Changing the current working directory in Java?

user_92222 0
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How can I change the current working directory from within a Java program? Everything I've been able to find about the issue claims that you simply can't do it, but I can't believe that that's really the case.

I have a piece of code that opens a file using a hard-coded relative file path from the directory it's normally started in, and I just want to be able to use that code from within a different Java program without having to start it from within a particular directory. It seems like you should just be able to call System.setProperty( "user.dir", "/path/to/dir" ), but as far as I can figure out, calling that line just silently fails and does nothing.

I would understand if Java didn't allow you to do this, if it weren't for the fact that it allows you to get the current working directory, and even allows you to open files using relative file paths....

working-directory java
14 Answers

There is no reliable way to do this in pure Java. Setting the user.dir property via System.setProperty() or java -Duser.dir=... does seem to affect subsequent creations of Files, but not e.g. FileOutputStreams.

The File(String parent, String child) constructor can help if you build up your directory path separately from your file path, allowing easier swapping.

An alternative is to set up a script to run Java from a different directory, or use JNI native code as suggested below.

The relevant OpenJDK bug was closed in 2008 as "will not fix".

user_135310
PhiLho_154590

If I understand correctly, a Java program starts with a copy of the current environment variables. Any changes via System.setProperty(String, String) are modifying the copy, not the original environment variables. Not that this provides a thorough reason as to why Sun chose this behavior, but perhaps it sheds a little light...

user_416190

The working directory is a operating system feature (set when the process starts). Why don't you just pass your own System property (-Dsomeprop=/my/path) and use that in your code as the parent of your File:

File f = new File ( System.getProperty("someprop"), myFilename)
user_1036050

The smarter/easier thing to do here is to just change your code so that instead of opening the file assuming that it exists in the current working directory (I assume you are doing something like new File("blah.txt"), just build the path to the file yourself.

Let the user pass in the base directory, read it from a config file, fall back to user.dir if the other properties can't be found, etc. But it's a whole lot easier to improve the logic in your program than it is to change how environment variables work.

user_42490

You can change the process's actual working directory using JNI or JNA.

With JNI, you can use native functions to set the directory. The POSIX method is chdir(). On Windows, you can use SetCurrentDirectory().

With JNA, you can wrap the native functions in Java binders.

For Windows:

private static interface MyKernel32 extends Library {
    public MyKernel32 INSTANCE = (MyKernel32) Native.loadLibrary("Kernel32", MyKernel32.class);

    /** BOOL SetCurrentDirectory( LPCTSTR lpPathName ); */
    int SetCurrentDirectoryW(char[] pathName);
}

For POSIX systems:

private interface MyCLibrary extends Library {
    MyCLibrary INSTANCE = (MyCLibrary) Native.loadLibrary("c", MyCLibrary.class);

    /** int chdir(const char *path); */
    int chdir( String path );
}
user_2020090

The other possible answer to this question may depend on the reason you are opening the file. Is this a property file or a file that has some configuration related to your application?

If this is the case you may consider trying to load the file through the classpath loader, this way you can load any file Java has access to.

Zorkerman_85630
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