How to list the files inside a JAR file?

问题:

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.

回答1:

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.

回答2:

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());
        }
    }
}

回答3:

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]));

回答4:

I would like to expand on acheron55's answer, since it is a very non-safe solution, for several reasons:

1. It doesn't close the FileSystem object.
2. It doesn't check if the FileSystem object already exists.
3. 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.

回答5:

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.

回答6:

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);
    }
}

回答7:

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.

回答8:

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

回答9:

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.

回答10:

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;
}

回答11:

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;
                }
            });
        }
    }
}

回答12:

There are two very useful utilities both called JarScan:

1. www.inetfeedback.com/jarscan

2. jarscan.dev.java.net

See also this question: JarScan, scan all JAR files in all subfolders for specific class

回答13:

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;
}

回答14:

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 
    }
});

回答15:

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.