I have created a plugin. It needs to access 2 files from res folder. Now the trick here is, res folder is not inside plugin. For example, i have published plugin locally, so it now visible under any project you create, inside Tools Menu. What i want is, any new project i create, it will have 2 files facility.json & groups.json inside res folder, now my plugin should be able to access these files. I have tried different approach like as shown below, but it always fails, saying "File not found" exception. If anyone can help me out in this. Its appreciated. Thank you.
private static File getFileFromResource(String fileName) throws URISyntaxException {
ClassLoader classLoader = Validator.class.getClassLoader();
URL resource = classLoader.getResource(fileName);
if (resource == null) {
throw new IllegalArgumentException("file not found! " + fileName);
} else {
// failed if files have whitespaces or special characters
//return new File(resource.getFile());
return new File(resource.toURI());
}
}
private static InputStream getFileFromResourceAsStream(String fileName) {
// The class loader that loaded the class
ClassLoader classLoader = Validator.class.getClassLoader();
InputStream inputStream = classLoader.getResourceAsStream(fileName);
// the stream holding the file content
if (inputStream == null) {
throw new IllegalArgumentException("file not found! " + fileName);
} else {
return inputStream;
}
}
That's the neat part, you don't.
you have to add them by yourself each time you make a new project since res folder isn't a plugin, instead res is for resource (assuming you use res as resource)
Try this by manually add each facility.json & groups.json with
InputStream IS = getClass().getResourceAsStream("/res/file name");
or
InputStream IS = getClass().getResourceAsStream("root/dir/file name");
your plugin already published locally so maybe adding address might be help
if still not, try to save your clean project that only have facility.json & groups.json inside res folder as your project template. it depends on your IDE,
Intelij IDEA -Jetbrain : https://www.jetbrains.com/help/idea/saving-project-as-template.html
NetBeans IDE -Apache : https://netbeans.apache.org/tutorials/nbm-filetemplates.html
as my experience, im creating my own netbeans template for new project
So we got this assignment in a basic java programming course and we're supposed to implement a kind of card deck. To help us with this they have given us resources that will present a GUI on the screen, but when running my program I get a IOException that says that it can't read the input file, most likely since the pathname is wrong. And I dont know how to fix it, we're not even supposed to be in meddling with this code. The error is thrown in this method:
private Image getImg(Card aCard) {
File pathToFile = null;
if (aCard == null) {
pathToFile = new File("cardset-oxymoron/shade.gif");
} else {
String suits = "cdhs";
char c = suits.charAt(aCard.getSuit());
String fileName = String.format("%s/%02d%c.gif", "cardset-oxymoron", aCard.getRank(), c);
pathToFile = new File(fileName);
}
Image img = null;
try {
img = ImageIO.read(pathToFile);
} catch (IOException ex) {
System.err.println("Failed to create image");
ex.printStackTrace();
}
return img;
}
And according to the error stack(?) it is at line 99, which is the
img = ImageIO.read(pathToFile);
line
The folder that the cards are in is inside the project folder, right in between bin and src. using IntelliJ debugger I can see that the the pathToFile is "cardset-oxymoron\02d.gif". The filename is correct as all the cards are "[01-13][c/d/h/s].gif". When I rightclicked and copied the path to the files inside IntelliJ it was using forwardsslashes and not backslashes. But then I checked in explorer and it was the other way around... I have no idea where this is going wrong, any input would be greatly appreciated!
According to your code your files are in directory cardset-oxymoron relative to your JVM run directory. I'm not sure about IntelliJ (I work all the time with Eclipse and Maven), but it could be bin directory.
You can check it by put those 2 lines to see what is it actually (somewhere before your actual code)
File currentDir = new File("./");
System.out.println(currentDir.getAbsolutePath());
Then your cardset-oxymoron must be in that directory. Or you can change file path appropriately.
E.g. if currentDir is bin then pathToFile will be
pathToFile = new File("../cardset-oxymoron/shade.gif");
as well as fileName for other case.
I want to access my current working directory using Java.
My code:
String currentPath = new java.io.File(".").getCanonicalPath();
System.out.println("Current dir:" + currentPath);
String currentDir = System.getProperty("user.dir");
System.out.println("Current dir using System:" + currentDir);
Output:
Current dir: C:\WINDOWS\system32
Current dir using System: C:\WINDOWS\system32
My output is not correct because the C drive is not my current directory.
How to get the current directory?
Code :
public class JavaApplication {
public static void main(String[] args) {
System.out.println("Working Directory = " + System.getProperty("user.dir"));
}
}
This will print the absolute path of the current directory from where your application was initialized.
Explanation:
From the documentation:
java.io package resolve relative pathnames using current user directory. The current directory is represented as system property, that is, user.dir and is the directory from where the JVM was invoked.
See: Path Operations (The Java™ Tutorials > Essential Classes > Basic I/O).
Using java.nio.file.Path and java.nio.file.Paths, you can do the following to show what Java thinks is your current path. This for 7 and on, and uses NIO.
Path currentRelativePath = Paths.get("");
String s = currentRelativePath.toAbsolutePath().toString();
System.out.println("Current absolute path is: " + s);
This outputs:
Current absolute path is: /Users/george/NetBeansProjects/Tutorials
that in my case is where I ran the class from.
Constructing paths in a relative way, by not using a leading separator to indicate you are constructing an absolute path, will use this relative path as the starting point.
The following works on Java 7 and up (see here for documentation).
import java.nio.file.Paths;
Paths.get(".").toAbsolutePath().normalize().toString();
This will give you the path of your current working directory:
Path path = FileSystems.getDefault().getPath(".");
And this will give you the path to a file called "Foo.txt" in the working directory:
Path path = FileSystems.getDefault().getPath("Foo.txt");
Edit :
To obtain an absolute path of current directory:
Path path = FileSystems.getDefault().getPath(".").toAbsolutePath();
* Update *
To get current working directory:
Path path = FileSystems.getDefault().getPath("").toAbsolutePath();
Java 11 and newer
This solution is better than others and more portable:
Path cwd = Path.of("").toAbsolutePath();
Or even
String cwd = Path.of("").toAbsolutePath().toString();
This is the solution for me
File currentDir = new File("");
What makes you think that c:\windows\system32 is not your current directory? The user.dir property is explicitly to be "User's current working directory".
To put it another way, unless you start Java from the command line, c:\windows\system32 probably is your CWD. That is, if you are double-clicking to start your program, the CWD is unlikely to be the directory that you are double clicking from.
Edit: It appears that this is only true for old windows and/or Java versions.
Use CodeSource#getLocation().
This works fine in JAR files as well. You can obtain CodeSource by ProtectionDomain#getCodeSource() and the ProtectionDomain in turn can be obtained by Class#getProtectionDomain().
public class Test {
public static void main(String... args) throws Exception {
URL location = Test.class.getProtectionDomain().getCodeSource().getLocation();
System.out.println(location.getFile());
}
}
this.getClass().getClassLoader().getResource("").getPath()
generally, as a File object:
File getCwd() {
return new File("").getAbsoluteFile();
}
you may want to have full qualified string like "D:/a/b/c" doing:
getCwd().getAbsolutePath()
I'm on Linux and get same result for both of these approaches:
#Test
public void aaa()
{
System.err.println(Paths.get("").toAbsolutePath().toString());
System.err.println(System.getProperty("user.dir"));
}
Paths.get("") docs
System.getProperty("user.dir") docs
I hope you want to access the current directory including the package i.e. If your Java program is in c:\myApp\com\foo\src\service\MyTest.java and you want to print until c:\myApp\com\foo\src\service then you can try the following code:
String myCurrentDir = System.getProperty("user.dir")
+ File.separator
+ System.getProperty("sun.java.command")
.substring(0, System.getProperty("sun.java.command").lastIndexOf("."))
.replace(".", File.separator);
System.out.println(myCurrentDir);
Note: This code is only tested in Windows with Oracle JRE.
On Linux when you run a jar file from terminal, these both will return the same String: "/home/CurrentUser", no matter, where youre jar file is. It depends just on what current directory are you using with your terminal, when you start the jar file.
Paths.get("").toAbsolutePath().toString();
System.getProperty("user.dir");
If your Class with main would be called MainClass, then try:
MainClass.class.getProtectionDomain().getCodeSource().getLocation().getFile();
This will return a String with absolute path of the jar file.
Using Windows user.dir returns the directory as expected, but NOT when you start your application with elevated rights (run as admin), in that case you get C:\WINDOWS\system32
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;
}
assume that you're trying to run your project inside eclipse, or netbean or stand alone from command line. I have write a method to fix it
public static final String getBasePathForClass(Class<?> clazz) {
File file;
try {
String basePath = null;
file = new File(clazz.getProtectionDomain().getCodeSource().getLocation().toURI().getPath());
if (file.isFile() || file.getPath().endsWith(".jar") || file.getPath().endsWith(".zip")) {
basePath = file.getParent();
} else {
basePath = file.getPath();
}
// 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 netbean
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;
} catch (URISyntaxException e) {
throw new RuntimeException("Cannot firgue out base path for class: " + clazz.getName());
}
}
To use, everywhere you want to get base path to read file, you can pass your anchor class to above method, result may be the thing you need :D
Best,
For Java 11 you could also use:
var path = Path.of(".").toRealPath();
This is a very confuse topic, and we need to understand some concepts before providing a real solution.
The File, and NIO File Api approaches with relative paths "" or "." uses internally the system parameter "user.dir" value to determine the return location.
The "user.dir" value is based on the USER working directory, and the behavior of that value depends on the operative system, and the way the jar is executed.
For example, executing a JAR from Linux using a File Explorer (opening it by double click) will set user.dir with the user home directory, regardless of the location of the jar. If the same jar is executed from command line, it will return the jar location, because each cd command to the jar location modified the working directory.
Having said that, the solutions using Java NIO, Files or "user.dir" property will work for all the scenarios in the way the "user.dir" has the correct value.
String userDirectory = System.getProperty("user.dir");
String userDirectory2 = new File("").getAbsolutePath();
String userDirectory3 = Paths.get("").toAbsolutePath().toString();
We could use the following code:
new File(MyApp.class.getProtectionDomain()
.getCodeSource()
.getLocation()
.toURI().getPath())
.getParent();
to get the current location of the executed JAR, and personally I used the following approach to get the expected location and overriding the "user.dir" system property at the very beginning of the application. So, later when the other approaches are used, I will get the expected values always.
More details here -> https://blog.adamgamboa.dev/getting-current-directory-path-in-java/
public class MyApp {
static {
//This static block runs at the very begin of the APP, even before the main method.
try{
File file = new File(MyApp.class.getProtectionDomain().getCodeSource()
.getLocation().toURI().getPath());
String basePath = file.getParent();
//Overrides the existing value of "user.dir"
System.getProperties().put("user.dir", basePath);
}catch(URISyntaxException ex){
//log the error
}
}
public static void main(String args []){
//Your app logic
//All these approaches should return the expected value
//regardless of the way the jar is executed.
String userDirectory = System.getProperty("user.dir");
String userDirectory2 = new File("").getAbsolutePath();
String userDirectory3 = Paths.get("").toAbsolutePath().toString();
}
}
I hope this explanation and details are helpful to others...
Current working directory is defined differently in different Java implementations. For certain version prior to Java 7 there was no consistent way to get the working directory. You could work around this by launching Java file with -D and defining a variable to hold the info
Something like
java -D com.mycompany.workingDir="%0"
That's not quite right, but you get the idea. Then System.getProperty("com.mycompany.workingDir")...
This is my silver bullet when ever the moment of confusion bubbles in.(Call it as first thing in main). Maybe for example JVM is slipped to be different version by IDE. This static function searches current process PID and opens VisualVM on that pid. Confusion stops right there because you want it all and you get it...
public static void callJVisualVM() {
System.out.println("USER:DIR!:" + System.getProperty("user.dir"));
//next search current jdk/jre
String jre_root = null;
String start = "vir";
try {
java.lang.management.RuntimeMXBean runtime =
java.lang.management.ManagementFactory.getRuntimeMXBean();
String jvmName = runtime.getName();
System.out.println("JVM Name = " + jvmName);
long pid = Long.valueOf(jvmName.split("#")[0]);
System.out.println("JVM PID = " + pid);
Runtime thisRun = Runtime.getRuntime();
jre_root = System.getProperty("java.home");
System.out.println("jre_root:" + jre_root);
start = jre_root.concat("\\..\\bin\\jvisualvm.exe " + "--openpid " + pid);
thisRun.exec(start);
} catch (Exception e) {
System.getProperties().list(System.out);
e.printStackTrace();
}
}
This isn't exactly what's asked, but here's an important note: When running Java on a Windows machine, the Oracle installer puts a "java.exe" into C:\Windows\system32, and this is what acts as the launcher for the Java application (UNLESS there's a java.exe earlier in the PATH, and the Java app is run from the command-line). This is why File(".") keeps returning C:\Windows\system32, and why running examples from macOS or *nix implementations keep coming back with different results from Windows.
Unfortunately, there's really no universally correct answer to this one, as far as I have found in twenty years of Java coding unless you want to create your own native launcher executable using JNI Invocation, and get the current working directory from the native launcher code when it's launched. Everything else is going to have at least some nuance that could break under certain situations.
Try something like this I know I am late for the answer but this obvious thing happened in java8 a new version from where this question is asked but..
The code
import java.io.File;
public class Find_this_dir {
public static void main(String[] args) {
//some sort of a bug in java path is correct but file dose not exist
File this_dir = new File("");
//but these both commands work too to get current dir
// File this_dir_2 = new File(this_dir.getAbsolutePath());
File this_dir_2 = new File(new File("").getAbsolutePath());
System.out.println("new File(" + "\"\"" + ")");
System.out.println(this_dir.getAbsolutePath());
System.out.println(this_dir.exists());
System.out.println("");
System.out.println("new File(" + "new File(" + "\"\"" + ").getAbsolutePath()" + ")");
System.out.println(this_dir_2.getAbsolutePath());
System.out.println(this_dir_2.exists());
}
}
This will work and show you the current path but I don't now why java fails to find current dir in new File(""); besides I am using Java8 compiler...
This works just fine I even tested it new File(new File("").getAbsolutePath());
Now you have current directory in a File object so (Example file object is f then),
f.getAbsolutePath() will give you the path in a String varaible type...
Tested in another directory that is not drive C works fine
My favorite method is to get it from the system environment variables attached to the current running process. In this case, your application is being managed by the JVM.
String currentDir = System.getenv("PWD");
/*
/home/$User/Documents/java
*/
To view other environment variables that you might find useful like, home dir, os version ........
//Home directory
String HomeDir = System.getEnv("HOME");
//Outputs for unix
/home/$USER
//Device user
String user = System.getEnv("USERNAME");
//Outputs for unix
$USER
The beautiful thing with this approach is that all paths will be resolved for all types of OS platform
You might use new File("./"). This way isDirectory() returns true (at least on Windows platform). On the other hand new File("") isDirectory() returns false.
None of the answers posted here worked for me. Here is what did work:
java.nio.file.Paths.get(
getClass().getProtectionDomain().getCodeSource().getLocation().toURI()
);
Edit: The final version in my code:
URL myURL = getClass().getProtectionDomain().getCodeSource().getLocation();
java.net.URI myURI = null;
try {
myURI = myURL.toURI();
} catch (URISyntaxException e1)
{}
return java.nio.file.Paths.get(myURI).toFile().toString()
System.getProperty("java.class.path")
I'm trying to load an image from an executable JAR file.
I've followed the information from here, then the information from here.
This is the function to retrieve the images:
public static ImageIcon loadImage(String fileName, Object o) {
BufferedImage buff = null;
try {
buff = ImageIO.read(o.getClass().getResource(fileName));
// Also tried getResourceAsStream
} catch (IOException e) {
e.printStackTrace();
return null;
}
if (buff == null) {
System.out.println("Image Null");
return null;
}
return new ImageIcon(buff);
}
And this is how it's being called:
logo = FileConverter.loadImage("/pictures/Logo1.png", this);
JFrame.setIconImage(logo.getImage());
With this being a simple Object.
I'm also not getting a NullPointerException unless it is being masked by the UI.
I checked the JAR file and the image is at:
/pictures/Logo1.png
This current code works both in eclipse and when it's been exported to a JAR and run in a terminal, but doesn't work when the JAR is double clicked, in which case the icon is the default JFrame icon.
Thanks for you're help. It's probably only me missing something obvious.
I had a similar problem once, which turned out to be down to issues relative addressing and my path being in the wrong place somehow. I dug this out of some old code I wrote that made it use an absolute path. That seemed to fix my problem; maybe it will work for you.
String basePath = (new File(".")).getAbsolutePath();
basePath = basePath.substring(0, basePath.length()-1);
FileConverter.loadImage(basePath+"/pictures/Logo1.png", this);
I'm trying to run a External Jar file, without actually inserting it into my jar itself. Because the jar file needs to be located in the same folder as the main jar file.
So, Within the main jar file I want to execute the other executable jar file, And I need to be able to know when the jar file is ended AND when you close the main jar file, the jar file that is started within the jar file needs to be closed to,
I currently do that by using this code:
public void LaunchLatestBuild()
{
try {
String path = new File(".").getCanonicalPath() +
"\\externaljar.jar";
List commands = new ArrayList();
commands.add(getJreExecutable().toString());
commands.add("-Xmx" + this.btd_serverram + "M");
commands.add("-Xms" + this.btd_serverram + "M");
commands.add("-jar");
commands.add(path);
int returnint = launch(commands); //It just waits and stops the tread here. And my Runtime.getRuntime().addShutdownHook doesn't get triggerd.
if (returnint != 201) //201 is a custom exit code I 'use' to know when the app needs a restart or not.
{
System.out.println("No restart needed! Closing...");
System.exit(1);
}
else
{
CloseCraftBukkit();
Launcher.main(new String[] { "" });
}
}
catch (Exception e)
{
System.out.println(e.toString());
e.printStackTrace();
}
}
public int launch(List<String> cmdarray) throws IOException, InterruptedException
{
byte[] buffer = new byte[1024];
ProcessBuilder processBuilder = new ProcessBuilder(cmdarray);
processBuilder.redirectErrorStream(true);
this.CBProcess = processBuilder.start();
InputStream in = this.CBProcess.getInputStream();
while (true) {
int r = in.read(buffer);
if (r <= 0) {
break;
}
System.out.write(buffer, 0, r);
}
return this.CBProcess.exitValue();
}
Limitations of this code:
Doesn't close my externaljar.jar java
process on exit of the main
application.
Cannot redirect input if main console, to external jar.
That are the most Important things I need.
I hope someone can tell me how I should do this.
Current source code is available at:
http://code.google.com/p/bukkit-to-date/
Why can't you just set your classpath so that it includes the second jar, and then you can simply use it as a library ? You can even invoke the MainClass.main() method manually, if you really want that to be executed, but from within the same VM and without spawning a separate process.
EDIT: If you don't know the name of the jar file when your application is launched, but you'll only figure that out at runtime, in order to invoke it, create a URLClassLoader provided with the path to your jar file and then:
URLClassLoader urlClassLoader = new URLClassLoader(
new File("/path/to/your/jar/file.jar").toURI().toURL() );
ClassLoader cl = Thread.currentThread().getContextClassLoader();
// switch to your custom CL
Thread.currentThread().setContextClassLoader(urlClassLoader);
// do your stuff with the other jar
// ....................
// now switch back to the original CL
Thread.currentThread().setContextClassLoader(cl);
Or simply grab a reference to a class in that other jar and make use of reflection:
Class<?> c = urlClassLoader.loadClass("org.ogher.packag.ClassFromExternalJar");