Как найти путь к папке java

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();

answered Jun 25, 2018 at 16:29

This is the solution for me

File currentDir = new File("");

answered May 28, 2013 at 23:09

What makes you think that c:windowssystem32 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:windowssystem32 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.

answered Feb 2, 2011 at 5:39

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()

answered Jan 26, 2015 at 12:13

answered Dec 28, 2013 at 3:11

I hope you want to access the current directory including the package i.e. If your Java program is in c:myAppcomfoosrcserviceMyTest.java and you want to print until c:myAppcomfoosrcservice 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.

answered Sep 27, 2012 at 10:09

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.

answered Dec 17, 2015 at 22:29

SijeDeHaanSijeDeHaan

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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;
    }

answered Nov 20, 2016 at 17:04

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 😀

Best,

answered Apr 16, 2015 at 4:39

Volodya LombrozoVolodya Lombrozo

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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();
    }
}

answered May 18, 2020 at 0:54

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:Windowssystem32, 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:Windowssystem32, 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.

answered Sep 15, 2020 at 22:20

Avon97Avon97

4712 bronze badges

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

WortigWortig

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Adam D.Adam D.

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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()

answered Mar 31, 2015 at 19:35

Marko StojkovicMarko Stojkovic

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Let’s say I have my main class in C:UsersJustianDocuments. How can I get my program to show that it’s in C:UsersJustianDocuments?

Hard-Coding is not an option- it needs to be adaptable if it’s moved to another location.

I want to dump a bunch of CSV files in a folder, have the program recognize all the files, then load the data and manipulate them. I really just want to know how to navigate to that folder.

If you want to get your current working directory then use the following line

System.out.println(new File("").getAbsolutePath());

answered Feb 16, 2018 at 3:28