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")
We recently had to set up one of the tomcat servers from scratch. Tomcat version is 8.0.20. Deploying a war file, now System.getProperty("mode") returns "null" where it should return PREPROD.
It should read this "mode" from a mode.properties file which is located in the webapps directory. The two lines commented out show another part of code that does not work anymore on the new tomcat server. I replaced it with code that should work.
//String pathOfWebInf = sce.getServletContext().getRealPath("WEB-INF");
//String pathOfLocalhostFile = pathOfWebInf + File.separator + "classes"
// + File.separator;
String pathOfLocalhostFile = this.getClass().getResource("/").getPath();
String mode = System.getProperty("mode");
String fileName = "localhost-oracle.properties." + mode;
StandardPBEStringEncryptor encryptor = new StandardPBEStringEncryptor();
encryptor.setPassword("xxx");
Properties dbProps = new EncryptableProperties(encryptor);
try
{
InputStream is = new FileInputStream(pathOfLocalhostFile + fileName);
dbProps.load(is);
} catch (Exception e)
{
throw new IOException("Could not read properties file " + pathOfLocalhostFile + fileName);
}
System.properties is related to all properties in the Computer where the JVM is running... there is no mode key defined there, that is why you get null as value....
check out all the properties in the pc by doing:
final Properties props = System.getProperties();
props.list(System.out);
and verify yourself, there is no mode key in that map...
You have to load mode.properties first, like this way
private Properties mode=null;
mode = new Properties();
mode.load(new FileInputStream(pathtoMODE));
String mode = mode.getProperty("mode");
I have to archive the HDFS files frequently. The files have to be compressed in the Bunzip format using Java code. Now, what I did is the following:
Move the input files to a local location hdfs.moveToLocalFile
bzip using the bzip2 command.
Move the .bz2 files to the HDFS to another locationhdfs.moveFromLocalFile.
I'm using Hadoop 1.1.2 version. Is there any API available to bzip the files directly, without local copy and BZip?
Also now I'm using the linux shell command to BZip the files. Can somebody help me how to do the BZip command using Java code?
public void addFile(String source, String destination, Configuration paramConfiguration) throws IOException, URISyntaxException {
FileSystem localFileSystem = FileSystem.get(paramConfiguration);
String str1 = paramString1.substring(source.lastIndexOf('/') + 1, source.length());
if (destination.charAt(destination.length() - 1) != '/') {
destination = destination + "/" + str1;
} else {
destination = destination + str1;
}
BZip2Codec localBZip2Codec = new BZip2Codec();
String str2 = localBZip2Codec.getDefaultExtension();
Path localPath = new Path(paramString2 + str2);
CompressionOutputStream localCompressionOutputStream = localBZip2Codec.createOutputStream(localFileSystem.create(localPath));
IOUtils.copyBytes(localFileSystem.open(new Path(paramString1)), localCompressionOutputStream, 4096, true);
}
I set-up a server with FreeSSHd and am able to Putty it, including changing directories and listing files. I have some example .txt files and a folder in the home directory. I set the home directory on my server to "C:\SFTP" using FreeSSHd (as opposed to defining a HOME variable with the directory being "$HOME\").
Apparently, when using JSch,
JSch jsch = new JSch();
session = jsch.getSession(username,host,port);
jsch.addIdentity(key.getAbsolutePath());
java.util.Properties config = new java.util.Properties();
config.put("StrictHostKeyChecking", "no");
session.setConfig(config);
session.setUserInfo(new MyUserInfo());
session.connect();
channel = session.openChannel("sftp");
channel.connect();
channelSftp = (ChannelSftp)channel;
System.out.println("Home: "+channelSftp.getHome());
the last line prints just "Home: /". Any attempts (made immediately after above code) to use
channelSftp.cd(WORKINGDIR);
results in
2: No such file
at com.jcraft.jsch.ChannelSftp.throwStatusError(ChannelSftp.java:2833)
at com.jcraft.jsch.ChannelSftp._stat(ChannelSftp.java:2185)
at com.jcraft.jsch.ChannelSftp.get(ChannelSftp.java:1295)
at com.jcraft.jsch.ChannelSftp.get(ChannelSftp.java:1267)
at test.SFTPTest.main(SFTPTest.java:71)
I think if I got to the root of why JSch does not have the correct home path (or any?) this would work. Also, strangely enough I upload and download files no problem using put() and get().
I've heard all kinds of things where people look into the source and find it does strange things in resolving paths and something with a method called "_realPath()" and superfluous leading/trailing "/" but I don't even have it telling me the home directory is correct after connecting.
Any ideas?
Thank´s everybody for your comments.
OS windows XP,I installed FreeSSHd and setting the default directory, then, when I tryng connect by console ssh, the directory was "/", I writing chdir but directory was: C:\Windows\system32\ this is very confused...
My Java code:
public void recursiveFolderUpload(String sourcePath, String destinationPath) throws FileNotFoundException {
if (c == null || session == null || !session.isConnected() || !c.isConnected()) {
log.debug("Connection to server is closed. Open it first.");
}
try {
// c.put(sourceFile, destinationFile);
// log.info("Upload successfull.");
File sourceFile = new File(sourcePath);
if (sourceFile.isFile()) {
// copy if it is a file
c.cd(destinationPath);
if (!sourceFile.getName().endsWith("."))
c.put(new FileInputStream(sourceFile), sourceFile.getName(), c.OVERWRITE);
} else {
log.info("Inside else " + sourceFile.getName());
File[] files = sourceFile.listFiles();
if (files != null && !sourceFile.getName().startsWith(".")) {
log.info("Directory remote server: " + c.pwd());
c.cd(destinationPath);
SftpATTRS attrs = null;
// check if the directory is already existing
try {
attrs = c.stat(destinationPath + sourceFile.getName());
} catch (Exception e) {
log.warn(destinationPath + sourceFile.getName() + " not found");
//e.printStackTrace();
}
// else create a directory
if (attrs != null) {
log.info("Directory exists IsDir : " + attrs.isDir());
} else {
log.info("Creating dir /" + sourceFile.getName());
c.mkdir(sourceFile.getName());
}
for (File f : files) {
if(!f.getName().contains(".dtd")){
log.info("Uploading file: " + f.getAbsoluteFile());
recursiveFolderUpload(f.getAbsolutePath(), destinationPath + sourceFile.getName() + "/");
}
}
}
}
} catch (SftpException e) {
e.printStackTrace();
}
}
My solution was only put the "/" en the input parameter destinationPath of method called recursiveFolderUpload
In other words, my properties file than so:
properties.host = IP
properties.user = user
properties.pass = pass
properties.port = port
properties.dir = / ---> This points to the directory configured by default in opensshd within windows
Thank so much everthing again.
SFTP directory should be the directory of the current user is not necessarily SFTP services with the directory, I also encountered the same problem, because I am with the directory and the user's default directory settings.
So I'm working on a game, and I need help with my file i/o for savefiles. Currently I have something like this setup to read from them:
public static void savesManagementMenu() {
for (int i = 1; i < 4; i++) {
fileURL = JRPG.class.getResource("Saves/save" + i + ".txt");
System.out.println(fileURL);
if (fileURL != null) {
saveMenuSaveFile = new File(fileURL.getPath());
try {
//System.out.println("File # " + i +" exists.");
saveMenuSaveFileReader = new FileReader(fileURL.getPath());
saveMenuFileScanner = new Scanner(saveMenuSaveFileReader);
saveMenuInfo[i - 1][0] = saveMenuFileScanner.nextLine();
saveMenuFileScanner.nextLine();
saveMenuFileScanner.nextLine();
saveMenuFileScanner.nextLine();
saveMenuInfo[i - 1][1] = saveMenuFileScanner.nextLine();
saveMenuInfo[i - 1][2] = saveMenuFileScanner.nextLine();
} catch (FileNotFoundException ex) {
Logger.getLogger(JRPG.class.getName()).log(Level.SEVERE, null, ex);
}
} else {
saveMenuInfo[i - 1][0] = null;
}
}...
And running/compiling using this method from Netbeans will make the application/game look in "E:\Copy\JRPG\build\classes\jrpg\Saves."
When I clean and build the project, and try to run it via the command line I get a response like this:
jar:file:/C:/Users/Adam/Desktop/New%20folder/JRPG.jar!/jrpg/Saves/save1.txt
Aug 22, 2013 11:54:18 PM jrpg.JRPG savesManagementMenu
SEVERE: null
java.io.FileNotFoundException: file:\C:\Users\Adam\Desktop\New%20folder\JRPG.jar
!\jrpg\Saves\save1.txt (The filename, directory name, or volume label syntax is
incorrect)
And the game just freezes up. The file path for the saves that I want it to look into when I run the compiled code is: C:\Users\Adam\Desktop\New folder\Saves
Which would be the relative file path right? How can I fix this problem so that my compiled code looks in the correct location no matter where I run the file from? (Lets say my friend wanted to run the game from his computer except he put the "New Folder" folder somwhere other than his desktop)
A embedded resource is not a File and can't be treated as one. Also, as far as output files, you shouldn't be trying to save inside your application anyway...
Instead, either save the file to the relative location of your application...
File saveMenuSaveFile = new File("./Saves/save" + i + ".txt");
Or to the users home directory...
String userHome = System.getProperty("user.home");
File saveMenuSaveFile = new File(userHome + "/.YouApplicationName/Saves/save" + i + ".txt");