java.nio.file.Path for a classpath resource - java

Is there an API to get a classpath resource (e.g. what I'd get from Class.getResource(String)) as a java.nio.file.Path? Ideally, I'd like to use the fancy new Path APIs with classpath resources.


This one works for me:
return Path.of(ClassLoader.getSystemResource(resourceName).toURI());


Guessing that what you want to do, is call Files.lines(...) on a resource that comes from the classpath - possibly from within a jar.
Since Oracle convoluted the notion of when a Path is a Path by not making getResource return a usable path if it resides in a jar file, what you need to do is something like this:
Stream<String> stream = new BufferedReader(new InputStreamReader(ClassLoader.getSystemResourceAsStream("/filename.txt"))).lines();


The most general solution is as follows:
interface IOConsumer<T> {
void accept(T t) throws IOException;
}
public static void processRessource(URI uri, IOConsumer<Path> action) throws IOException{
try {
Path p=Paths.get(uri);
action.accept(p);
}
catch(FileSystemNotFoundException ex) {
try(FileSystem fs = FileSystems.newFileSystem(
uri, Collections.<String,Object>emptyMap())) {
Path p = fs.provider().getPath(uri);
action.accept(p);
}
}
}
The main obstacle is to deal with the two possibilities, either, having an existing filesystem that we should use, but not close (like with file URIs or the Java 9’s module storage), or having to open and thus safely close the filesystem ourselves (like zip/jar files).
Therefore, the solution above encapsulates the actual action in an interface, handles both cases, safely closing afterwards in the second case, and works from Java 7 to Java 18. It probes whether there is already an open filesystem before opening a new one, so it also works in the case that another component of your application has already opened a filesystem for the same zip/jar file.
It can be used in all Java versions named above, e.g. to list the contents of a package (java.lang in the example) as Paths, like this:
processRessource(Object.class.getResource("Object.class").toURI(),new IOConsumer<Path>(){
public void accept(Path path) throws IOException {
try(DirectoryStream<Path> ds = Files.newDirectoryStream(path.getParent())) {
for(Path p: ds)
System.out.println(p);
}
}
});
With Java 8 or newer, you can use lambda expressions or method references to represent the actual action, e.g.
processRessource(Object.class.getResource("Object.class").toURI(), path -> {
try(Stream<Path> stream = Files.list(path.getParent())) {
stream.forEach(System.out::println);
}
});
to do the same.
The final release of Java 9’s module system has broken the above code example. The Java versions from 9 to 12 inconsistently return the path /java.base/java/lang/Object.class for Paths.get(Object.class.getResource("Object.class")) whereas it should be /modules/java.base/java/lang/Object.class. This can be fixed by prepending the missing /modules/ when the parent path is reported as non-existent:
processRessource(Object.class.getResource("Object.class").toURI(), path -> {
Path p = path.getParent();
if(!Files.exists(p))
p = p.resolve("/modules").resolve(p.getRoot().relativize(p));
try(Stream<Path> stream = Files.list(p)) {
stream.forEach(System.out::println);
}
});
Then, it will again work with all versions and storage methods. Starting with JDK 13, this work-around is not necessary anymore.


It turns out you can do this, with the help of the built-in Zip File System provider. However, passing a resource URI directly to Paths.get won't work; instead, one must first create a zip filesystem for the jar URI without the entry name, then refer to the entry in that filesystem:
static Path resourceToPath(URL resource)
throws IOException,
URISyntaxException {
Objects.requireNonNull(resource, "Resource URL cannot be null");
URI uri = resource.toURI();
String scheme = uri.getScheme();
if (scheme.equals("file")) {
return Paths.get(uri);
}
if (!scheme.equals("jar")) {
throw new IllegalArgumentException("Cannot convert to Path: " + uri);
}
String s = uri.toString();
int separator = s.indexOf("!/");
String entryName = s.substring(separator + 2);
URI fileURI = URI.create(s.substring(0, separator));
FileSystem fs = FileSystems.newFileSystem(fileURI,
Collections.<String, Object>emptyMap());
return fs.getPath(entryName);
}
Update:
It’s been rightly pointed out that the above code contains a resource leak, since the code opens a new FileSystem object but never closes it. The best approach is to pass a Consumer-like worker object, much like how Holger’s answer does it. Open the ZipFS FileSystem just long enough for the worker to do whatever it needs to do with the Path (as long as the worker doesn’t try to store the Path object for later use), then close the FileSystem.


I wrote a small helper method to read Paths from your class resources. It is quite handy to use as it only needs a reference of the class you have stored your resources as well as the name of the resource itself.
public static Path getResourcePath(Class<?> resourceClass, String resourceName) throws URISyntaxException {
URL url = resourceClass.getResource(resourceName);
return Paths.get(url.toURI());
}


Read a File from resources folder using NIO, in java8
public static String read(String fileName) {
Path path;
StringBuilder data = new StringBuilder();
Stream<String> lines = null;
try {
path = Paths.get(Thread.currentThread().getContextClassLoader().getResource(fileName).toURI());
lines = Files.lines(path);
} catch (URISyntaxException | IOException e) {
logger.error("Error in reading propertied file " + e);
throw new RuntimeException(e);
}
lines.forEach(line -> data.append(line));
lines.close();
return data.toString();
}


You can not create URI from resources inside of the jar file. You can simply write it to the temp file and then use it (java8):
Path path = File.createTempFile("some", "address").toPath();
Files.copy(ClassLoader.getSystemResourceAsStream("/path/to/resource"), path, StandardCopyOption.REPLACE_EXISTING);


You need to define the Filesystem to read resource from jar file as mentioned in https://docs.oracle.com/javase/8/docs/technotes/guides/io/fsp/zipfilesystemprovider.html. I success to read resource from jar file with below codes:
Map<String, Object> env = new HashMap<>();
try (FileSystem fs = FileSystems.newFileSystem(uri, env)) {
Path path = fs.getPath("/path/myResource");
try (Stream<String> lines = Files.lines(path)) {
....
}
}

Related

How do you list all files in the Resources folder (java/scala)

I am writing a function which needs to access a folder in the resources, and loop through all filenames, and if those match the criteria, those files are loaded.
new File(getClass.getResource("/images/sprites").getPath).listFiles()
returns a null pointer exception, where the directory tree follows Resources -> images -> sprites ->
Please can somebody point me in the right direction?

Joop Eggen's answer is awesome, however it can only do one of two things:
Read the resources content when running from a IDE
Read the resources content when running the JAR via the command line
So here's a example (Kotlin, but it should be easy to migrate it to Java) that allows you to have both: Reading the resources content when running from a IDE or via the command line!
val uri = MainApp::class.java.getResource("/locales/").toURI()
val dirPath = try {
Paths.get(uri)
} catch (e: FileSystemNotFoundException) {
// If this is thrown, then it means that we are running the JAR directly (example: not from an IDE)
val env = mutableMapOf<String, String>()
FileSystems.newFileSystem(uri, env).getPath("/locales/")
}
Files.list(dirPath).forEach {
println(it.fileName)
if (it.fileName.toString().endsWith("txt")) {
println("Result:")
println(Files.readString(it))
}
}

A zip file system using jar:file: URIs would be something like this:
URI uri = MainApp.class.getResource("/images/sprites").toURI();
Map<String, String> env = new HashMap<>();
try (FileSystem zipfs = FileSystems.newFileSystem(uri, env)) {
//Path path = zipfs.getPath("/images/icons16");
for (Path path : zipfs.getRootDirectories()) {
Files.list(path.resolve("/images/sprites"))
.forEach(p -> System.out.println("* " + p));
}
}
Here I show getRootDirectories to possibly iterate over all resources.
Using the Files.copy one may copy them etcetera.
After comment of #MrPowerGamerBR:
The solution above deals with a jar. A more general solution, not exposing the jar character, is:
URI uri = MAinApp.class.getResource("/images/sprites").toURI();
Path dirPath = Paths.get(uri);
Files.list(dirPath)
.forEach(p -> System.out.println("* " + p));
(In fact one might even read lines from the directory itself, but this
is the correct abstraction, using Path.)

How to get the location of the program you're running and store it in a File object [duplicate]

Want to improve this post? Provide detailed answers to this question, including citations and an explanation of why your answer is correct. Answers without enough detail may be edited or deleted.
My code runs inside a JAR file, say foo.jar, and I need to know, in the code, in which folder the running foo.jar is.
So, if foo.jar is in C:\FOO\, I want to get that path no matter what my current working directory is.

return new File(MyClass.class.getProtectionDomain().getCodeSource().getLocation()
.toURI()).getPath();
Replace "MyClass" with the name of your class.
Obviously, this will do odd things if your class was loaded from a non-file location.

Best solution for me:
String path = Test.class.getProtectionDomain().getCodeSource().getLocation().getPath();
String decodedPath = URLDecoder.decode(path, "UTF-8");
This should solve the problem with spaces and special characters.

To obtain the File for a given Class, there are two steps:
Convert the Class to a URL
Convert the URL to a File
It is important to understand both steps, and not conflate them.
Once you have the File, you can call getParentFile to get the containing folder, if that is what you need.
Step 1: Class to URL
As discussed in other answers, there are two major ways to find a URL relevant to a Class.
URL url = Bar.class.getProtectionDomain().getCodeSource().getLocation();
URL url = Bar.class.getResource(Bar.class.getSimpleName() + ".class");
Both have pros and cons.
The getProtectionDomain approach yields the base location of the class (e.g., the containing JAR file). However, it is possible that the Java runtime's security policy will throw SecurityException when calling getProtectionDomain(), so if your application needs to run in a variety of environments, it is best to test in all of them.
The getResource approach yields the full URL resource path of the class, from which you will need to perform additional string manipulation. It may be a file: path, but it could also be jar:file: or even something nastier like bundleresource://346.fwk2106232034:4/foo/Bar.class when executing within an OSGi framework. Conversely, the getProtectionDomain approach correctly yields a file: URL even from within OSGi.
Note that both getResource("") and getResource(".") failed in my tests, when the class resided within a JAR file; both invocations returned null. So I recommend the #2 invocation shown above instead, as it seems safer.
Step 2: URL to File
Either way, once you have a URL, the next step is convert to a File. This is its own challenge; see Kohsuke Kawaguchi's blog post about it for full details, but in short, you can use new File(url.toURI()) as long as the URL is completely well-formed.
Lastly, I would highly discourage using URLDecoder. Some characters of the URL, : and / in particular, are not valid URL-encoded characters. From the URLDecoder Javadoc:
It is assumed that all characters in the encoded string are one of the following: "a" through "z", "A" through "Z", "0" through "9", and "-", "_", ".", and "*". The character "%" is allowed but is interpreted as the start of a special escaped sequence.
...
There are two possible ways in which this decoder could deal with illegal strings. It could either leave illegal characters alone or it could throw an IllegalArgumentException. Which approach the decoder takes is left to the implementation.
In practice, URLDecoder generally does not throw IllegalArgumentException as threatened above. And if your file path has spaces encoded as %20, this approach may appear to work. However, if your file path has other non-alphameric characters such as + you will have problems with URLDecoder mangling your file path.
Working code
To achieve these steps, you might have methods like the following:
/**
* Gets the base location of the given class.
* <p>
* If the class is directly on the file system (e.g.,
* "/path/to/my/package/MyClass.class") then it will return the base directory
* (e.g., "file:/path/to").
* </p>
* <p>
* If the class is within a JAR file (e.g.,
* "/path/to/my-jar.jar!/my/package/MyClass.class") then it will return the
* path to the JAR (e.g., "file:/path/to/my-jar.jar").
* </p>
*
* #param c The class whose location is desired.
* #see FileUtils#urlToFile(URL) to convert the result to a {#link File}.
*/
public static URL getLocation(final Class<?> c) {
if (c == null) return null; // could not load the class
// try the easy way first
try {
final URL codeSourceLocation =
c.getProtectionDomain().getCodeSource().getLocation();
if (codeSourceLocation != null) return codeSourceLocation;
}
catch (final SecurityException e) {
// NB: Cannot access protection domain.
}
catch (final NullPointerException e) {
// NB: Protection domain or code source is null.
}
// NB: The easy way failed, so we try the hard way. We ask for the class
// itself as a resource, then strip the class's path from the URL string,
// leaving the base path.
// get the class's raw resource path
final URL classResource = c.getResource(c.getSimpleName() + ".class");
if (classResource == null) return null; // cannot find class resource
final String url = classResource.toString();
final String suffix = c.getCanonicalName().replace('.', '/') + ".class";
if (!url.endsWith(suffix)) return null; // weird URL
// strip the class's path from the URL string
final String base = url.substring(0, url.length() - suffix.length());
String path = base;
// remove the "jar:" prefix and "!/" suffix, if present
if (path.startsWith("jar:")) path = path.substring(4, path.length() - 2);
try {
return new URL(path);
}
catch (final MalformedURLException e) {
e.printStackTrace();
return null;
}
}
/**
* Converts the given {#link URL} to its corresponding {#link File}.
* <p>
* This method is similar to calling {#code new File(url.toURI())} except that
* it also handles "jar:file:" URLs, returning the path to the JAR file.
* </p>
*
* #param url The URL to convert.
* #return A file path suitable for use with e.g. {#link FileInputStream}
* #throws IllegalArgumentException if the URL does not correspond to a file.
*/
public static File urlToFile(final URL url) {
return url == null ? null : urlToFile(url.toString());
}
/**
* Converts the given URL string to its corresponding {#link File}.
*
* #param url The URL to convert.
* #return A file path suitable for use with e.g. {#link FileInputStream}
* #throws IllegalArgumentException if the URL does not correspond to a file.
*/
public static File urlToFile(final String url) {
String path = url;
if (path.startsWith("jar:")) {
// remove "jar:" prefix and "!/" suffix
final int index = path.indexOf("!/");
path = path.substring(4, index);
}
try {
if (PlatformUtils.isWindows() && path.matches("file:[A-Za-z]:.*")) {
path = "file:/" + path.substring(5);
}
return new File(new URL(path).toURI());
}
catch (final MalformedURLException e) {
// NB: URL is not completely well-formed.
}
catch (final URISyntaxException e) {
// NB: URL is not completely well-formed.
}
if (path.startsWith("file:")) {
// pass through the URL as-is, minus "file:" prefix
path = path.substring(5);
return new File(path);
}
throw new IllegalArgumentException("Invalid URL: " + url);
}
You can find these methods in the SciJava Common library:
org.scijava.util.ClassUtils
org.scijava.util.FileUtils.

You can also use:
CodeSource codeSource = YourMainClass.class.getProtectionDomain().getCodeSource();
File jarFile = new File(codeSource.getLocation().toURI().getPath());
String jarDir = jarFile.getParentFile().getPath();

Use ClassLoader.getResource() to find the URL for your current class.
For example:
package foo;
public class Test
{
public static void main(String[] args)
{
ClassLoader loader = Test.class.getClassLoader();
System.out.println(loader.getResource("foo/Test.class"));
}
}
(This example taken from a similar question.)
To find the directory, you'd then need to take apart the URL manually. See the JarClassLoader tutorial for the format of a jar URL.

I'm surprised to see that none recently proposed to use Path. Here follows a citation: "The Path class includes various methods that can be used to obtain information about the path, access elements of the path, convert the path to other forms, or extract portions of a path"
Thus, a good alternative is to get the Path objest as:
Path path = Paths.get(Test.class.getProtectionDomain().getCodeSource().getLocation().toURI());

The only solution that works for me on Linux, Mac and Windows:
public static String getJarContainingFolder(Class aclass) throws Exception {
CodeSource codeSource = aclass.getProtectionDomain().getCodeSource();
File jarFile;
if (codeSource.getLocation() != null) {
jarFile = new File(codeSource.getLocation().toURI());
}
else {
String path = aclass.getResource(aclass.getSimpleName() + ".class").getPath();
String jarFilePath = path.substring(path.indexOf(":") + 1, path.indexOf("!"));
jarFilePath = URLDecoder.decode(jarFilePath, "UTF-8");
jarFile = new File(jarFilePath);
}
return jarFile.getParentFile().getAbsolutePath();
}

If you are really looking for a simple way to get the folder in which your JAR is located you should use this implementation.
Solutions like this are hard to find and many solutions are no longer supported, many others provide the path of the file instead of the actual directory. This is easier than other solutions you are going to find and works for java version 1.12.
new File(".").getCanonicalPath()
Gathering the Input from other answers this is a simple one too:
String localPath=new File(getClass().getProtectionDomain().getCodeSource().getLocation().toURI()).getParentFile().getPath()+"\\";
Both will return a String with this format:
"C:\Users\User\Desktop\Folder\"
In a simple and concise line.

I had the the same problem and I solved it that way:
File currentJavaJarFile = new File(Main.class.getProtectionDomain().getCodeSource().getLocation().getPath());
String currentJavaJarFilePath = currentJavaJarFile.getAbsolutePath();
String currentRootDirectoryPath = currentJavaJarFilePath.replace(currentJavaJarFile.getName(), "");
I hope I was of help to you.

Here's upgrade to other comments, that seem to me incomplete for the specifics of
using a relative "folder" outside .jar file (in the jar's same
location):
String path =
YourMainClassName.class.getProtectionDomain().
getCodeSource().getLocation().getPath();
path =
URLDecoder.decode(
path,
"UTF-8");
BufferedImage img =
ImageIO.read(
new File((
new File(path).getParentFile().getPath()) +
File.separator +
"folder" +
File.separator +
"yourfile.jpg"));

For getting the path of running jar file I have studied the above solutions and tried all methods which exist some difference each other. If these code are running in Eclipse IDE they all should be able to find the path of the file including the indicated class and open or create an indicated file with the found path.
But it is tricky, when run the runnable jar file directly or through the command line, it will be failed as the path of jar file gotten from the above methods will give an internal path in the jar file, that is it always gives a path as
rsrc:project-name (maybe I should say that it is the package name of the main class file - the indicated class)
I can not convert the rsrc:... path to an external path, that is when run the jar file outside the Eclipse IDE it can not get the path of jar file.
The only possible way for getting the path of running jar file outside Eclipse IDE is
System.getProperty("java.class.path")
this code line may return the living path (including the file name) of the running jar file (note that the return path is not the working directory), as the java document and some people said that it will return the paths of all class files in the same directory, but as my tests if in the same directory include many jar files, it only return the path of running jar (about the multiple paths issue indeed it happened in the Eclipse).

Other answers seem to point to the code source which is Jar file location which is not a directory.
Use
return new File(MyClass.class.getProtectionDomain().getCodeSource().getLocation().toURI().getPath()).getParentFile();

the selected answer above is not working if you run your jar by click on it from Gnome desktop environment (not from any script or terminal).
Instead, I have fond that the following solution is working everywhere:
try {
return URLDecoder.decode(ClassLoader.getSystemClassLoader().getResource(".").getPath(), "UTF-8");
} catch (UnsupportedEncodingException e) {
return "";
}

I had to mess around a lot before I finally found a working (and short) solution.
It is possible that the jarLocation comes with a prefix like file:\ or jar:file\, which can be removed by using String#substring().
URL jarLocationUrl = MyClass.class.getProtectionDomain().getCodeSource().getLocation();
String jarLocation = new File(jarLocationUrl.toString()).getParent();

For the jar file path:
String jarPath = new File(MyClass.class.getProtectionDomain().getCodeSource().getLocation()
.toURI()).getPath();
For getting the directory path of that jar file:
String dirPath = new File(MyClass.class.getProtectionDomain().getCodeSource().getLocation()
.toURI()).getParent();
The results of the two lines above are like this:
/home/user/MyPrograms/myapp/myjar.jar (value of jarPath)
/home/user/MyPrograms/myapp (value of dirPath)

public static String dir() throws URISyntaxException
{
URI path=Main.class.getProtectionDomain().getCodeSource().getLocation().toURI();
String name= Main.class.getPackage().getName()+".jar";
String path2 = path.getRawPath();
path2=path2.substring(1);
if (path2.contains(".jar"))
{
path2=path2.replace(name, "");
}
return path2;}
Works good on Windows

I tried to get the jar running path using
String folder = MyClassName.class.getProtectionDomain().getCodeSource().getLocation().getPath();
c:\app>java -jar application.jar
Running the jar application named "application.jar", on Windows in the folder "c:\app", the value of the String variable "folder" was "\c:\app\application.jar" and I had problems testing for path's correctness
File test = new File(folder);
if(file.isDirectory() && file.canRead()) { //always false }
So I tried to define "test" as:
String fold= new File(folder).getParentFile().getPath()
File test = new File(fold);
to get path in a right format like "c:\app" instead of "\c:\app\application.jar" and I noticed that it work.

The simplest solution is to pass the path as an argument when running the jar.
You can automate this with a shell script (.bat in Windows, .sh anywhere else):
java -jar my-jar.jar .
I used . to pass the current working directory.
UPDATE
You may want to stick the jar file in a sub-directory so users don't accidentally click it. Your code should also check to make sure that the command line arguments have been supplied, and provide a good error message if the arguments are missing.

Actually here is a better version - the old one failed if a folder name had a space in it.
private String getJarFolder() {
// get name and path
String name = getClass().getName().replace('.', '/');
name = getClass().getResource("/" + name + ".class").toString();
// remove junk
name = name.substring(0, name.indexOf(".jar"));
name = name.substring(name.lastIndexOf(':')-1, name.lastIndexOf('/')+1).replace('%', ' ');
// remove escape characters
String s = "";
for (int k=0; k<name.length(); k++) {
s += name.charAt(k);
if (name.charAt(k) == ' ') k += 2;
}
// replace '/' with system separator char
return s.replace('/', File.separatorChar);
}
As for failing with applets, you wouldn't usually have access to local files anyway. I don't know much about JWS but to handle local files might it not be possible to download the app.?

String path = getClass().getResource("").getPath();
The path always refers to the resource within the jar file.

Try this:
String path = new File("").getAbsolutePath();

This code worked for me to identify if the program is being executed inside a JAR file or IDE:
private static boolean isRunningOverJar() {
try {
String pathJar = Application.class.getResource(Application.class.getSimpleName() + ".class").getFile();
if (pathJar.toLowerCase().contains(".jar")) {
return true;
} else {
return false;
}
} catch (Exception e) {
return false;
}
}
If I need to get the Windows full path of JAR file I am using this method:
private static String getPathJar() {
try {
final URI jarUriPath =
Application.class.getResource(Application.class.getSimpleName() + ".class").toURI();
String jarStringPath = jarUriPath.toString().replace("jar:", "");
String jarCleanPath = Paths.get(new URI(jarStringPath)).toString();
if (jarCleanPath.toLowerCase().contains(".jar")) {
return jarCleanPath.substring(0, jarCleanPath.lastIndexOf(".jar") + 4);
} else {
return null;
}
} catch (Exception e) {
log.error("Error getting JAR path.", e);
return null;
}
}
My complete code working with a Spring Boot application using CommandLineRunner implementation, to ensure that the application always be executed within of a console view (Double clicks by mistake in JAR file name), I am using the next code:
#SpringBootApplication
public class Application implements CommandLineRunner {
public static void main(String[] args) throws IOException {
Console console = System.console();
if (console == null && !GraphicsEnvironment.isHeadless() && isRunningOverJar()) {
Runtime.getRuntime().exec(new String[]{"cmd", "/c", "start", "cmd", "/k",
"java -jar \"" + getPathJar() + "\""});
} else {
SpringApplication.run(Application.class, args);
}
}
#Override
public void run(String... args) {
/*
Additional code here...
*/
}
private static boolean isRunningOverJar() {
try {
String pathJar = Application.class.getResource(Application.class.getSimpleName() + ".class").getFile();
if (pathJar.toLowerCase().contains(".jar")) {
return true;
} else {
return false;
}
} catch (Exception e) {
return false;
}
}
private static String getPathJar() {
try {
final URI jarUriPath =
Application.class.getResource(Application.class.getSimpleName() + ".class").toURI();
String jarStringPath = jarUriPath.toString().replace("jar:", "");
String jarCleanPath = Paths.get(new URI(jarStringPath)).toString();
if (jarCleanPath.toLowerCase().contains(".jar")) {
return jarCleanPath.substring(0, jarCleanPath.lastIndexOf(".jar") + 4);
} else {
return null;
}
} catch (Exception e) {
return null;
}
}
}

Something that is frustrating is that when you are developing in Eclipse MyClass.class.getProtectionDomain().getCodeSource().getLocation() returns the /bin directory which is great, but when you compile it to a jar, the path includes the /myjarname.jar part which gives you illegal file names.
To have the code work both in the ide and once it is compiled to a jar, I use the following piece of code:
URL applicationRootPathURL = getClass().getProtectionDomain().getCodeSource().getLocation();
File applicationRootPath = new File(applicationRootPathURL.getPath());
File myFile;
if(applicationRootPath.isDirectory()){
myFile = new File(applicationRootPath, "filename");
}
else{
myFile = new File(applicationRootPath.getParentFile(), "filename");
}

Not really sure about the others but in my case it didn't work with a "Runnable jar" and i got it working by fixing codes together from phchen2 answer and another from this link :How to get the path of a running JAR file?
The code:
String path=new java.io.File(Server.class.getProtectionDomain()
.getCodeSource()
.getLocation()
.getPath())
.getAbsolutePath();
path=path.substring(0, path.lastIndexOf("."));
path=path+System.getProperty("java.class.path");

Have tried several of the solutions up there but none yielded correct results for the (probably special) case that the runnable jar has been exported with "Packaging external libraries" in Eclipse. For some reason all solutions based on the ProtectionDomain do result in null in that case.
From combining some solutions above I managed to achieve the following working code:
String surroundingJar = null;
// gets the path to the jar file if it exists; or the "bin" directory if calling from Eclipse
String jarDir = new File(ClassLoader.getSystemClassLoader().getResource(".").getPath()).getAbsolutePath();
// gets the "bin" directory if calling from eclipse or the name of the .jar file alone (without its path)
String jarFileFromSys = System.getProperty("java.class.path").split(";")[0];
// If both are equal that means it is running from an IDE like Eclipse
if (jarFileFromSys.equals(jarDir))
{
System.out.println("RUNNING FROM IDE!");
// The path to the jar is the "bin" directory in that case because there is no actual .jar file.
surroundingJar = jarDir;
}
else
{
// Combining the path and the name of the .jar file to achieve the final result
surroundingJar = jarDir + jarFileFromSys.substring(1);
}
System.out.println("JAR File: " + surroundingJar);

The above methods didn't work for me in my Spring environment, since Spring shades the actual classes into a package called BOOT-INF, thus not the actual location of the running file. I found another way to retrieve the running file through the Permissions object which have been granted to the running file:
public static Path getEnclosingDirectory() {
return Paths.get(FileUtils.class.getProtectionDomain().getPermissions()
.elements().nextElement().getName()).getParent();
}

Mention that it is checked only in Windows but i think it works perfect on other Operating Systems [Linux,MacOs,Solaris] :).
I had 2 .jar files in the same directory . I wanted from the one .jar file to start the other .jar file which is in the same directory.
The problem is that when you start it from the cmd the current directory is system32.
Warnings!
The below seems to work pretty well in all the test i have done even
with folder name ;][[;'57f2g34g87-8+9-09!2##!$%^^&() or ()%&$%^##
it works well.
I am using the ProcessBuilder with the below as following:
🍂..
//The class from which i called this was the class `Main`
String path = getBasePathForClass(Main.class);
String applicationPath= new File(path + "application.jar").getAbsolutePath();
System.out.println("Directory Path is : "+applicationPath);
//Your know try catch here
//Mention that sometimes it doesn't work for example with folder `;][[;'57f2g34g87-8+9-09!2##!$%^^&()`
ProcessBuilder builder = new ProcessBuilder("java", "-jar", applicationPath);
builder.redirectErrorStream(true);
Process process = builder.start();
//...code
🍂getBasePathForClass(Class<?> classs):
/**
* Returns the absolute path of the current directory in which the given
* class
* file is.
*
* #param classs
* #return The absolute path of the current directory in which the class
* file is.
* #author GOXR3PLUS[StackOverFlow user] + bachden [StackOverFlow user]
*/
public static final String getBasePathForClass(Class<?> classs) {
// Local variables
File file;
String basePath = "";
boolean failed = false;
// Let's give a first try
try {
file = new File(classs.getProtectionDomain().getCodeSource().getLocation().toURI().getPath());
if (file.isFile() || file.getPath().endsWith(".jar") || file.getPath().endsWith(".zip")) {
basePath = file.getParent();
} else {
basePath = file.getPath();
}
} catch (URISyntaxException ex) {
failed = true;
Logger.getLogger(classs.getName()).log(Level.WARNING,
"Cannot firgue out base path for class with way (1): ", ex);
}
// The above failed?
if (failed) {
try {
file = new File(classs.getClassLoader().getResource("").toURI().getPath());
basePath = file.getAbsolutePath();
// the below is for testing purposes...
// starts with File.separator?
// String l = local.replaceFirst("[" + File.separator +
// "/\\\\]", "")
} catch (URISyntaxException ex) {
Logger.getLogger(classs.getName()).log(Level.WARNING,
"Cannot firgue out base path for class with way (2): ", ex);
}
}
// fix to run inside eclipse
if (basePath.endsWith(File.separator + "lib") || basePath.endsWith(File.separator + "bin")
|| basePath.endsWith("bin" + File.separator) || basePath.endsWith("lib" + File.separator)) {
basePath = basePath.substring(0, basePath.length() - 4);
}
// fix to run inside netbeans
if (basePath.endsWith(File.separator + "build" + File.separator + "classes")) {
basePath = basePath.substring(0, basePath.length() - 14);
}
// end fix
if (!basePath.endsWith(File.separator)) {
basePath = basePath + File.separator;
}
return basePath;
}

This code worked for me:
private static String getJarPath() throws IOException, URISyntaxException {
File f = new File(LicensingApp.class.getProtectionDomain().().getLocation().toURI());
String jarPath = f.getCanonicalPath().toString();
String jarDir = jarPath.substring( 0, jarPath.lastIndexOf( File.separator ));
return jarDir;
}

The getProtectionDomain approach might not work sometimes e.g. when you have to find the jar for some of the core java classes (e.g in my case StringBuilder class within IBM JDK), however following works seamlessly:
public static void main(String[] args) {
System.out.println(findSource(MyClass.class));
// OR
System.out.println(findSource(String.class));
}
public static String findSource(Class<?> clazz) {
String resourceToSearch = '/' + clazz.getName().replace(".", "/") + ".class";
java.net.URL location = clazz.getResource(resourceToSearch);
String sourcePath = location.getPath();
// Optional, Remove junk
return sourcePath.replace("file:", "").replace("!" + resourceToSearch, "");
}

I have another way to get the String location of a class.
URL path = Thread.currentThread().getContextClassLoader().getResource("");
Path p = Paths.get(path.toURI());
String location = p.toString();
The output String will have the form of
C:\Users\Administrator\new Workspace\...
The spaces and other characters are handled, and in the form without file:/. So will be easier to use.

Get folder from Resources folder JAVA

Hi everyone I can't figure out with this problem : this line of code should work
File[] file = (new File(getClass().getResource("resources/images_resultats"))).listFiles();
I want a list of File, these Files are under "images_resultats" under "resources".

It won't work if resources/images_resultats is not in your classpath and/or if it is in a jar file.
Your code is not even correct it should something like:
File[] file = (new File(getClass().getResource("/my/path").toURI())).listFiles();

You can determine what files are in a folder in resources (even if its in a jar) using the FileSystem class.
public static void doSomethingWithResourcesFolder(String inResourcesPath) throws URISyntaxException {
URI uri = ResourcesFolderUts.class.getResource(inResourcesPath).toURI();
try( FileSystem fileSystem = FileSystems.newFileSystem(uri, Collections.emptyMap() ) ){
Path folderRootPath = fileSystem.getPath(inResourcesPath);
Stream<Path> walk = Files.walk(folderRootPath, 1);
walk.forEach(childFileOrFolder -> {
//do something with the childFileOrFolder
});
} catch (IOException e) {
throw new RuntimeException(e);
}
}
inResourcesPath should be something like "/images_resultats"
Note that the childFileOrFolder paths can only be used while the FileSystem remains open, if you try to (for example) return the paths then use them later you've get a file system closed exception.
Change ResourcesFolderUts for one of your own classes

Assuming that resources folder is in classpath, this might work.
String folder = getClass().getResource("images_resultats").getFile();
File[] test = new File(folder).listFiles();

How to get the absolute path of the Jar file? [duplicate]

Want to improve this post? Provide detailed answers to this question, including citations and an explanation of why your answer is correct. Answers without enough detail may be edited or deleted.
My code runs inside a JAR file, say foo.jar, and I need to know, in the code, in which folder the running foo.jar is.
So, if foo.jar is in C:\FOO\, I want to get that path no matter what my current working directory is.

return new File(MyClass.class.getProtectionDomain().getCodeSource().getLocation()
.toURI()).getPath();
Replace "MyClass" with the name of your class.
Obviously, this will do odd things if your class was loaded from a non-file location.

Best solution for me:
String path = Test.class.getProtectionDomain().getCodeSource().getLocation().getPath();
String decodedPath = URLDecoder.decode(path, "UTF-8");
This should solve the problem with spaces and special characters.

To obtain the File for a given Class, there are two steps:
Convert the Class to a URL
Convert the URL to a File
It is important to understand both steps, and not conflate them.
Once you have the File, you can call getParentFile to get the containing folder, if that is what you need.
Step 1: Class to URL
As discussed in other answers, there are two major ways to find a URL relevant to a Class.
URL url = Bar.class.getProtectionDomain().getCodeSource().getLocation();
URL url = Bar.class.getResource(Bar.class.getSimpleName() + ".class");
Both have pros and cons.
The getProtectionDomain approach yields the base location of the class (e.g., the containing JAR file). However, it is possible that the Java runtime's security policy will throw SecurityException when calling getProtectionDomain(), so if your application needs to run in a variety of environments, it is best to test in all of them.
The getResource approach yields the full URL resource path of the class, from which you will need to perform additional string manipulation. It may be a file: path, but it could also be jar:file: or even something nastier like bundleresource://346.fwk2106232034:4/foo/Bar.class when executing within an OSGi framework. Conversely, the getProtectionDomain approach correctly yields a file: URL even from within OSGi.
Note that both getResource("") and getResource(".") failed in my tests, when the class resided within a JAR file; both invocations returned null. So I recommend the #2 invocation shown above instead, as it seems safer.
Step 2: URL to File
Either way, once you have a URL, the next step is convert to a File. This is its own challenge; see Kohsuke Kawaguchi's blog post about it for full details, but in short, you can use new File(url.toURI()) as long as the URL is completely well-formed.
Lastly, I would highly discourage using URLDecoder. Some characters of the URL, : and / in particular, are not valid URL-encoded characters. From the URLDecoder Javadoc:
It is assumed that all characters in the encoded string are one of the following: "a" through "z", "A" through "Z", "0" through "9", and "-", "_", ".", and "*". The character "%" is allowed but is interpreted as the start of a special escaped sequence.
...
There are two possible ways in which this decoder could deal with illegal strings. It could either leave illegal characters alone or it could throw an IllegalArgumentException. Which approach the decoder takes is left to the implementation.
In practice, URLDecoder generally does not throw IllegalArgumentException as threatened above. And if your file path has spaces encoded as %20, this approach may appear to work. However, if your file path has other non-alphameric characters such as + you will have problems with URLDecoder mangling your file path.
Working code
To achieve these steps, you might have methods like the following:
/**
* Gets the base location of the given class.
* <p>
* If the class is directly on the file system (e.g.,
* "/path/to/my/package/MyClass.class") then it will return the base directory
* (e.g., "file:/path/to").
* </p>
* <p>
* If the class is within a JAR file (e.g.,
* "/path/to/my-jar.jar!/my/package/MyClass.class") then it will return the
* path to the JAR (e.g., "file:/path/to/my-jar.jar").
* </p>
*
* #param c The class whose location is desired.
* #see FileUtils#urlToFile(URL) to convert the result to a {#link File}.
*/
public static URL getLocation(final Class<?> c) {
if (c == null) return null; // could not load the class
// try the easy way first
try {
final URL codeSourceLocation =
c.getProtectionDomain().getCodeSource().getLocation();
if (codeSourceLocation != null) return codeSourceLocation;
}
catch (final SecurityException e) {
// NB: Cannot access protection domain.
}
catch (final NullPointerException e) {
// NB: Protection domain or code source is null.
}
// NB: The easy way failed, so we try the hard way. We ask for the class
// itself as a resource, then strip the class's path from the URL string,
// leaving the base path.
// get the class's raw resource path
final URL classResource = c.getResource(c.getSimpleName() + ".class");
if (classResource == null) return null; // cannot find class resource
final String url = classResource.toString();
final String suffix = c.getCanonicalName().replace('.', '/') + ".class";
if (!url.endsWith(suffix)) return null; // weird URL
// strip the class's path from the URL string
final String base = url.substring(0, url.length() - suffix.length());
String path = base;
// remove the "jar:" prefix and "!/" suffix, if present
if (path.startsWith("jar:")) path = path.substring(4, path.length() - 2);
try {
return new URL(path);
}
catch (final MalformedURLException e) {
e.printStackTrace();
return null;
}
}
/**
* Converts the given {#link URL} to its corresponding {#link File}.
* <p>
* This method is similar to calling {#code new File(url.toURI())} except that
* it also handles "jar:file:" URLs, returning the path to the JAR file.
* </p>
*
* #param url The URL to convert.
* #return A file path suitable for use with e.g. {#link FileInputStream}
* #throws IllegalArgumentException if the URL does not correspond to a file.
*/
public static File urlToFile(final URL url) {
return url == null ? null : urlToFile(url.toString());
}
/**
* Converts the given URL string to its corresponding {#link File}.
*
* #param url The URL to convert.
* #return A file path suitable for use with e.g. {#link FileInputStream}
* #throws IllegalArgumentException if the URL does not correspond to a file.
*/
public static File urlToFile(final String url) {
String path = url;
if (path.startsWith("jar:")) {
// remove "jar:" prefix and "!/" suffix
final int index = path.indexOf("!/");
path = path.substring(4, index);
}
try {
if (PlatformUtils.isWindows() && path.matches("file:[A-Za-z]:.*")) {
path = "file:/" + path.substring(5);
}
return new File(new URL(path).toURI());
}
catch (final MalformedURLException e) {
// NB: URL is not completely well-formed.
}
catch (final URISyntaxException e) {
// NB: URL is not completely well-formed.
}
if (path.startsWith("file:")) {
// pass through the URL as-is, minus "file:" prefix
path = path.substring(5);
return new File(path);
}
throw new IllegalArgumentException("Invalid URL: " + url);
}
You can find these methods in the SciJava Common library:
org.scijava.util.ClassUtils
org.scijava.util.FileUtils.

You can also use:
CodeSource codeSource = YourMainClass.class.getProtectionDomain().getCodeSource();
File jarFile = new File(codeSource.getLocation().toURI().getPath());
String jarDir = jarFile.getParentFile().getPath();

Use ClassLoader.getResource() to find the URL for your current class.
For example:
package foo;
public class Test
{
public static void main(String[] args)
{
ClassLoader loader = Test.class.getClassLoader();
System.out.println(loader.getResource("foo/Test.class"));
}
}
(This example taken from a similar question.)
To find the directory, you'd then need to take apart the URL manually. See the JarClassLoader tutorial for the format of a jar URL.

I'm surprised to see that none recently proposed to use Path. Here follows a citation: "The Path class includes various methods that can be used to obtain information about the path, access elements of the path, convert the path to other forms, or extract portions of a path"
Thus, a good alternative is to get the Path objest as:
Path path = Paths.get(Test.class.getProtectionDomain().getCodeSource().getLocation().toURI());

The only solution that works for me on Linux, Mac and Windows:
public static String getJarContainingFolder(Class aclass) throws Exception {
CodeSource codeSource = aclass.getProtectionDomain().getCodeSource();
File jarFile;
if (codeSource.getLocation() != null) {
jarFile = new File(codeSource.getLocation().toURI());
}
else {
String path = aclass.getResource(aclass.getSimpleName() + ".class").getPath();
String jarFilePath = path.substring(path.indexOf(":") + 1, path.indexOf("!"));
jarFilePath = URLDecoder.decode(jarFilePath, "UTF-8");
jarFile = new File(jarFilePath);
}
return jarFile.getParentFile().getAbsolutePath();
}

If you are really looking for a simple way to get the folder in which your JAR is located you should use this implementation.
Solutions like this are hard to find and many solutions are no longer supported, many others provide the path of the file instead of the actual directory. This is easier than other solutions you are going to find and works for java version 1.12.
new File(".").getCanonicalPath()
Gathering the Input from other answers this is a simple one too:
String localPath=new File(getClass().getProtectionDomain().getCodeSource().getLocation().toURI()).getParentFile().getPath()+"\\";
Both will return a String with this format:
"C:\Users\User\Desktop\Folder\"
In a simple and concise line.

I had the the same problem and I solved it that way:
File currentJavaJarFile = new File(Main.class.getProtectionDomain().getCodeSource().getLocation().getPath());
String currentJavaJarFilePath = currentJavaJarFile.getAbsolutePath();
String currentRootDirectoryPath = currentJavaJarFilePath.replace(currentJavaJarFile.getName(), "");
I hope I was of help to you.

Here's upgrade to other comments, that seem to me incomplete for the specifics of
using a relative "folder" outside .jar file (in the jar's same
location):
String path =
YourMainClassName.class.getProtectionDomain().
getCodeSource().getLocation().getPath();
path =
URLDecoder.decode(
path,
"UTF-8");
BufferedImage img =
ImageIO.read(
new File((
new File(path).getParentFile().getPath()) +
File.separator +
"folder" +
File.separator +
"yourfile.jpg"));

For getting the path of running jar file I have studied the above solutions and tried all methods which exist some difference each other. If these code are running in Eclipse IDE they all should be able to find the path of the file including the indicated class and open or create an indicated file with the found path.
But it is tricky, when run the runnable jar file directly or through the command line, it will be failed as the path of jar file gotten from the above methods will give an internal path in the jar file, that is it always gives a path as
rsrc:project-name (maybe I should say that it is the package name of the main class file - the indicated class)
I can not convert the rsrc:... path to an external path, that is when run the jar file outside the Eclipse IDE it can not get the path of jar file.
The only possible way for getting the path of running jar file outside Eclipse IDE is
System.getProperty("java.class.path")
this code line may return the living path (including the file name) of the running jar file (note that the return path is not the working directory), as the java document and some people said that it will return the paths of all class files in the same directory, but as my tests if in the same directory include many jar files, it only return the path of running jar (about the multiple paths issue indeed it happened in the Eclipse).

Other answers seem to point to the code source which is Jar file location which is not a directory.
Use
return new File(MyClass.class.getProtectionDomain().getCodeSource().getLocation().toURI().getPath()).getParentFile();

the selected answer above is not working if you run your jar by click on it from Gnome desktop environment (not from any script or terminal).
Instead, I have fond that the following solution is working everywhere:
try {
return URLDecoder.decode(ClassLoader.getSystemClassLoader().getResource(".").getPath(), "UTF-8");
} catch (UnsupportedEncodingException e) {
return "";
}

I had to mess around a lot before I finally found a working (and short) solution.
It is possible that the jarLocation comes with a prefix like file:\ or jar:file\, which can be removed by using String#substring().
URL jarLocationUrl = MyClass.class.getProtectionDomain().getCodeSource().getLocation();
String jarLocation = new File(jarLocationUrl.toString()).getParent();

For the jar file path:
String jarPath = new File(MyClass.class.getProtectionDomain().getCodeSource().getLocation()
.toURI()).getPath();
For getting the directory path of that jar file:
String dirPath = new File(MyClass.class.getProtectionDomain().getCodeSource().getLocation()
.toURI()).getParent();
The results of the two lines above are like this:
/home/user/MyPrograms/myapp/myjar.jar (value of jarPath)
/home/user/MyPrograms/myapp (value of dirPath)

public static String dir() throws URISyntaxException
{
URI path=Main.class.getProtectionDomain().getCodeSource().getLocation().toURI();
String name= Main.class.getPackage().getName()+".jar";
String path2 = path.getRawPath();
path2=path2.substring(1);
if (path2.contains(".jar"))
{
path2=path2.replace(name, "");
}
return path2;}
Works good on Windows

I tried to get the jar running path using
String folder = MyClassName.class.getProtectionDomain().getCodeSource().getLocation().getPath();
c:\app>java -jar application.jar
Running the jar application named "application.jar", on Windows in the folder "c:\app", the value of the String variable "folder" was "\c:\app\application.jar" and I had problems testing for path's correctness
File test = new File(folder);
if(file.isDirectory() && file.canRead()) { //always false }
So I tried to define "test" as:
String fold= new File(folder).getParentFile().getPath()
File test = new File(fold);
to get path in a right format like "c:\app" instead of "\c:\app\application.jar" and I noticed that it work.

The simplest solution is to pass the path as an argument when running the jar.
You can automate this with a shell script (.bat in Windows, .sh anywhere else):
java -jar my-jar.jar .
I used . to pass the current working directory.
UPDATE
You may want to stick the jar file in a sub-directory so users don't accidentally click it. Your code should also check to make sure that the command line arguments have been supplied, and provide a good error message if the arguments are missing.

Actually here is a better version - the old one failed if a folder name had a space in it.
private String getJarFolder() {
// get name and path
String name = getClass().getName().replace('.', '/');
name = getClass().getResource("/" + name + ".class").toString();
// remove junk
name = name.substring(0, name.indexOf(".jar"));
name = name.substring(name.lastIndexOf(':')-1, name.lastIndexOf('/')+1).replace('%', ' ');
// remove escape characters
String s = "";
for (int k=0; k<name.length(); k++) {
s += name.charAt(k);
if (name.charAt(k) == ' ') k += 2;
}
// replace '/' with system separator char
return s.replace('/', File.separatorChar);
}
As for failing with applets, you wouldn't usually have access to local files anyway. I don't know much about JWS but to handle local files might it not be possible to download the app.?

String path = getClass().getResource("").getPath();
The path always refers to the resource within the jar file.

Try this:
String path = new File("").getAbsolutePath();

This code worked for me to identify if the program is being executed inside a JAR file or IDE:
private static boolean isRunningOverJar() {
try {
String pathJar = Application.class.getResource(Application.class.getSimpleName() + ".class").getFile();
if (pathJar.toLowerCase().contains(".jar")) {
return true;
} else {
return false;
}
} catch (Exception e) {
return false;
}
}
If I need to get the Windows full path of JAR file I am using this method:
private static String getPathJar() {
try {
final URI jarUriPath =
Application.class.getResource(Application.class.getSimpleName() + ".class").toURI();
String jarStringPath = jarUriPath.toString().replace("jar:", "");
String jarCleanPath = Paths.get(new URI(jarStringPath)).toString();
if (jarCleanPath.toLowerCase().contains(".jar")) {
return jarCleanPath.substring(0, jarCleanPath.lastIndexOf(".jar") + 4);
} else {
return null;
}
} catch (Exception e) {
log.error("Error getting JAR path.", e);
return null;
}
}
My complete code working with a Spring Boot application using CommandLineRunner implementation, to ensure that the application always be executed within of a console view (Double clicks by mistake in JAR file name), I am using the next code:
#SpringBootApplication
public class Application implements CommandLineRunner {
public static void main(String[] args) throws IOException {
Console console = System.console();
if (console == null && !GraphicsEnvironment.isHeadless() && isRunningOverJar()) {
Runtime.getRuntime().exec(new String[]{"cmd", "/c", "start", "cmd", "/k",
"java -jar \"" + getPathJar() + "\""});
} else {
SpringApplication.run(Application.class, args);
}
}
#Override
public void run(String... args) {
/*
Additional code here...
*/
}
private static boolean isRunningOverJar() {
try {
String pathJar = Application.class.getResource(Application.class.getSimpleName() + ".class").getFile();
if (pathJar.toLowerCase().contains(".jar")) {
return true;
} else {
return false;
}
} catch (Exception e) {
return false;
}
}
private static String getPathJar() {
try {
final URI jarUriPath =
Application.class.getResource(Application.class.getSimpleName() + ".class").toURI();
String jarStringPath = jarUriPath.toString().replace("jar:", "");
String jarCleanPath = Paths.get(new URI(jarStringPath)).toString();
if (jarCleanPath.toLowerCase().contains(".jar")) {
return jarCleanPath.substring(0, jarCleanPath.lastIndexOf(".jar") + 4);
} else {
return null;
}
} catch (Exception e) {
return null;
}
}
}

Something that is frustrating is that when you are developing in Eclipse MyClass.class.getProtectionDomain().getCodeSource().getLocation() returns the /bin directory which is great, but when you compile it to a jar, the path includes the /myjarname.jar part which gives you illegal file names.
To have the code work both in the ide and once it is compiled to a jar, I use the following piece of code:
URL applicationRootPathURL = getClass().getProtectionDomain().getCodeSource().getLocation();
File applicationRootPath = new File(applicationRootPathURL.getPath());
File myFile;
if(applicationRootPath.isDirectory()){
myFile = new File(applicationRootPath, "filename");
}
else{
myFile = new File(applicationRootPath.getParentFile(), "filename");
}

Not really sure about the others but in my case it didn't work with a "Runnable jar" and i got it working by fixing codes together from phchen2 answer and another from this link :How to get the path of a running JAR file?
The code:
String path=new java.io.File(Server.class.getProtectionDomain()
.getCodeSource()
.getLocation()
.getPath())
.getAbsolutePath();
path=path.substring(0, path.lastIndexOf("."));
path=path+System.getProperty("java.class.path");

Have tried several of the solutions up there but none yielded correct results for the (probably special) case that the runnable jar has been exported with "Packaging external libraries" in Eclipse. For some reason all solutions based on the ProtectionDomain do result in null in that case.
From combining some solutions above I managed to achieve the following working code:
String surroundingJar = null;
// gets the path to the jar file if it exists; or the "bin" directory if calling from Eclipse
String jarDir = new File(ClassLoader.getSystemClassLoader().getResource(".").getPath()).getAbsolutePath();
// gets the "bin" directory if calling from eclipse or the name of the .jar file alone (without its path)
String jarFileFromSys = System.getProperty("java.class.path").split(";")[0];
// If both are equal that means it is running from an IDE like Eclipse
if (jarFileFromSys.equals(jarDir))
{
System.out.println("RUNNING FROM IDE!");
// The path to the jar is the "bin" directory in that case because there is no actual .jar file.
surroundingJar = jarDir;
}
else
{
// Combining the path and the name of the .jar file to achieve the final result
surroundingJar = jarDir + jarFileFromSys.substring(1);
}
System.out.println("JAR File: " + surroundingJar);

The above methods didn't work for me in my Spring environment, since Spring shades the actual classes into a package called BOOT-INF, thus not the actual location of the running file. I found another way to retrieve the running file through the Permissions object which have been granted to the running file:
public static Path getEnclosingDirectory() {
return Paths.get(FileUtils.class.getProtectionDomain().getPermissions()
.elements().nextElement().getName()).getParent();
}

Mention that it is checked only in Windows but i think it works perfect on other Operating Systems [Linux,MacOs,Solaris] :).
I had 2 .jar files in the same directory . I wanted from the one .jar file to start the other .jar file which is in the same directory.
The problem is that when you start it from the cmd the current directory is system32.
Warnings!
The below seems to work pretty well in all the test i have done even
with folder name ;][[;'57f2g34g87-8+9-09!2##!$%^^&() or ()%&$%^##
it works well.
I am using the ProcessBuilder with the below as following:
🍂..
//The class from which i called this was the class `Main`
String path = getBasePathForClass(Main.class);
String applicationPath= new File(path + "application.jar").getAbsolutePath();
System.out.println("Directory Path is : "+applicationPath);
//Your know try catch here
//Mention that sometimes it doesn't work for example with folder `;][[;'57f2g34g87-8+9-09!2##!$%^^&()`
ProcessBuilder builder = new ProcessBuilder("java", "-jar", applicationPath);
builder.redirectErrorStream(true);
Process process = builder.start();
//...code
🍂getBasePathForClass(Class<?> classs):
/**
* Returns the absolute path of the current directory in which the given
* class
* file is.
*
* #param classs
* #return The absolute path of the current directory in which the class
* file is.
* #author GOXR3PLUS[StackOverFlow user] + bachden [StackOverFlow user]
*/
public static final String getBasePathForClass(Class<?> classs) {
// Local variables
File file;
String basePath = "";
boolean failed = false;
// Let's give a first try
try {
file = new File(classs.getProtectionDomain().getCodeSource().getLocation().toURI().getPath());
if (file.isFile() || file.getPath().endsWith(".jar") || file.getPath().endsWith(".zip")) {
basePath = file.getParent();
} else {
basePath = file.getPath();
}
} catch (URISyntaxException ex) {
failed = true;
Logger.getLogger(classs.getName()).log(Level.WARNING,
"Cannot firgue out base path for class with way (1): ", ex);
}
// The above failed?
if (failed) {
try {
file = new File(classs.getClassLoader().getResource("").toURI().getPath());
basePath = file.getAbsolutePath();
// the below is for testing purposes...
// starts with File.separator?
// String l = local.replaceFirst("[" + File.separator +
// "/\\\\]", "")
} catch (URISyntaxException ex) {
Logger.getLogger(classs.getName()).log(Level.WARNING,
"Cannot firgue out base path for class with way (2): ", ex);
}
}
// fix to run inside eclipse
if (basePath.endsWith(File.separator + "lib") || basePath.endsWith(File.separator + "bin")
|| basePath.endsWith("bin" + File.separator) || basePath.endsWith("lib" + File.separator)) {
basePath = basePath.substring(0, basePath.length() - 4);
}
// fix to run inside netbeans
if (basePath.endsWith(File.separator + "build" + File.separator + "classes")) {
basePath = basePath.substring(0, basePath.length() - 14);
}
// end fix
if (!basePath.endsWith(File.separator)) {
basePath = basePath + File.separator;
}
return basePath;
}

This code worked for me:
private static String getJarPath() throws IOException, URISyntaxException {
File f = new File(LicensingApp.class.getProtectionDomain().().getLocation().toURI());
String jarPath = f.getCanonicalPath().toString();
String jarDir = jarPath.substring( 0, jarPath.lastIndexOf( File.separator ));
return jarDir;
}

The getProtectionDomain approach might not work sometimes e.g. when you have to find the jar for some of the core java classes (e.g in my case StringBuilder class within IBM JDK), however following works seamlessly:
public static void main(String[] args) {
System.out.println(findSource(MyClass.class));
// OR
System.out.println(findSource(String.class));
}
public static String findSource(Class<?> clazz) {
String resourceToSearch = '/' + clazz.getName().replace(".", "/") + ".class";
java.net.URL location = clazz.getResource(resourceToSearch);
String sourcePath = location.getPath();
// Optional, Remove junk
return sourcePath.replace("file:", "").replace("!" + resourceToSearch, "");
}

I have another way to get the String location of a class.
URL path = Thread.currentThread().getContextClassLoader().getResource("");
Path p = Paths.get(path.toURI());
String location = p.toString();
The output String will have the form of
C:\Users\Administrator\new Workspace\...
The spaces and other characters are handled, and in the form without file:/. So will be easier to use.

How can I access a folder inside of a resource folder from inside my jar File?

I have a resources folder/package in the root of my project, I "don't" want to load a certain File. If I wanted to load a certain File, I would use class.getResourceAsStream and I would be fine!! What I actually want to do is to load a "Folder" within the resources folder, loop on the Files inside that Folder and get a Stream to each file and read in the content... Assume that the File names are not determined before runtime... What should I do? Is there a way to get a list of the files inside a Folder in your jar File?
Notice that the Jar file with the resources is the same jar file from which the code is being run...

Finally, I found the solution:
final String path = "sample/folder";
final File jarFile = new File(getClass().getProtectionDomain().getCodeSource().getLocation().getPath());
if(jarFile.isFile()) { // Run with JAR file
final JarFile jar = new JarFile(jarFile);
final Enumeration<JarEntry> entries = jar.entries(); //gives ALL entries in jar
while(entries.hasMoreElements()) {
final String name = entries.nextElement().getName();
if (name.startsWith(path + "/")) { //filter according to the path
System.out.println(name);
}
}
jar.close();
} else { // Run with IDE
final URL url = Launcher.class.getResource("/" + path);
if (url != null) {
try {
final File apps = new File(url.toURI());
for (File app : apps.listFiles()) {
System.out.println(app);
}
} catch (URISyntaxException ex) {
// never happens
}
}
}
The second block just work when you run the application on IDE (not with jar file), You can remove it if you don't like that.

Try the following.
Make the resource path "<PathRelativeToThisClassFile>/<ResourceDirectory>" E.g. if your class path is com.abc.package.MyClass and your resoure files are within src/com/abc/package/resources/:
URL url = MyClass.class.getResource("resources/");
if (url == null) {
// error - missing folder
} else {
File dir = new File(url.toURI());
for (File nextFile : dir.listFiles()) {
// Do something with nextFile
}
}
You can also use
URL url = MyClass.class.getResource("/com/abc/package/resources/");

The following code returns the wanted "folder" as Path regardless of if it is inside a jar or not.
private Path getFolderPath() throws URISyntaxException, IOException {
URI uri = getClass().getClassLoader().getResource("folder").toURI();
if ("jar".equals(uri.getScheme())) {
FileSystem fileSystem = FileSystems.newFileSystem(uri, Collections.emptyMap(), null);
return fileSystem.getPath("path/to/folder/inside/jar");
} else {
return Paths.get(uri);
}
}
Requires java 7+.

I know this is many years ago . But just for other people come across this topic.
What you could do is to use getResourceAsStream() method with the directory path, and the input Stream will have all the files name from that dir. After that you can concat the dir path with each file name and call getResourceAsStream for each file in a loop.

I had the same problem at hands while i was attempting to load some hadoop configurations from resources packed in the jar... on both the IDE and on jar (release version).
I found java.nio.file.DirectoryStream to work the best to iterate over directory contents over both local filesystem and jar.
String fooFolder = "/foo/folder";
....
ClassLoader classLoader = foofClass.class.getClassLoader();
try {
uri = classLoader.getResource(fooFolder).toURI();
} catch (URISyntaxException e) {
throw new FooException(e.getMessage());
} catch (NullPointerException e){
throw new FooException(e.getMessage());
}
if(uri == null){
throw new FooException("something is wrong directory or files missing");
}
/** i want to know if i am inside the jar or working on the IDE*/
if(uri.getScheme().contains("jar")){
/** jar case */
try{
URL jar = FooClass.class.getProtectionDomain().getCodeSource().getLocation();
//jar.toString() begins with file:
//i want to trim it out...
Path jarFile = Paths.get(jar.toString().substring("file:".length()));
FileSystem fs = FileSystems.newFileSystem(jarFile, null);
DirectoryStream<Path> directoryStream = Files.newDirectoryStream(fs.getPath(fooFolder));
for(Path p: directoryStream){
InputStream is = FooClass.class.getResourceAsStream(p.toString()) ;
performFooOverInputStream(is);
/** your logic here **/
}
}catch(IOException e) {
throw new FooException(e.getMessage());
}
}
else{
/** IDE case */
Path path = Paths.get(uri);
try {
DirectoryStream<Path> directoryStream = Files.newDirectoryStream(path);
for(Path p : directoryStream){
InputStream is = new FileInputStream(p.toFile());
performFooOverInputStream(is);
}
} catch (IOException _e) {
throw new FooException(_e.getMessage());
}
}

Another solution, you can do it using ResourceLoader like this:
import org.springframework.core.io.Resource;
import org.apache.commons.io.FileUtils;
#Autowire
private ResourceLoader resourceLoader;
...
Resource resource = resourceLoader.getResource("classpath:/path/to/you/dir");
File file = resource.getFile();
Iterator<File> fi = FileUtils.iterateFiles(file, null, true);
while(fi.hasNext()) {
load(fi.next())
}

If you are using Spring you can use org.springframework.core.io.support.PathMatchingResourcePatternResolver and deal with Resource objects rather than files. This works when running inside and outside of a Jar file.
PathMatchingResourcePatternResolver r = new PathMatchingResourcePatternResolver();
Resource[] resources = r.getResources("/myfolder/*");
Then you can access the data using getInputStream and the filename from getFilename.
Note that it will still fail if you try to use the getFile while running from a Jar.

As the other answers point out, once the resources are inside a jar file, things get really ugly. In our case, this solution:
https://stackoverflow.com/a/13227570/516188
works very well in the tests (since when the tests are run the code is not packed in a jar file), but doesn't work when the app actually runs normally. So what I've done is... I hardcode the list of the files in the app, but I have a test which reads the actual list from disk (can do it since that works in tests) and fails if the actual list doesn't match with the list the app returns.
That way I have simple code in my app (no tricks), and I'm sure I didn't forget to add a new entry in the list thanks to the test.

Below code gets .yaml files from a custom resource directory.
ClassLoader classLoader = this.getClass().getClassLoader();
URI uri = classLoader.getResource(directoryPath).toURI();
if("jar".equalsIgnoreCase(uri.getScheme())){
Pattern pattern = Pattern.compile("^.+" +"/classes/" + directoryPath + "/.+.yaml$");
log.debug("pattern {} ", pattern.pattern());
ApplicationHome home = new ApplicationHome(SomeApplication.class);
JarFile file = new JarFile(home.getSource());
Enumeration<JarEntry> jarEntries = file.entries() ;
while(jarEntries.hasMoreElements()){
JarEntry entry = jarEntries.nextElement();
Matcher matcher = pattern.matcher(entry.getName());
if(matcher.find()){
InputStream in =
file.getInputStream(entry);
//work on the stream
}
}
}else{
//When Spring boot application executed through Non-Jar strategy like through IDE or as a War.
String path = uri.getPath();
File[] files = new File(path).listFiles();
for(File file: files){
if(file != null){
try {
InputStream is = new FileInputStream(file);
//work on stream
} catch (Exception e) {
log.error("Exception while parsing file yaml file {} : {} " , file.getAbsolutePath(), e.getMessage());
}
}else{
log.warn("File Object is null while parsing yaml file");
}
}
}

Took me 2-3 days to get this working, in order to have the same url that work for both Jar or in local, the url (or path) needs to be a relative path from the repository root.
..meaning, the location of your file or folder from your src folder.
could be "/main/resources/your-folder/" or "/client/notes/somefile.md"
Whatever it is, in order for your JAR file to find it, the url must be a relative path from the repository root.
it must be "src/main/resources/your-folder/" or "src/client/notes/somefile.md"
Now you get the drill, and luckily for Intellij Idea users, you can get the correct path with a right-click on the folder or file -> copy Path/Reference.. -> Path From Repository Root (this is it)
Last, paste it and do your thing.

Simple ... use OSGi. In OSGi you can iterate over your Bundle's entries with findEntries and findPaths.

Inside my jar file I had a folder called Upload, this folder had three other text files inside it and I needed to have an exactly the same folder and files outside of the jar file, I used the code below:
URL inputUrl = getClass().getResource("/upload/blabla1.txt");
File dest1 = new File("upload/blabla1.txt");
FileUtils.copyURLToFile(inputUrl, dest1);
URL inputUrl2 = getClass().getResource("/upload/blabla2.txt");
File dest2 = new File("upload/blabla2.txt");
FileUtils.copyURLToFile(inputUrl2, dest2);
URL inputUrl3 = getClass().getResource("/upload/blabla3.txt");
File dest3 = new File("upload/Bblabla3.txt");
FileUtils.copyURLToFile(inputUrl3, dest3);

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