[java] Get a list of resources from classpath directory

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.

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...

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

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>

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> fileNames = reflections.getResources(Pattern.compile(".*\\.csv"));

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.

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().whitelistPaths("x/y/z").scan()) {
    resourceNames = scanResult.getAllResources().getNames();
}

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;

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");
}

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

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;
}