a JAR
file = new File(res.getFile());
}
if (file != null && !file.exists()) {
throw new RuntimeException("Error: File " + file + " not found!");
}

The one line answer is -
String path = this.getClass().getClassLoader().getResource(<resourceFileName>).toExternalForm()
Basically getResource method gives the URL. From this URL you can extract the path by calling toExternalForm().

I spent a while messing around with this problem, because no solution I found actually worked, strangely enough! The working directory is frequently not the directory of the JAR, especially if a JAR (or any program, for that matter) is run from the Start Menu under Windows. So here is what I did, and it works for .class files run from outside a JAR just as well as it works for a JAR. (I only tested it under Windows 7.)
try {
//Attempt to get the path of the actual JAR file, because the working directory is frequently not where the file is.
//Example: file:/D:/all/Java/TitanWaterworks/TitanWaterworks-en.jar!/TitanWaterworks.class
//Another example: /D:/all/Java/TitanWaterworks/TitanWaterworks.class
PROGRAM_DIRECTORY = getClass().getClassLoader().getResource("TitanWaterworks.class").getPath(); // Gets the path of the class or jar.
//Find the last ! and cut it off at that location. If this isn't being run from a jar, there is no !, so it'll cause an exception, which is fine.
try {
PROGRAM_DIRECTORY = PROGRAM_DIRECTORY.substring(0, PROGRAM_DIRECTORY.lastIndexOf('!'));
} catch (Exception e) { }
//Find the last / and cut it off at that location.
PROGRAM_DIRECTORY = PROGRAM_DIRECTORY.substring(0, PROGRAM_DIRECTORY.lastIndexOf('/') + 1);
//If it starts with /, cut it off.
if (PROGRAM_DIRECTORY.startsWith("/")) PROGRAM_DIRECTORY = PROGRAM_DIRECTORY.substring(1, PROGRAM_DIRECTORY.length());
//If it starts with file:/, cut that off, too.
if (PROGRAM_DIRECTORY.startsWith("file:/")) PROGRAM_DIRECTORY = PROGRAM_DIRECTORY.substring(6, PROGRAM_DIRECTORY.length());
} catch (Exception e) {
PROGRAM_DIRECTORY = ""; //Current working directory instead.
}

This is same code from user Tombart with stream flush and close to avoid incomplete temporary file content copy from jar resource and to have unique temp file names.
File file = null;
String resource = "/view/Trial_main.html" ;
URL res = getClass().getResource(resource);
if (res.toString().startsWith("jar:")) {
try {
InputStream input = getClass().getResourceAsStream(resource);
file = File.createTempFile(new Date().getTime()+"", ".html");
OutputStream out = new FileOutputStream(file);
int read;
byte[] bytes = new byte[1024];
while ((read = input.read(bytes)) != -1) {
out.write(bytes, 0, read);
}
out.flush();
out.close();
input.close();
file.deleteOnExit();
} catch (IOException ex) {
ex.printStackTrace();
}
} else {
//this will probably work in your IDE, but not from a JAR
file = new File(res.getFile());
}

if netclient.p is inside a JAR file, it won't have a path because that file is located inside other file. in that case, the best path you can have is really file:/path/to/jarfile/bot.jar!/config/netclient.p.

You need to understand the path within the jar file.
Simply refer to it relative. So if you have a file (myfile.txt), located in foo.jar under the \src\main\resources directory (maven style). You would refer to it like:
src/main/resources/myfile.txt
If you dump your jar using jar -tvf myjar.jar you will see the output and the relative path within the jar file, and use the FORWARD SLASHES.

In my case, I have used a URL object instead Path.
File
File file = new File("my_path");
URL url = file.toURI().toURL();
Resource in classpath using classloader
URL url = MyClass.class.getClassLoader().getResource("resource_name")
When I need to read the content, I can use the following code:
InputStream stream = url.openStream();
And you can access the content using an InputStream.

follow code!
/src/main/resources/file
streamToFile(getClass().getClassLoader().getResourceAsStream("file"))
public static File streamToFile(InputStream in) {
if (in == null) {
return null;
}
try {
File f = File.createTempFile(String.valueOf(in.hashCode()), ".tmp");
f.deleteOnExit();
FileOutputStream out = new FileOutputStream(f);
byte[] buffer = new byte[1024];
int bytesRead;
while ((bytesRead = in.read(buffer)) != -1) {
out.write(buffer, 0, bytesRead);
}
return f;
} catch (IOException e) {
LOGGER.error(e.getMessage(), e);
return null;
}
}

A File is an abstraction for a file in a filesystem, and the filesystems don't know anything about what are the contents of a JAR.
Try with an URI, I think there's a jar:// protocol that might be useful for your purpouses.

The following path worked for me: classpath:/path/to/resource/in/jar

private static final String FILE_LOCATION = "com/input/file/somefile.txt";
//Method Body
InputStream invalidCharacterInputStream = URLClassLoader.getSystemResourceAsStream(FILE_LOCATION);
Getting this from getSystemResourceAsStream is the best option. By getting the inputstream rather than the file or the URL, works in a JAR file and as stand alone.

It may be a little late but you may use my library KResourceLoader to get a resource from your jar:
File resource = getResource("file.txt")

Inside your ressources folder (java/main/resources) of your jar add your file (we assume that you have added an xml file named imports.xml), after that you inject ResourceLoader if you use spring like bellow
#Autowired
private ResourceLoader resourceLoader;
inside tour function write the bellow code in order to load file:
Resource resource = resourceLoader.getResource("classpath:imports.xml");
try{
File file;
file = resource.getFile();//will load the file
...
}catch(IOException e){e.printStackTrace();}

Maybe this method can be used for quick solution.
public class TestUtility
{
public static File getInternalResource(String relativePath)
{
File resourceFile = null;
URL location = TestUtility.class.getProtectionDomain().getCodeSource().getLocation();
String codeLocation = location.toString();
try{
if (codeLocation.endsWith(".jar"){
//Call from jar
Path path = Paths.get(location.toURI()).resolve("../classes/" + relativePath).normalize();
resourceFile = path.toFile();
}else{
//Call from IDE
resourceFile = new File(TestUtility.class.getClassLoader().getResource(relativePath).getPath());
}
}catch(URISyntaxException ex){
ex.printStackTrace();
}
return resourceFile;
}
}

When in a jar file, the resource is located absolutely in the package hierarchy (not file system hierarchy). So if you have class com.example.Sweet loading a resource named ./default.conf then the resource's name is specified as /com/example/default.conf.
But if it's in a jar then it's not a File ...

This Class function can get absolute file path by relative file path in .jar file.
public class Utility {
public static void main(String[] args) throws Exception {
Utility utility = new Utility();
String absolutePath = utility.getAbsolutePath("./absolute/path/to/file");
}
public String getAbsolutePath(String relativeFilePath) throws IOException {
URL url = this.getClass().getResource(relativeFilePath);
return url.getPath();
}
}
If target file is same directory with Utility.java, your input will be ./file.txt.
When you input only / on getAbsolutePath(), it returns /Users/user/PROJECT_NAME/target/classes/. This means you can select the file like this /com/example/file.txt.