Get Random File from Folder - Java [duplicate] - java
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.
Related
Can't get list of files using PathMatcher and regex
I'm using PathMatcher and SimpleFileVisitor to iterate over directory and find all the files that start only with a certain prefix. However, I can't get any files although there are some files that match my preference. Example of file required: Prefix_some_text.csv Here is the Main code that invokes the call for SimpleFileVisitor class, and it uses regex pattern with the prefix and suppose to find all files starting with the certain pattern: String directoryAsString = "C:/Users"; String pattern = "Prefix"; SearchFileByWildcard sfbw = new SearchFileByWildcard(); try { List<String> actual = sfbw.searchWithWc(Paths.get(directoryAsString),pattern); } catch (IOException e) { e.printStackTrace(); } The implementation of SearchFileByWildcard class that uses SimpleFileVisitor : static class SearchFileByWildcard { List<String> matchesList = new ArrayList<String>(); List<String> searchWithWc(Path rootDir, String pattern) throws IOException { matchesList.clear(); FileVisitor<Path> matcherVisitor = new SimpleFileVisitor<Path>() { #Override public FileVisitResult visitFile(Path file, BasicFileAttributes attribs) throws IOException { FileSystem fs = FileSystems.getDefault(); PathMatcher matcher = fs.getPathMatcher("regex:" + pattern); Path name = file.getFileName(); //takes the filename from the full path if (matcher.matches(name)) { matchesList.add(name.toString()); } return FileVisitResult.CONTINUE; } }; Files.walkFileTree(rootDir, matcherVisitor); return matchesList; } } I'm debating whether to use glob instead of regex? Or maybe something with my regex is flawed.
It seems like the pattern is wrong. It matches only files named "Prefix". Try to change it in String pattern = "Prefix.*";. Otherwise you can scan for files which name starts by the string "Prefix". String name = file.getFileName().toString(); if (name.startsWith(pattern)) { matchesList.add(name); }
detect main inside a jar using java code.
I am trying to detect which class inside a jar contains main or a supplied method name (if possible). At the moment I have the following code public static void getFromJars(String pathToAppJar) throws IOException{ FileInputStream jar = new FileInputStream(pathToAppJar); ZipInputStream zipSteam = new ZipInputStream(jar); ZipEntry ze; while ((ze = zipSteam.getNextEntry()) != null) { System.out.println(ze.toString()); } zipSteam.close(); } This will allow me to get packages and classes under these packages, but I do not know if it is possible to even get methods inside classes. Further, I do not know if this approach is even good for a case of several pkgs inside the jar, since each package can have a class with main in it. I would appreciate any ideas.
Thanks to fvu's comments, I ended up with the following code. public static void getFromJars(String pathToAppJar) throws IOException, ClassNotFoundException { FileInputStream jar = new FileInputStream(pathToAppJar); ZipInputStream zipSteam = new ZipInputStream(jar); ZipEntry ze; URL[] urls = { new URL("jar:file:" + pathToAppJar+"!/") }; URLClassLoader cl = URLClassLoader.newInstance(urls); while ((ze = zipSteam.getNextEntry()) != null) { // Is this a class? if (ze.getName().endsWith(".class")) { // Relative path of file into the jar. String className = ze.getName(); // Complete class name className = className.replace(".class", "").replace("/", "."); Class<?> klazz = cl.loadClass(className); Method[] methodsArray = klazz.getMethods(); } } zipSteam.close(); } I removed the code that uses the methods found, since it is not important for this answer
Detect file type based on content
Tried the following: import java.io.IOException; import java.nio.file.Path; import java.nio.file.Paths; import java.nio.file.spi.FileTypeDetector; import org.apache.tika.Tika; import org.apache.tika.mime.MimeTypes; /** * * #author kiriti.k */ public class TikaFileTypeDetector { private final Tika tika = new Tika(); public TikaFileTypeDetector() { super(); } public String probeContentType(Path path) throws IOException { // Try to detect based on the file name only for efficiency String fileNameDetect = tika.detect(path.toString()); if (!fileNameDetect.equals(MimeTypes.OCTET_STREAM)) { return fileNameDetect; } // Then check the file content if necessary String fileContentDetect = tika.detect(path.toFile()); if (!fileContentDetect.equals(MimeTypes.OCTET_STREAM)) { return fileContentDetect; } // Specification says to return null if we could not // conclusively determine the file type return null; } public static void main(String[] args) throws IOException { Tika tika = new Tika(); // expects file path as the program argument if (args.length != 1) { printUsage(); return; } Path path = Paths.get(args[0]); TikaFileTypeDetector detector = new TikaFileTypeDetector(); // Analyse the file - first based on file name for efficiency. // If cannot determine based on name and then analyse content String contentType = detector.probeContentType(path); System.out.println("File is of type - " + contentType); } public static void printUsage() { System.out.print("Usage: java -classpath ... " + TikaFileTypeDetector.class.getName() + " "); } } The above program is checking based on file extension only. How do I make it to check content type also(mime) and then determine the type. I am using tika-app-1.8.jar in netbean 8.0.2. What am I missing?
The code checks the file extension first and returns the MIME type based on that, if it finds a result. If you want it to check the content first, just switch the two statements: public String probeContentType(Path path) throws IOException { // Check contents first String fileContentDetect = tika.detect(path.toFile()); if (!fileContentDetect.equals(MimeTypes.OCTET_STREAM)) { return fileContentDetect; } // Try file name only if content search was not successful String fileNameDetect = tika.detect(path.toString()); if (!fileNameDetect.equals(MimeTypes.OCTET_STREAM)) { return fileNameDetect; } // Specification says to return null if we could not // conclusively determine the file type return null; } Be aware that this may have huge performance impact.
You can use Files.probeContentType(path)
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 } });
Spring - validate that all message resources are well configured
We need to add some code to be executed on application load in order to validate that all the messages.properties elements are well defined for all languages. Is this possible? Steps: dynamically read on application load all the spring message codes from JSP or java classes then pass through all message resources properties files and validate that nothing is missing from them.
We ended up doing this manually but without using any library. Steps: have all the keys used in Java classes or JSP defined in a constant file Read them using Java .class properties: Field[] fields = Constants.class.getFields(); String filed[i].get(Constants.class); Read all messageResources.properties file names from the project using: String pathToThisClass = MessageResourcesValidator.class.getProtectionDomain().getCodeSource().getLocatin().getPath(); File filePath = new File(pathToThisClass); String[] list = filePath.list(new DirFilter("(messages).*\\.(properties)")); DirFilter is a normal class implementing Java's FileNameFilter Create a class that read the properties from a file using its file name: public class PropertiesFile{ private Properties prop; public PropertiesFile(String fileName) throws Exception { init(fileName); } private void init(String fileName) throws Exception { prop = new Properties(); try (InputStream input = getClass().getClassLoader().getResourceAsStream(fileName);) { if(input == null) { throw new Exception("Enable to load properties file " + fileName); } prop.load(input); } catch(IOException e) { throw new Exception("Error loading properties file " + fileName); } } public List<String> getPropertiesKeysList() { List<String> result = new ArrayList<>(); Enumeration<?> e = prop.propertyNames(); while(e.hasMoreElements()) { result.add((String) e.nextElement()); // String value = prop.getProperty(key); } return result; } } The part that does the comparison should be something as the following code that calls the above methods: List<String> msgResourcesFiles = getMessageResourcesFileNames(); List<String> codeKeys = getListOfCodeMessageResources(); PropertiesFile file = null; List<String> propKeys = null; for(String fileName : msgResourcesFiles) { file = new PropertiesFile(fileName); propKeys = file.getPropertiesKeysList(); for(String key : codeKeys) { if(!propKeys.contains(key)) { throw new Exception("Missing key " + key); } } } Note: or another workaround would be to compare all message resources files to a default one and this way we minimize the code needed from the above explanation.