How to get current working directory in Java?

One way would be to use the system property System.getProperty("user.dir"); this will give you "The current working directory when the properties were initialized". This is probably what you want. to find out where the java command was issued, in your case in the directory with the files to process, even though the actual .jar file might reside somewhere else on the machine. Having the directory of the actual .jar file isn't that useful in most cases.

The following will print out the current directory from where the command was invoked regardless where the .class or .jar file the .class file is in.

public class Test
{
    public static void main(final String[] args)
    {
        final String dir = System.getProperty("user.dir");
        System.out.println("current dir = " + dir);
    }
}  

if you are in /User/me/ and your .jar file containing the above code is in /opt/some/nested/dir/ the command java -jar /opt/some/nested/dir/test.jar Test will output current dir = /User/me.

You should also as a bonus look at using a good object oriented command line argument parser. I highly recommend JSAP, the Java Simple Argument Parser. This would let you use System.getProperty("user.dir") and alternatively pass in something else to over-ride the behavior. A much more maintainable solution. This would make passing in the directory to process very easy to do, and be able to fall back on user.dir if nothing was passed in.

File currentDirectory = new File(new File(".").getAbsolutePath());
System.out.println(currentDirectory.getCanonicalPath());
System.out.println(currentDirectory.getAbsolutePath());

Prints something like:

/path/to/current/directory
/path/to/current/directory/.

Note that File.getCanonicalPath() throws a checked IOException but it will remove things like ../../../

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());
    }
}

---

Update as per the comment of the OP:

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.

That would require hardcoding/knowing their relative path in your program. Rather consider adding its path to the classpath so that you can use ClassLoader#getResource()

File classpathRoot = new File(classLoader.getResource("").getPath());
File[] csvFiles = classpathRoot.listFiles(new FilenameFilter() {
    @Override public boolean accept(File dir, String name) {
        return name.endsWith(".csv");
    }
});

Or to pass its path as main() argument.