How to get the Desktop path in java

I think this will work only on an English language Windows installation:

How can I make this work for non English Windows?

8 Answers 8

I use a french version of Windows and with it the instruction:

works fine for me.

I think this is the same question. but I’m not sure!:

Reading it I would expect that solution to return the user.home, but apparently not, and the link in the answer comments back that up. Haven’t tried it myself.

I guess by using JFileChooser the solution will require a non-headless JVM, but you are probably running one of them.

This is for Windows only. Launch REG.EXE and capture its output :

Seems not that easy.

But you could try to find an anwser browsing the code of some open-source projects, e.g. on Koders. I guess all the solutions boil down to checking the HKEY_CURRENT_USER\Software\Microsoft\Windows\CurrentVersion\Explorer\Shell Folders\Desktop path in the Windows registry. And probably are Windows-specific.

If you need a more general solution I would try to find an open-source application you know is working properly on different platforms and puts some icons on the user’s Desktop.

there are 2 things.

1. you are using the wrong slash. for windows it’s \ not / .
2. i’m using RandomAccesFile and File to manage fles and folders, and it requires double slash ( \\ ) to separate the folders name.

What is the alternative for

I’m trying to use the command prompt to move some files, I am used to the linux terminal where I use

to specify the my home directory I’ve looked everywhere but I couldn’t seem to find it for windows command prompt ( Documents and Settings\[user] )

11 Answers 11

You’re going to be disappointed: %userprofile%

You can use other terminals, though. Powershell, which I believe you can get on XP and later (and comes preinstalled with Win7), allows you to use

for home directory.

You can %HOMEDRIVE%%HOMEPATH% for the drive + \docs settings\username or \users\username .

You can use %systemdrive%%homepath% environment variable to accomplish this.

The two command variables when concatenated gives you the desired user’s home directory path as below:

Running echo %systemdrive% on command prompt gives:

Running echo %homepath% on command prompt gives:

When used together it becomes:

Update — better version 18th July 2019.

Final summary, even though I’ve moved on to powershell for most windows console work anyway, but I decided to wrap this old cmd issue up, I had to get on a cmd console today, and the lack of this feature really struck me. This one finally works with spaces as well, where my previous answer would fail.

In addition, this one now is also able to use

as a prefix for other home sub-folders too, and it swaps forward-slashes to back-slashes as well. So here it is;

Step 1. Create these doskey macros, somewhere they get picked up every time cmd starts up.

Step 2. Create the cdtilde.bat file and put it somewhere in your PATH

Tested fine with;

Oh, also it allows lazy quoting, which I found useful, even when spaces are in the folder path names, since it wraps all of the arguments as if it was one long string. Which means just an initial quote also works, or completely without quotes also works.

All other stuff below may be ignored now, it is left for historical reasons — so I dont make the same mistakes again

old update 19th Oct 2018.
In case anyone else tried my approach, my original answer below didn’t handle spaces, eg, the following failed.

I think there must be a way to solve that. Will post again if I can improve my answer. (see above, I finally got it all working the way I wanted it to.)

My Original Answer, still needed work. 7th Oct 2018.
I was just trying to do it today, and I think I got it, this is what I think works well;

First, some doskey macros;

How to get the current working directory in Java?

I want to access my current working directory using java.

My output is not correct because C drive is not my current directory.

How to get the current directory ?

24 Answers 24

Code :

This will print the absolute path of the current directory from where your application was initialized.

Explanation:

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.

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.

This outputs Current relative 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).

This will give you the path of your current working directory:

And this will give you the path to a file called «Foo.txt» in the working directory:

Edit : To obtain an absolute path of current directory:

* Update * To get current working directory:

Java 11 and newer

This solution is better than others and more portable:

This is the solution for me

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.

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() .

generally, as a File object:

you may want to have full qualified string like «D:/a/b/c» doing:

I’m on Linux and get same result for both of these approaches:

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.

If your Class with main would be called MainClass , then try:

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

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:

Note: This code is only tested in Windows with Oracle JRE.

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 .

• 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:

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

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 😀

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

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.

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..

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

What is the home directory for JDK?

I am having a problem in selecting home directory for JDK. Even though I have selected the following path: «C:\Program Files (x86)\Java\jre1.8.0_91», it keeps showing the error: «The selected directory is not a valid home for JDK». My java folder has two folders- bin and lib. Window for selecting home directory

Kindly help me in selecting home directory. Thank you!

4 Answers 4

You have currently located a JRE, or Java Runtime Environment. That does indeed contain a java executable and can launch JVMs, but it’s not made for development, rather more as a «Java client». For example if you look into it’s /bin folder, you won’t find the javac executable needed to compile.

What you need is a JDK, or Java Development Toolkit. It contains all the JRE contains, with additional tools for development. You can find it for download on the Oracle website.

Your home directory will jdk not jre. If your jdk is installed properly then there will be a jdk folder. Try setting C:\Program Files (x86)\Java\jdk1.8.0_91\bin

Your home directory for JDK would be «C:\Program Files (x86)\Java\jre1.8.0_91\bin»

try this on windows to know the home of your jdk

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Getting My Documents path in Java

I need to find my documents path using Java. The following code doesn’t give me «accurate» loation

What should be the other way around?

P.S: I don’t want to use the JFileChooser Dirty trick.

8 Answers 8

You can get it using a registry query, no need for JNA or admin rights for that.

Obviously this will fail on anything other than Windows, and I am not certain whether this works for Windows XP.

EDIT: Put this in a working sequence of code:

Note this will lock the process until «reg query» is done, which might cause trouble dependeing on what you are doing.

That’s easy, JFileChooser finds it for you

I hope this helps someone

Since the most upvoted answer from @xchiltonx uses JFileChooser I would like to add that, regarding performance, this is faster than using JFileChooser :

In my PC, JFileChooser aproach needed 300ms, and calling FileSystemView directly needed less than 100ms.

Note: The question is a possible duplicate of How to find “My Documents” folder in Java

Note in 2020: The ShellFolder class seems to be missing on Linux (tested with openjdk8), so IvanRF’s answer is likely better.

Original answer:

The best way I’ve found is to use AWTs:

I have redirected my Documents folder to the D: drive, and it successfully fetches this directory. It also does so in about 40 ms (on my machine). Using FileSystemView takes about 48 ms, and new JFileChooser() about 250 ms.

All three of these methods actually use ShellFolder under the hood, and the difference with FileSystemView is negligible, but calling it directly avoids the overhead of the other two.

Note: You can also cast this directly to File instead of implicitly getting the toString() method of it, which can help you further: