We Keep Coding

How to list folders and files from different paths stored in a json file - Java

I want to list all the folders and files from different paths stored in a json file in order to delete them later with a filter. In the folowing code I list the folders using the FILE class and only one path.
public class Depurador {
public static void main(String[] args) {
File ruta = new File("C:\\Users\\erick\\Desktop\\prueba");
ImprimirFicheros1(ruta, 0);
}
public static void ImprimirFicheros1(File file, int nivelArbol){
for(int i = 0; i < nivelArbol; i++){
System.out.print("-");
}
System.out.println(file.getName());
if(file.isDirectory()){
File[] listaFicheros = file.listFiles();
for(File fichero : listaFicheros){
ImprimirFicheros1(fichero, nivelArbol + 2);
}
}
}
I would like to list all the folders and files from this json file and not only from one path.
{
"Directorios":[
{
"ruta": "C:\\Users\\erick\\Desktop\\prueba",
"periodo": 15,
"archivos":["yyyy-mm-dd_reportesdetalles_1.csv, yyyy-mm-dd_reportesdetalles_2.csv"]
},
{
"ruta": "C:\\Users\\erick\\Desktop\\test",
"periodo": 20,
"archivos":["yyyy-mm-dd_test_1.csv, yyyy-mm-dd_test_2.csv"]
}
]
}

You will need to deserialize the JSON string into object. You can use a library like jackson ObjectMapper
class Directory {
private String ruta;
private int periodo;
private List<String> archivos;
// getters and setters
}
or use a record (java 14+)
public record Directory(String ruta, int periodo, List<String> archivos) {}
public record Input(List<Directory> Directorios) {}
In your main function you can do this now.
// triple quote text block is java 13+
String jsonString = """
{
"Directorios":[
{
"ruta": "C:\\Users\\erick\\Desktop\\prueba",
"periodo": 15,
"archivos":["yyyy-mm-dd_reportesdetalles_1.csv, yyyy-mm-dd_reportesdetalles_2.csv"]
},
{
"ruta": "C:\\Users\\erick\\Desktop\\test",
"periodo": 20,
"archivos":["yyyy-mm-dd_test_1.csv, yyyy-mm-dd_test_2.csv"]
}
]
}
""";
List<Directory> directories = new com.fasterxml.jackson.databind.ObjectMapper().readValue(jsonString, Input.class).getDirectorios();
for (Directory d : directories) {
File ruta = new File(d.getRuta());
ImprimirFicheros1(ruta, 0);
}

Get Random File from Folder - Java [duplicate]

I am looking for a way to get a list of all resource names from a given classpath directory, something like a method List<String> getResourceNames (String directoryName).
For example, given a classpath directory x/y/z containing files a.html, b.html, c.html and a subdirectory d, getResourceNames("x/y/z") should return a List<String> containing the following strings:['a.html', 'b.html', 'c.html', 'd'].
It should work both for resources in filesystem and jars.
I know that I can write a quick snippet with Files, JarFiles and URLs, but I do not want to reinvent the wheel. My question is, given existing publicly available libraries, what is the quickest way to implement getResourceNames? Spring and Apache Commons stacks are both feasible.

Custom Scanner
Implement your own scanner. For example:
(limitations of this solution are mentioned in the comments)
private List<String> getResourceFiles(String path) throws IOException {
List<String> filenames = new ArrayList<>();
try (
InputStream in = getResourceAsStream(path);
BufferedReader br = new BufferedReader(new InputStreamReader(in))) {
String resource;
while ((resource = br.readLine()) != null) {
filenames.add(resource);
}
}
return filenames;
}
private InputStream getResourceAsStream(String resource) {
final InputStream in
= getContextClassLoader().getResourceAsStream(resource);
return in == null ? getClass().getResourceAsStream(resource) : in;
}
private ClassLoader getContextClassLoader() {
return Thread.currentThread().getContextClassLoader();
}
Spring Framework
Use PathMatchingResourcePatternResolver from Spring Framework.
Ronmamo Reflections
The other techniques might be slow at runtime for huge CLASSPATH values. A faster solution is to use ronmamo's Reflections API, which precompiles the search at compile time.

Here is the code
Source: forums.devx.com/showthread.php?t=153784
import java.io.File;
import java.io.IOException;
import java.util.ArrayList;
import java.util.Collection;
import java.util.Enumeration;
import java.util.regex.Pattern;
import java.util.zip.ZipEntry;
import java.util.zip.ZipException;
import java.util.zip.ZipFile;
/**
* list resources available from the classpath # *
*/
public class ResourceList{
/**
* for all elements of java.class.path get a Collection of resources Pattern
* pattern = Pattern.compile(".*"); gets all resources
*
* #param pattern
* the pattern to match
* #return the resources in the order they are found
*/
public static Collection<String> getResources(
final Pattern pattern){
final ArrayList<String> retval = new ArrayList<String>();
final String classPath = System.getProperty("java.class.path", ".");
final String[] classPathElements = classPath.split(System.getProperty("path.separator"));
for(final String element : classPathElements){
retval.addAll(getResources(element, pattern));
}
return retval;
}
private static Collection<String> getResources(
final String element,
final Pattern pattern){
final ArrayList<String> retval = new ArrayList<String>();
final File file = new File(element);
if(file.isDirectory()){
retval.addAll(getResourcesFromDirectory(file, pattern));
} else{
retval.addAll(getResourcesFromJarFile(file, pattern));
}
return retval;
}
private static Collection<String> getResourcesFromJarFile(
final File file,
final Pattern pattern){
final ArrayList<String> retval = new ArrayList<String>();
ZipFile zf;
try{
zf = new ZipFile(file);
} catch(final ZipException e){
throw new Error(e);
} catch(final IOException e){
throw new Error(e);
}
final Enumeration e = zf.entries();
while(e.hasMoreElements()){
final ZipEntry ze = (ZipEntry) e.nextElement();
final String fileName = ze.getName();
final boolean accept = pattern.matcher(fileName).matches();
if(accept){
retval.add(fileName);
}
}
try{
zf.close();
} catch(final IOException e1){
throw new Error(e1);
}
return retval;
}
private static Collection<String> getResourcesFromDirectory(
final File directory,
final Pattern pattern){
final ArrayList<String> retval = new ArrayList<String>();
final File[] fileList = directory.listFiles();
for(final File file : fileList){
if(file.isDirectory()){
retval.addAll(getResourcesFromDirectory(file, pattern));
} else{
try{
final String fileName = file.getCanonicalPath();
final boolean accept = pattern.matcher(fileName).matches();
if(accept){
retval.add(fileName);
}
} catch(final IOException e){
throw new Error(e);
}
}
}
return retval;
}
/**
* list the resources that match args[0]
*
* #param args
* args[0] is the pattern to match, or list all resources if
* there are no args
*/
public static void main(final String[] args){
Pattern pattern;
if(args.length < 1){
pattern = Pattern.compile(".*");
} else{
pattern = Pattern.compile(args[0]);
}
final Collection<String> list = ResourceList.getResources(pattern);
for(final String name : list){
System.out.println(name);
}
}
}
If you are using Spring Have a look at PathMatchingResourcePatternResolver

Using Reflections
Get everything on the classpath:
Reflections reflections = new Reflections(null, new ResourcesScanner());
Set<String> resourceList = reflections.getResources(x -> true);
Another example - get all files with extension .csv from some.package:
Reflections reflections = new Reflections("some.package", new ResourcesScanner());
Set<String> resourceList = reflections.getResources(Pattern.compile(".*\\.csv"));

So in terms of the PathMatchingResourcePatternResolver this is what is needed in the code:
#Autowired
ResourcePatternResolver resourceResolver;
public void getResources() {
resourceResolver.getResources("classpath:config/*.xml");
}

If you use apache commonsIO you can use for the filesystem (optionally with extension filter):
Collection<File> files = FileUtils.listFiles(new File("directory/"), null, false);
and for resources/classpath:
List<String> files = IOUtils.readLines(MyClass.class.getClassLoader().getResourceAsStream("directory/"), Charsets.UTF_8);
If you don't know if "directoy/" is in the filesystem or in resources you may add a
if (new File("directory/").isDirectory())
or
if (MyClass.class.getClassLoader().getResource("directory/") != null)
before the calls and use both in combination...

The most robust mechanism for listing all resources in the classpath is currently to use this pattern with ClassGraph, because it handles the widest possible array of classpath specification mechanisms, including the new JPMS module system. (I am the author of ClassGraph.)
List<String> resourceNames;
try (ScanResult scanResult = new ClassGraph().acceptPaths("x/y/z").scan()) {
resourceNames = scanResult.getAllResources().getNames();
}

The Spring framework's PathMatchingResourcePatternResolver is really awesome for these things:
private Resource[] getXMLResources() throws IOException
{
ClassLoader classLoader = MethodHandles.lookup().getClass().getClassLoader();
PathMatchingResourcePatternResolver resolver = new PathMatchingResourcePatternResolver(classLoader);
return resolver.getResources("classpath:x/y/z/*.xml");
}
Maven dependency:
<dependency>
<groupId>org.springframework</groupId>
<artifactId>spring-core</artifactId>
<version>LATEST</version>
</dependency>

This should work (if spring is not an option):
public static List<String> getFilenamesForDirnameFromCP(String directoryName) throws URISyntaxException, UnsupportedEncodingException, IOException {
List<String> filenames = new ArrayList<>();
URL url = Thread.currentThread().getContextClassLoader().getResource(directoryName);
if (url != null) {
if (url.getProtocol().equals("file")) {
File file = Paths.get(url.toURI()).toFile();
if (file != null) {
File[] files = file.listFiles();
if (files != null) {
for (File filename : files) {
filenames.add(filename.toString());
}
}
}
} else if (url.getProtocol().equals("jar")) {
String dirname = directoryName + "/";
String path = url.getPath();
String jarPath = path.substring(5, path.indexOf("!"));
try (JarFile jar = new JarFile(URLDecoder.decode(jarPath, StandardCharsets.UTF_8.name()))) {
Enumeration<JarEntry> entries = jar.entries();
while (entries.hasMoreElements()) {
JarEntry entry = entries.nextElement();
String name = entry.getName();
if (name.startsWith(dirname) && !dirname.equals(name)) {
URL resource = Thread.currentThread().getContextClassLoader().getResource(name);
filenames.add(resource.toString());
}
}
}
}
}
return filenames;
}

My way, no Spring, used during a unit test:
URI uri = TestClass.class.getResource("/resources").toURI();
Path myPath = Paths.get(uri);
Stream<Path> walk = Files.walk(myPath, 1);
for (Iterator<Path> it = walk.iterator(); it.hasNext(); ) {
Path filename = it.next();
System.out.println(filename);
}

With Spring it's easy. Be it a file, or folder, or even multiple files, there are chances, you can do it via injection.
This example demonstrates the injection of multiple files located in x/y/z folder.
import org.springframework.beans.factory.annotation.Value;
import org.springframework.core.io.Resource;
import org.springframework.stereotype.Service;
#Service
public class StackoverflowService {
#Value("classpath:x/y/z/*")
private Resource[] resources;
public List<String> getResourceNames() {
return Arrays.stream(resources)
.map(Resource::getFilename)
.collect(Collectors.toList());
}
}
It does work for resources in the filesystem as well as in JARs.

Used a combination of Rob's response.
final String resourceDir = "resourceDirectory/";
List<String> files = IOUtils.readLines(Thread.currentThread().getClass().getClassLoader().getResourceAsStream(resourceDir), Charsets.UTF_8);
for (String f : files) {
String data = IOUtils.toString(Thread.currentThread().getClass().getClassLoader().getResourceAsStream(resourceDir + f));
// ... process data
}

I think you can leverage the [Zip File System Provider][1] to achieve this. When using FileSystems.newFileSystem it looks like you can treat the objects in that ZIP as a "regular" file.
In the linked documentation above:
Specify the configuration options for the zip file system in the java.util.Map object passed to the FileSystems.newFileSystem method. See the [Zip File System Properties][2] topic for information about the provider-specific configuration properties for the zip file system.
Once you have an instance of a zip file system, you can invoke the methods of the [java.nio.file.FileSystem][3] and [java.nio.file.Path][4] classes to perform operations such as copying, moving, and renaming files, as well as modifying file attributes.
The documentation for the jdk.zipfs module in [Java 11 states][5]:
The zip file system provider treats a zip or JAR file as a file system and provides the ability to manipulate the contents of the file. The zip file system provider can be created by [FileSystems.newFileSystem][6] if installed.
Here is a contrived example I did using your example resources. Note that a .zip is a .jar, but you could adapt your code to instead use classpath resources:
Setup
cd /tmp
mkdir -p x/y/z
touch x/y/z/{a,b,c}.html
echo 'hello world' > x/y/z/d
zip -r example.zip x
Java
import java.io.IOException;
import java.net.URI;
import java.nio.file.FileSystem;
import java.nio.file.FileSystems;
import java.nio.file.Files;
import java.util.Collections;
import java.util.stream.Collectors;
public class MkobitZipRead {
public static void main(String[] args) throws IOException {
final URI uri = URI.create("jar:file:/tmp/example.zip");
try (
final FileSystem zipfs = FileSystems.newFileSystem(uri, Collections.emptyMap());
) {
Files.walk(zipfs.getPath("/")).forEach(path -> System.out.println("Files in zip:" + path));
System.out.println("-----");
final String manifest = Files.readAllLines(
zipfs.getPath("x", "y", "z").resolve("d")
).stream().collect(Collectors.joining(System.lineSeparator()));
System.out.println(manifest);
}
}
}
Output
Files in zip:/
Files in zip:/x/
Files in zip:/x/y/
Files in zip:/x/y/z/
Files in zip:/x/y/z/c.html
Files in zip:/x/y/z/b.html
Files in zip:/x/y/z/a.html
Files in zip:/x/y/z/d
-----
hello world

Neither of answers worked for me even though I had my resources put in resources folders and followed the above answers. What did make a trick was:
#Value("file:*/**/resources/**/schema/*.json")
private Resource[] resources;

Expanding on Luke Hutchinsons answer above, using his ClassGraph library, I was able to easily get a list of all files in a Resource folder with almost no effort at all.
Let's say that in your resource folder, you have a folder called MyImages. This is how easy it is to get a URL list of all the files in that folder:
import io.github.classgraph.ClassGraph;
import io.github.classgraph.ResourceList;
import io.github.classgraph.ScanResult;
public static LinkedList<URL> getURLList (String folder) {
LinkedList<URL> urlList = new LinkedList<>();
ScanResult scanResult = new ClassGraph().enableAllInfo().scan();
ResourceList resources = scanResult.getAllResources();
for (URL url : resources.getURLs()) {
if (url.toString().contains(folder)) {
urlList.addLast(url);
}
}
return urlList;
}
Then you simply do this:
LinkedList<URL> myURLFileList = getURLList("MyImages");
The URLs can then be loaded into streams or use Apache's FileUtils to copy the files somewhere else like this:
String outPath = "/My/Output/Path";
for(URL url : myURLFileList) {
FileUtils.copyURLToFile(url, new File(outPath, url.getFile()));
}
I think ClassGraph is a pretty slick library for making tasks like this very simple and easy to comprehend.

Based on #rob 's information above, I created the implementation which I am releasing to the public domain:
private static List<String> getClasspathEntriesByPath(String path) throws IOException {
InputStream is = Main.class.getClassLoader().getResourceAsStream(path);
StringBuilder sb = new StringBuilder();
while (is.available()>0) {
byte[] buffer = new byte[1024];
sb.append(new String(buffer, Charset.defaultCharset()));
}
return Arrays
.asList(sb.toString().split("\n")) // Convert StringBuilder to individual lines
.stream() // Stream the list
.filter(line -> line.trim().length()>0) // Filter out empty lines
.collect(Collectors.toList()); // Collect remaining lines into a List again
}
While I would not have expected getResourcesAsStream to work like that on a directory, it really does and it works well.

SWT file drop: identify folders

Code
I'm trying to add drop functionality to my program so users can drag and drop files from Windows. Therefore I implemented the SWT DropTargetListener interface and added a SWT DropTarget to my SWT control.
public class MyFileDrop implements DropTargetListener {
public MyFileDrop(final Control control) {
addDropTarget(control);
}
private void addDropTarget(final Control control) {
final DropTarget dropTarget = new DropTarget(control, DND.DROP_COPY | DND.DROP_DEFAULT);
dropTarget.setTransfer(new Transfer[] { FileTransfer.getInstance() });
dropTarget.addDropListener(this);
}
#Override
public void drop(final DropTargetEvent event) {
String[] fileNames = (String[]) event.data;
for (final String fileName : fileNames) {
System.out.println(fileName);
}
}
// other DropTargetListener methods [...]
}
If I drag and drop a folder from Windows Explorer to my control, the folder name gets printed out. But all files inside that dropped folder aren't listed.
Example
folder1
- subfile1.png
- subfile2.exe
file1.png
file2.exe
If I select folder1, file1 and file2 and drag and drop it to my program, this is the output:
path/to/folder1
path/to/file1.png
path/to/file2.exe
Questions
Why aren't the files inside the folder dropped to the program? How can I archive that these files are also dropped?
How can I set the files apart from the folder?
Regards winklerrr

Code
So this is my solution
#Override
public void drop(DropTargetEvent event) {
String[] fileNames = (String[]) event.data;
for (String fileName : fileNames) {
final File file = new File(fileName);
if (file.isDirectory()) {
Collection<File> subFiles = FileUtils.listFiles(file, TrueFileFilter.INSTANCE, TrueFileFilter.INSTANCE);
for (File subFile : subFiles) {
System.out.println(subFile.getAbsolutePath());
}
} else {
System.out.println(file.getAbsolutePath());
}
}
}
I used the FileUtils from the commons-io package.
Example
folder1
- subfile1.png
- subfile2.exe
file1.png
file2.exe
With the new code, this is now the output:
path/to/folder1/subfile1.png
path/to/folder1/subfile2.exe
path/to/file1.png
path/to/file2.exe
Answers
How can I archive that these files are also dropped?
FileUtils.listFiles(File, IOFileFilter, IOFileFilter)
How can I set the files apart from the folder?
File.isDirectory()

How to get all files from directory and sub directory with a given pattern

I have a directory containing sub directories each one of this directories has a file called
*.properties
I want to search for these files with java
Thanks

See the java tutorial Walking the File Tree:
Do you need to create an application that will recursively visit all the files in a file tree? Perhaps you need to delete every .class file in a tree, or find every file that hasn't been accessed in the last year. You can do so with the FileVisitor interface.
In particular see the Finding Files example:
import java.io.*;
import java.nio.file.*;
import java.nio.file.attribute.*;
import static java.nio.file.FileVisitResult.*;
import static java.nio.file.FileVisitOption.*;
import java.util.*;
/**
* Sample code that finds files that
* match the specified glob pattern.
* For more information on what
* constitutes a glob pattern, see
* http://docs.oracle.com/javase/javatutorials/tutorial/essential/io/fileOps.html#glob
*
* The file or directories that match
* the pattern are printed to
* standard out. The number of
* matches is also printed.
*
* When executing this application,
* you must put the glob pattern
* in quotes, so the shell will not
* expand any wild cards:
* java Find . -name "*.java"
*/
public class Find {
/**
* A {#code FileVisitor} that finds
* all files that match the
* specified pattern.
*/
public static class Finder
extends SimpleFileVisitor<Path> {
private final PathMatcher matcher;
private int numMatches = 0;
Finder(String pattern) {
matcher = FileSystems.getDefault()
.getPathMatcher("glob:" + pattern);
}
// Compares the glob pattern against
// the file or directory name.
void find(Path file) {
Path name = file.getFileName();
if (name != null && matcher.matches(name)) {
numMatches++;
System.out.println(file);
}
}
// Prints the total number of
// matches to standard out.
void done() {
System.out.println("Matched: "
+ numMatches);
}
// Invoke the pattern matching
// method on each file.
#Override
public FileVisitResult visitFile(Path file,
BasicFileAttributes attrs) {
find(file);
return CONTINUE;
}
// Invoke the pattern matching
// method on each directory.
#Override
public FileVisitResult preVisitDirectory(Path dir,
BasicFileAttributes attrs) {
find(dir);
return CONTINUE;
}
#Override
public FileVisitResult visitFileFailed(Path file,
IOException exc) {
System.err.println(exc);
return CONTINUE;
}
}
static void usage() {
System.err.println("java Find <path>" +
" -name \"<glob_pattern>\"");
System.exit(-1);
}
public static void main(String[] args)
throws IOException {
if (args.length < 3 || !args[1].equals("-name"))
usage();
Path startingDir = Paths.get(args[0]);
String pattern = args[2];
Finder finder = new Finder(pattern);
Files.walkFileTree(startingDir, finder);
finder.done();
}
}

You can recursively call this method
File dir = new File("C:/");
File [] files = dir.listFiles(new FilenameFilter() {
#Override
public boolean accept(File dir, String name) {
return name.endsWith(".properties");
}
});
Alternately you use Java 7 Files.walkTree method and filter as described here.

In addition to other answers:
Guava (since v15):
for (File f : Files.fileTreeTraverser().preOrderTraversal(rootDir)) {
// filter and process
}
Commons IO:
for (File f : FileUtils.listFiles(rootDir, fileFilter, dirFilter)) {
// process
}
Java 7:
Files.walkFileTree(rootDir, fileVisitor);

How to list the files inside a JAR file?

I have this code which reads all the files from a directory.
File textFolder = new File("text_directory");
File [] texFiles = textFolder.listFiles( new FileFilter() {
public boolean accept( File file ) {
return file.getName().endsWith(".txt");
}
});
It works great. It fills the array with all the files that end with ".txt" from directory "text_directory".
How can I read the contents of a directory in a similar fashion within a JAR file?
So what I really want to do is, to list all the images inside my JAR file, so I can load them with:
ImageIO.read(this.getClass().getResource("CompanyLogo.png"));
(That one works because the "CompanyLogo" is "hardcoded" but the number of images inside the JAR file could be from 10 to 200 variable length.)
EDIT
So I guess my main problem would be: How to know the name of the JAR file where my main class lives?
Granted I could read it using java.util.Zip.
My Structure is like this:
They are like:
my.jar!/Main.class
my.jar!/Aux.class
my.jar!/Other.class
my.jar!/images/image01.png
my.jar!/images/image02a.png
my.jar!/images/imwge034.png
my.jar!/images/imagAe01q.png
my.jar!/META-INF/manifest
Right now I'm able to load for instance "images/image01.png" using:
ImageIO.read(this.getClass().getResource("images/image01.png));
But only because I know the file name, for the rest I have to load them dynamically.

CodeSource src = MyClass.class.getProtectionDomain().getCodeSource();
if (src != null) {
URL jar = src.getLocation();
ZipInputStream zip = new ZipInputStream(jar.openStream());
while(true) {
ZipEntry e = zip.getNextEntry();
if (e == null)
break;
String name = e.getName();
if (name.startsWith("path/to/your/dir/")) {
/* Do something with this entry. */
...
}
}
}
else {
/* Fail... */
}
Note that in Java 7, you can create a FileSystem from the JAR (zip) file, and then use NIO's directory walking and filtering mechanisms to search through it. This would make it easier to write code that handles JARs and "exploded" directories.

Code that works for both IDE's and .jar files:
import java.io.*;
import java.net.*;
import java.nio.file.*;
import java.util.*;
import java.util.stream.*;
public class ResourceWalker {
public static void main(String[] args) throws URISyntaxException, IOException {
URI uri = ResourceWalker.class.getResource("/resources").toURI();
Path myPath;
if (uri.getScheme().equals("jar")) {
FileSystem fileSystem = FileSystems.newFileSystem(uri, Collections.<String, Object>emptyMap());
myPath = fileSystem.getPath("/resources");
} else {
myPath = Paths.get(uri);
}
Stream<Path> walk = Files.walk(myPath, 1);
for (Iterator<Path> it = walk.iterator(); it.hasNext();){
System.out.println(it.next());
}
}
}

erickson's answer worked perfectly:
Here's the working code.
CodeSource src = MyClass.class.getProtectionDomain().getCodeSource();
List<String> list = new ArrayList<String>();
if( src != null ) {
URL jar = src.getLocation();
ZipInputStream zip = new ZipInputStream( jar.openStream());
ZipEntry ze = null;
while( ( ze = zip.getNextEntry() ) != null ) {
String entryName = ze.getName();
if( entryName.startsWith("images") && entryName.endsWith(".png") ) {
list.add( entryName );
}
}
}
webimages = list.toArray( new String[ list.size() ] );
And I have just modify my load method from this:
File[] webimages = ...
BufferedImage image = ImageIO.read(this.getClass().getResource(webimages[nextIndex].getName() ));
To this:
String [] webimages = ...
BufferedImage image = ImageIO.read(this.getClass().getResource(webimages[nextIndex]));

I would like to expand on acheron55's answer, since it is a very non-safe solution, for several reasons:
It doesn't close the FileSystem object.
It doesn't check if the FileSystem object already exists.
It isn't thread-safe.
This is somewhat a safer solution:
private static ConcurrentMap<String, Object> locks = new ConcurrentHashMap<>();
public void walk(String path) throws Exception {
URI uri = getClass().getResource(path).toURI();
if ("jar".equals(uri.getScheme()) {
safeWalkJar(path, uri);
} else {
Files.walk(Paths.get(path));
}
}
private void safeWalkJar(String path, URI uri) throws Exception {
synchronized (getLock(uri)) {
// this'll close the FileSystem object at the end
try (FileSystem fs = getFileSystem(uri)) {
Files.walk(fs.getPath(path));
}
}
}
private Object getLock(URI uri) {
String fileName = parseFileName(uri);
locks.computeIfAbsent(fileName, s -> new Object());
return locks.get(fileName);
}
private String parseFileName(URI uri) {
String schemeSpecificPart = uri.getSchemeSpecificPart();
return schemeSpecificPart.substring(0, schemeSpecificPart.indexOf("!"));
}
private FileSystem getFileSystem(URI uri) throws IOException {
try {
return FileSystems.getFileSystem(uri);
} catch (FileSystemNotFoundException e) {
return FileSystems.newFileSystem(uri, Collections.<String, String>emptyMap());
}
}
There's no real need to synchronize over the file name; one could simply synchronize on the same object every time (or make the method synchronized), it's purely an optimization.
I would say that this is still a problematic solution, since there might be other parts in the code that use the FileSystem interface over the same files, and it could interfere with them (even in a single threaded application).
Also, it doesn't check for nulls (for instance, on getClass().getResource().
This particular Java NIO interface is kind of horrible, since it introduces a global/singleton non thread-safe resource, and its documentation is extremely vague (a lot of unknowns due to provider specific implementations). Results may vary for other FileSystem providers (not JAR). Maybe there's a good reason for it being that way; I don't know, I haven't researched the implementations.

So I guess my main problem would be, how to know the name of the jar where my main class lives.
Assuming that your project is packed in a Jar (not necessarily true!), you can use ClassLoader.getResource() or findResource() with the class name (followed by .class) to get the jar that contains a given class. You'll have to parse the jar name from the URL that gets returned (not that tough), which I will leave as an exercise for the reader :-)
Be sure to test for the case where the class is not part of a jar.

I've ported acheron55's answer to Java 7 and closed the FileSystem object. This code works in IDE's, in jar files and in a jar inside a war on Tomcat 7; but note that it does not work in a jar inside a war on JBoss 7 (it gives FileSystemNotFoundException: Provider "vfs" not installed, see also this post). Furthermore, like the original code, it is not thread safe, as suggested by errr. For these reasons I have abandoned this solution; however, if you can accept these issues, here is my ready-made code:
import java.io.IOException;
import java.net.*;
import java.nio.file.*;
import java.nio.file.attribute.BasicFileAttributes;
import java.util.Collections;
public class ResourceWalker {
public static void main(String[] args) throws URISyntaxException, IOException {
URI uri = ResourceWalker.class.getResource("/resources").toURI();
System.out.println("Starting from: " + uri);
try (FileSystem fileSystem = (uri.getScheme().equals("jar") ? FileSystems.newFileSystem(uri, Collections.<String, Object>emptyMap()) : null)) {
Path myPath = Paths.get(uri);
Files.walkFileTree(myPath, new SimpleFileVisitor<Path>() {
#Override
public FileVisitResult visitFile(Path file, BasicFileAttributes attrs) throws IOException {
System.out.println(file);
return FileVisitResult.CONTINUE;
}
});
}
}
}

Here is an example of using Reflections library to recursively scan classpath by regex name pattern augmented with a couple of Guava perks to to fetch resources contents:
Reflections reflections = new Reflections("com.example.package", new ResourcesScanner());
Set<String> paths = reflections.getResources(Pattern.compile(".*\\.template$"));
Map<String, String> templates = new LinkedHashMap<>();
for (String path : paths) {
log.info("Found " + path);
String templateName = Files.getNameWithoutExtension(path);
URL resource = getClass().getClassLoader().getResource(path);
String text = Resources.toString(resource, StandardCharsets.UTF_8);
templates.put(templateName, text);
}
This works with both jars and exploded classes.

Here's a method I wrote for a "run all JUnits under a package". You should be able to adapt it to your needs.
private static void findClassesInJar(List<String> classFiles, String path) throws IOException {
final String[] parts = path.split("\\Q.jar\\\\E");
if (parts.length == 2) {
String jarFilename = parts[0] + ".jar";
String relativePath = parts[1].replace(File.separatorChar, '/');
JarFile jarFile = new JarFile(jarFilename);
final Enumeration<JarEntry> entries = jarFile.entries();
while (entries.hasMoreElements()) {
final JarEntry entry = entries.nextElement();
final String entryName = entry.getName();
if (entryName.startsWith(relativePath)) {
classFiles.add(entryName.replace('/', File.separatorChar));
}
}
}
}
Edit:
Ah, in that case, you might want this snippet as well (same use case :) )
private static File findClassesDir(Class<?> clazz) {
try {
String path = clazz.getProtectionDomain().getCodeSource().getLocation().getFile();
final String codeSourcePath = URLDecoder.decode(path, "UTF-8");
final String thisClassPath = new File(codeSourcePath, clazz.getPackage().getName().repalce('.', File.separatorChar));
} catch (UnsupportedEncodingException e) {
throw new AssertionError("impossible", e);
}
}

Just to mention that if you are already using Spring, you can take advantage of the PathMatchingResourcePatternResolver.
For instance to get all the PNG files from a images folder in resources
ClassLoader cl = this.getClass().getClassLoader();
ResourcePatternResolver resolver = new PathMatchingResourcePatternResolver(cl);
Resource[] resources = resolver.getResources("images/*.png");
for (Resource r: resources){
logger.info(r.getFilename());
// From your example
// ImageIO.read(cl.getResource("images/" + r.getFilename()));
}

A jar file is just a zip file with a structured manifest. You can open the jar file with the usual java zip tools and scan the file contents that way, inflate streams, etc. Then use that in a getResourceAsStream call, and it should be all hunky dory.
EDIT / after clarification
It took me a minute to remember all the bits and pieces and I'm sure there are cleaner ways to do it, but I wanted to see that I wasn't crazy. In my project image.jpg is a file in some part of the main jar file. I get the class loader of the main class (SomeClass is the entry point) and use it to discover the image.jpg resource. Then some stream magic to get it into this ImageInputStream thing and everything is fine.
InputStream inputStream = SomeClass.class.getClassLoader().getResourceAsStream("image.jpg");
JPEGImageReaderSpi imageReaderSpi = new JPEGImageReaderSpi();
ImageReader ir = imageReaderSpi.createReaderInstance();
ImageInputStream iis = new MemoryCacheImageInputStream(inputStream);
ir.setInput(iis);
....
ir.read(0); //will hand us a buffered image

Given an actual JAR file, you can list the contents using JarFile.entries(). You will need to know the location of the JAR file though - you can't just ask the classloader to list everything it could get at.
You should be able to work out the location of the JAR file based on the URL returned from ThisClassName.class.getResource("ThisClassName.class"), but it may be a tiny bit fiddly.

Some time ago I made a function that gets classess from inside JAR:
public static Class[] getClasses(String packageName)
throws ClassNotFoundException{
ArrayList<Class> classes = new ArrayList<Class> ();
packageName = packageName.replaceAll("\\." , "/");
File f = new File(jarName);
if(f.exists()){
try{
JarInputStream jarFile = new JarInputStream(
new FileInputStream (jarName));
JarEntry jarEntry;
while(true) {
jarEntry=jarFile.getNextJarEntry ();
if(jarEntry == null){
break;
}
if((jarEntry.getName ().startsWith (packageName)) &&
(jarEntry.getName ().endsWith (".class")) ) {
classes.add(Class.forName(jarEntry.getName().
replaceAll("/", "\\.").
substring(0, jarEntry.getName().length() - 6)));
}
}
}
catch( Exception e){
e.printStackTrace ();
}
Class[] classesA = new Class[classes.size()];
classes.toArray(classesA);
return classesA;
}else
return null;
}

public static ArrayList<String> listItems(String path) throws Exception{
InputStream in = ClassLoader.getSystemClassLoader().getResourceAsStream(path);
byte[] b = new byte[in.available()];
in.read(b);
String data = new String(b);
String[] s = data.split("\n");
List<String> a = Arrays.asList(s);
ArrayList<String> m = new ArrayList<>(a);
return m;
}

There are two very useful utilities both called JarScan:
www.inetfeedback.com/jarscan
jarscan.dev.java.net
See also this question: JarScan, scan all JAR files in all subfolders for specific class

The most robust mechanism for listing all resources in the classpath is currently to use this pattern with ClassGraph, because it handles the widest possible array of classpath specification mechanisms, including the new JPMS module system. (I am the author of ClassGraph.)
How to know the name of the JAR file where my main class lives?
URI mainClasspathElementURI;
try (ScanResult scanResult = new ClassGraph().whitelistPackages("x.y.z")
.enableClassInfo().scan()) {
mainClasspathElementURI =
scanResult.getClassInfo("x.y.z.MainClass").getClasspathElementURI();
}
How can I read the contents of a directory in a similar fashion within a JAR file?
List<String> classpathElementResourcePaths;
try (ScanResult scanResult = new ClassGraph().overrideClasspath(mainClasspathElementURI)
.scan()) {
classpathElementResourcePaths = scanResult.getAllResources().getPaths();
}
There are lots of other ways to deal with resources too.

One more for the road that's a bit more flexible for matching specific filenames because it uses wildcard globbing. In a functional style this could resemble:
import java.io.IOException;
import java.net.URISyntaxException;
import java.nio.file.FileSystem;
import java.nio.file.Files;
import java.nio.file.Path;
import java.nio.file.Paths;
import java.util.function.Consumer;
import static java.nio.file.FileSystems.getDefault;
import static java.nio.file.FileSystems.newFileSystem;
import static java.util.Collections.emptyMap;
/**
* Responsible for finding file resources.
*/
public class ResourceWalker {
/**
* Globbing pattern to match font names.
*/
public static final String GLOB_FONTS = "**.{ttf,otf}";
/**
* #param directory The root directory to scan for files matching the glob.
* #param c The consumer function to call for each matching path
* found.
* #throws URISyntaxException Could not convert the resource to a URI.
* #throws IOException Could not walk the tree.
*/
public static void walk(
final String directory, final String glob, final Consumer<Path> c )
throws URISyntaxException, IOException {
final var resource = ResourceWalker.class.getResource( directory );
final var matcher = getDefault().getPathMatcher( "glob:" + glob );
if( resource != null ) {
final var uri = resource.toURI();
final Path path;
FileSystem fs = null;
if( "jar".equals( uri.getScheme() ) ) {
fs = newFileSystem( uri, emptyMap() );
path = fs.getPath( directory );
}
else {
path = Paths.get( uri );
}
try( final var walk = Files.walk( path, 10 ) ) {
for( final var it = walk.iterator(); it.hasNext(); ) {
final Path p = it.next();
if( matcher.matches( p ) ) {
c.accept( p );
}
}
} finally {
if( fs != null ) { fs.close(); }
}
}
}
}
Consider parameterizing the file extensions, left an exercise for the reader.
Be careful with Files.walk. According to the documentation:
This method must be used within a try-with-resources statement or similar control structure to ensure that the stream's open directories are closed promptly after the stream's operations have completed.
Likewise, newFileSystem must be closed, but not before the walker has had a chance to visit the file system paths.

Just a different way of listing/reading files from a jar URL and it does it recursively for nested jars
https://gist.github.com/trung/2cd90faab7f75b3bcbaa
URL urlResource = Thead.currentThread().getContextClassLoader().getResource("foo");
JarReader.read(urlResource, new InputStreamCallback() {
#Override
public void onFile(String name, InputStream is) throws IOException {
// got file name and content stream
}
});