Related
How should I effectively check for the availability of particular folder(myfolder) recursively and if available, then create a tmp directory parallel to it
Example:
#ls -l
--parent folder
--projects
-- sub folders (further depth is possible)
-- myfolder
-- tmp
I'm from python background and yet to get used to java. Below is what I could come up with.
import java.io.File;
class Main
{
public static void main(String[] args)
{
String currentDir = System.getProperty("user.dir");
String projectDir = currentDir + "/projects"; // under this I have to search the for the `myfolder` recursively.
File file = new File(projectDir);
if (file.isDirectory()) {
new File("tmp").mkdirs();
}
else {
System.out.println("Directory doesn't exist!!");
}
}
note: I use java 8
Below is a method that recursively searches through all the sub-directories that might be contained within the provided local directory path for a specific directory (folder) name. When the first instance of that directory name is found the search halts and the full path to that directory is returned.
From that point, the returned path string should be parsed to get the parent path. Something like that could be done something like this:
String foundFolderParentPath = foundDirectory.substring(0,
foundDirectory.lastIndexOf(File.separator));
Now you would want to check and see if the tmp directory already exists there. Maybe you don't need to create it or, you may want to carry out some other action based on that fact:
if (new File(foundFolderParentPath + File.separator + "tmp").exists()) {
// tmp already exists...Do whatever...
}
else {
// Otherwise Create the tmp directory...
new File(foundFolderParentPath + File.separator + "tmp").mkdir();
}
Here is the recursive findDirectory() method:
/**
* This method recursively searches through all sub-directories beginning
* from the supplied searchStartPath until the supplied folder to find is
* found.<br>
*
* #param searchStartPath (String) The full path to start the search from.<br>
*
* #param folderToSearchFor (String) The directory (folder) or sub-directory
* (sub-folder) name to search for. Just a single name should be supplied, not multiple directory names..<br>
*
* #return (String) If the search is successful then the full path to that
* found folder is returned. If the search was unsuccessful then Null String
* (""), an empty string is returned.
*/
public static String findDirectory(String searchStartPath, String folderToSearchFor) {
String foundPath = "";
File[] folders = new File(searchStartPath).listFiles(File::isDirectory);
if (folders.length == 0) {
return "";
}
String tmp;
for (int i = 0; i < folders.length; i++) {
String currentPath = folders[i].getAbsolutePath();
if (currentPath.equals(folderToSearchFor) ||
currentPath.substring(currentPath.lastIndexOf(File.separator) + 1)
.equals(folderToSearchFor)) {
foundPath = currentPath;
break;
}
tmp = "";
// The recursive call...
tmp = findDirectory(folders[i].getAbsolutePath(), folderToSearchFor);
if (!tmp.isEmpty()) {
// Directory is found...
foundPath = tmp;
break; // Get out of loop. It's No longer needed.
}
}
return foundPath;
}
How you might use this method:
String currentDir = System.getProperty("user.dir");
String projectDir = currentDir + "/Projects"; // under this I have to search the for the `myfolder` recursively.
String searchForDirectory = "mySpecialFolder";
String foundDirectory = findDirectory(currentDir, searchForDirectory);
if (foundDirectory.isEmpty()) {
System.out.println("The folder to find (" + searchForDirectory
+ ") could not be found!");
}
else {
System.out.println("The ' " + searchForDirectory +
"' Folder is found at: --> " + foundDirectory);
/* Create the 'tmp' folder within the same parent folder where
mySpecialFolder resides in. */
String foundFolderParentPath = foundDirectory.substring(0,
foundDirectory.lastIndexOf(File.separator));
// Is there a 'tmp' folder already there?
if (new File(foundFolderParentPath + File.separator + "tmp").exists()) {
// Yes there is..
System.out.println("\nThere is no need to create the 'tmp' folder! It already"
+ "exists within the\nparent path of: --> " + foundFolderParentPath);
}
else {
// No here isn't so create it...
new File(foundFolderParentPath + File.separator + "tmp").mkdir();
System.out.println("The 'tmp' folder was created within the parent path indicated below:");
System.out.println(foundFolderParentPath);
System.out.println();
}
// Display a File-Chooser to prove it just for the heck of it.
javax.swing.JFileChooser fc = new javax.swing.JFileChooser(foundFolderParentPath);
fc.showDialog(null, "Just A Test");
}
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.
I am trying to achieve the following: From a given Class object, I want to be able to retrieve the folder or file in which it is located. This should also work for System classes like java.lang.String (which would return the location of rt.jar). For 'source' classes, the method should return the root folder:
- bin
- com
- test
- Test.class
would return the location of the bin folder for file(com.test.Test.class). This is my implementation so far:
public static File getFileLocation(Class<?> klass)
{
String classLocation = '/' + klass.getName().replace('.', '/') + ".class";
URL url = klass.getResource(classLocation);
String path = url.getPath();
int index = path.lastIndexOf(classLocation);
if (index < 0)
{
return null;
}
// Jar Handling
if (path.charAt(index - 1) == '!')
{
index--;
}
else
{
index++;
}
int index1 = path.lastIndexOf(':', index);
String newPath = path.substring(index1 + 1, index);
System.out.println(url.toExternalForm());
URI uri = URI.create(newPath).normalize();
return new File(uri);
}
However, this code fails because the File(URI) constructor throws an IllegalArgumentException - "URI is not absolute". I already tried to use the newPath to construct the file, but this failed for directory structures with spaces, like this one:
- Eclipse Workspace
- MyProgram
- bin
- Test.class
This is due to the fact that the URL representation uses %20 to denote a whitespace, which is not recognized by the file constructor.
Is there an efficient and reliable way to get the (classpath) location of a Java class, which works for both directory structures and Jar files?
Note that I don't need the exact file of the exact class - only the container! I use this code to locate rt.jar and the language library for using them in a compiler.
Slight modification in your code should work here. You can try below code:
public static File getFileLocation(Class<?> klass)
{
String classLocation = '/' + klass.getName().replace('.', '/') + ".class";
URL url = klass.getResource(classLocation);
String path = url.getPath();
int index = path.lastIndexOf(classLocation);
if (index < 0)
{
return null;
}
String fileCol = "file:";
//add "file:" for local files
if (path.indexOf(fileCol) == -1)
{
path = fileCol + path;
index+=fileCol.length();
}
// Jar Handling
if (path.charAt(index - 1) == '!')
{
index--;
}
else
{
index++;
}
String newPath = path.substring(0, index);
System.out.println(url.toExternalForm());
URI uri = URI.create(newPath).normalize();
return new File(uri);
}
Hope this will help.
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 reflectively get all of the packages in the project? I started with Package.getPackages(), but that only got all the packages associated to the current package. Is there a way to do this?
#PhilippWendler's comment led me to a method of accomplishing what I needed. I tweaked the method a little to make it recursive.
/**
* Recursively fetches a list of all the classes in a given
* directory (and sub-directories) that have the #UnitTestable
* annotation.
* #param packageName The top level package to search.
* #param loader The class loader to use. May be null; we'll
* just grab the current threads.
* #return The list of all #UnitTestable classes.
*/
public List<Class<?>> getTestableClasses(String packageName, ClassLoader loader) {
// State what package we are exploring
System.out.println("Exploring package: " + packageName);
// Create the list that will hold the testable classes
List<Class<?>> ret = new ArrayList<Class<?>>();
// Create the list of immediately accessible directories
List<File> directories = new ArrayList<File>();
// If we don't have a class loader, get one.
if (loader == null)
loader = Thread.currentThread().getContextClassLoader();
// Convert the package path to file path
String path = packageName.replace('.', '/');
// Try to get all of nested directories.
try {
// Get all of the resources for the given path
Enumeration<URL> res = loader.getResources(path);
// While we have directories to look at, recursively
// get all their classes.
while (res.hasMoreElements()) {
// Get the file path the the directory
String dirPath = URLDecoder.decode(res.nextElement()
.getPath(), "UTF-8");
// Make a file handler for easy managing
File dir = new File(dirPath);
// Check every file in the directory, if it's a
// directory, recursively add its viable files
for (File file : dir.listFiles()) {
if (file.isDirectory())
ret.addAll(getTestableClasses(packageName + '.' + file.getName(), loader));
}
}
} catch (IOException e) {
// We failed to get any nested directories. State
// so and continue; this directory may still have
// some UnitTestable classes.
System.out.println("Failed to load resources for [" + packageName + ']');
}
// We need access to our directory, so we can pull
// all the classes.
URL tmp = loader.getResource(path);
System.out.println(tmp);
if (tmp == null)
return ret;
File currDir = new File(tmp.getPath());
// Now we iterate through all of the classes we find
for (String classFile : currDir.list()) {
// Ensure that we only find class files; can't load gif's!
if (classFile.endsWith(".class")) {
// Attempt to load the class or state the issue
try {
// Try loading the class
Class<?> add = Class.forName(packageName + '.' +
classFile.substring(0, classFile.length() - 6));
// If the class has the correct annotation, add it
if (add.isAnnotationPresent(UnitTestable.class))
ret.add(add);
else
System.out.println(add.getName() + " is not a UnitTestable class");
} catch (NoClassDefFoundError e) {
// The class loader could not load the class
System.out.println("We have found class [" + classFile + "], and couldn't load it.");
} catch (ClassNotFoundException e) {
// We couldn't even find the damn class
System.out.println("We could not find class [" + classFile + ']');
}
}
}
return ret;
}
It's possible but tricky and expensive, since you need to walk the classpath yourself. Here is how TestNG does it, you can probably extract the important part for yourself:
https://github.com/cbeust/testng/blob/master/src/main/java/org/testng/internal/PackageUtils.java
This approach prints all packages only (at least a root "packageName" has to be given first).
It is derived from above.
package devTools;
import java.io.File;
import java.io.IOException;
import java.net.URL;
import java.net.URLDecoder;
import java.util.ArrayList;
import java.util.Enumeration;
import java.util.List;
public class DevToolUtil {
/**
* Method prints all packages (at least a root "packageName" has to be given first).
*
* #see http://stackoverflow.com/questions/9316726/reflectively-get-all-packages-in-a-project
* #since 2016-12-05
* #param packageName
* #param loader
* #return List of classes.
*/
public List<Class<?>> getTestableClasses(final String packageName, ClassLoader loader) {
System.out.println("Exploring package: " + packageName);
final List<Class<?>> ret = new ArrayList<Class<?>>();
if (loader == null) {
loader = Thread.currentThread().getContextClassLoader();
}
final String path = packageName.replace('.', '/');
try {
final Enumeration<URL> res = loader.getResources(path);
while (res.hasMoreElements()) {
final String dirPath = URLDecoder.decode(res.nextElement().getPath(), "UTF-8");
final File dir = new File(dirPath);
if (dir.listFiles() != null) {
for (final File file : dir.listFiles()) {
if (file.isDirectory()) {
final String packageNameAndFile = packageName + '.' + file.getName();
ret.addAll(getTestableClasses(packageNameAndFile, loader));
}
}
}
}
} catch (final IOException e) {
System.out.println("Failed to load resources for [" + packageName + ']');
}
return ret;
}
public static void main(final String[] args) {
new DevToolUtil().getTestableClasses("at", null);
}
}
May be off-topic (because it is not exactly in terms of java "reflection") ...
However, how about the following solution:
Java packages can be treated as folders (or directories on Linux\UNIX).
Assuming that you have a root package and its absolute path is known, you can recursively print all sub-folders that have *.java or *.class files using the following batch as the basis:
#echo off
rem List all the subfolders under current dir
FOR /R "." %%G in (.) DO (
Pushd %%G
Echo now in %%G
Popd )
Echo "back home"
You can wrap this batch in java or if you run on Linux\Unix rewrite it in some shell script.
Good Luck!
Aviad.