How to get the current working directory in Java?

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

The following works on Java 7 and up (see here for documentation).

import java.nio.file.Paths;

Paths.get(".").toAbsolutePath().normalize().toString();

Answer 4

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();

Answer 5

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();

Answer 6

This is the solution for me

File currentDir = new File("");

Answer 7

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.

Answer 8

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());
    }
}

Answer 9

this.getClass().getClassLoader().getResource("").getPath()

Answer 10

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()

Answer 11

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

Answer 12

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.

Answer 13

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.

Answer 14

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

Answer 15

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;
    }

Answer 16

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,

Answer 17

For Java 11 you could also use:

var path = Path.of(".").toRealPath();

Answer 18

This is a very confuse topic, and we need to understand some concepts before providing a real solution.

1. The File, and NIO File Api approaches with relative paths "" or "." uses internally the system parameter "user.dir" value to determine the return location.

2. 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.

3. 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...

Answer 19

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")...

Answer 20

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();
    }
}

Answer 21

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.

Answer 22

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

Answer 23

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

Answer 24

You might use new File("./"). This way isDirectory() returns true (at least on Windows platform). On the other hand new File("") isDirectory() returns false.

Answer 25

Not exactly sure what you're trying to accomplish but the above answers work on the current working directory where you are asking from not from where the program is actually located.

For instance:

$> mkdir -p /tmp/paths/check 
$> cat > /tmp/paths/check/GetPath.java
import java.nio.file.Paths;
import java.nio.file.FileSystems;

public class GetPath {
// Note: these methods work slightly differently, the last two will provide the working dir
// for where this program was launched, while the first will get the cwd from where this
// program resides.
public static void main(String[] args) {
        String rootPath = Thread.currentThread().getContextClassLoader().getResource("").getPath();
        System.out.println("Thread rootPath is "+rootPath);
        System.out.println("System property user.dir is "+System.getProperty("user.dir"));
        System.out.println("NIO Paths way is "+Paths.get("").toAbsolutePath());
        System.out.println("NIO FileSystems way is "+FileSystems.getDefault().getPath("").toAbsolutePath());
}
}

Then if you run it from /tmp, you will see:

$> cd /tmp/paths/check; javac GetPath.java
$> cd /tmp
$> java -cp paths/check GetPath
Thread rootPath is /tmp/paths/check/
system property user.dir is /tmp
NIO Paths way is/tmp
NIO FileSystems way is/tmp

So the first method using the ContextClassLoader may be more appropriate depending on what you are trying to accomplish.

Answer 26

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()

Answer 27

System.getProperty("java.class.path")