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System.getProperty("user.dir") alternative

I would like to write JSon files in my Java Application, it works perfectly in my IDE but as soon as I move the directory, the system fails to write the file. I eventually found the problem : The line System.getProperty("user.dir") returns a bad path.
The files are located in C:\wamp\www\myProject during development. Here is my code at the moment :
String url = System.getProperty("user.dir") + "\\src\\json\\Crc.json";
//Returns "C:\wamp\www\myProject\src\json\Crc.json"
After moving my project to C:\Users\myUser\Desktop :
String url = System.getProperty("user.dir") + "\\src\\json\\Crc.json";
//Returns "C:\Users\myUser\src\json\Crc.json"
I would like to have a way to find where my project directory is on the computer at anytime. Do you guys have a solution ?

The C:\wamp\www\myProject folder is just a random place on your hard disk.
Java supports 3 generic places to look for resources:
the current working directory. this is where your command prompt is and what System.getProperty("user.dir") returns, but you cannot rely on that beeing somehow related to any cretain place in the file system, especially not related to the project structure on your own system.
You should only use that if your program has a command line interface and looks for some default file name to work with.
the user home This is what you get when calling System.getProperty("user.home"). On Unix this resoves to $HOME and on Windows to %USERPROFILE%.
This is the best place for files changed at runtime or holding user specific content.
the own code location. Resources in the same package as your class are accessed with getClass().getResource("filenameWithoutPath") But usually you place resources in a special folder in the application root and access it like this: getClass().getResource("/relative/path/from/src/root/filenameWithoutPath").
In your IDE this special folder should be Project/src/main/resources (according to the maven Standard Directory Layout
This is appropriate for some global configurations that you change when creating the delivery package.

Java get current file name EXE

This is my first Stackoverflow question.
I searched across Google for getting the current file name in Java. Most of the sources tell users how to find the current file name if the file is a JAR file, but I'm asking for if the current file is an EXE file.
I saw one EXE answer in Get name of running Jar or Exe, but I don't think it worked.
I use the JSmooth EXE wrapper (launch4j somehow didn't work for me), and Java 8. Is there a straightforward solution to my question? Also, it's nice to explain how it works, or provide a link to the Java documentary about it.
EDIT: To clarify, let's say that I made a Java program and used a JAR wrapper, and I named the resulting EXE "test.exe". I want the Java program be able to give me the current directory of "test.exe", including the filename itself (test.exe).
ANOTHER EDIT: Just to clarify more, go onto your desktop, create a text file, and put some text in it. Save it, and change the text file to an EXE file. Then, try to open it. Windows will give an error. Notice how the title of the message dialog is the file path of the opened file. That is the type of output I want.
Thanks.

Per the JSmooth documentation,
JSmooth also makes some special variable accessible for your application.
Form Meaning
${EXECUTABLEPATH} Replaced by the path to the executable binary. For
instance, if the executable binary launched is located
at c:/program files/jsmooth/test.exe, this variable
is replaced with c:/program files/jsmooth
${EXECUTABLENAME} Replaced by the name of the executable binary. For
instance, if the executable binary launched is located
at c:/program files/jsmooth/test.exe, this variable is
replaced with test.exe
You set these in JSmooth under the "Environment Settings" (the last panel), which allows you to map the variable name. So,
MY_EXECUTABLEPATH=${EXECUTABLEPATH}
MY_EXECUTABLENAME=${EXECUTABLENAME}
In your application, you can get those with
String execPath = System.getProperty("MY_EXECUTABLEPATH");
String execName = System.getProperty("MY_EXECUTABLENAME");

public class JavaApplication1 {
public static void main(String[] args) {
System.out.println("Working Directory = " +
System.getProperty("user.dir"));
}
}
This will print a complete absolute path from where your application has initialized. It is used to get only the DIRECTORY.
If you are using a .exe why do you not create a installer? You can use Inno Setup, this way you are able to specify where do you want to store your .exe, and get it from your application just passing your custom directory

Where is java saving files on Linux?

so I have this problem:
I need to know where is Java saving the files when you create them with new, like this File file = new File ("file.txt"); on Linux?

Linux on not File file = new File ("file.txt") does not create a file on the file system. File is just a file path holder. You need to call file.createNewFile to create a file. Relative paths like file.txt are resolved against the current user directory, typically the directory in which the Java virtual machine was invoked

Assuming the file is actually being created by additional code (as Evgeniy mentioned), you could try checking to see if your current working directory isn't what you expect it to be. To find that directory you could try:
String cwd = System.getProperty("user.dir"));
System.out.println("Current working directory: " + cwd);
or just
System.out.println ("Path to file: " + file.getAbsolutePath());
To see where it should end up.

same place like in Windows: in the current directory

Your mentioned code doesn't create new file physically on drive, but logically yes. If you alter the code to create a new file like file.createNewFile then the directory where you executed the code, a file would be created there.

How do I set a working directory for Java application?

I've written a simple java app, say, with the following code:
String currentDir = new java.io.File(".").getCanonicalPath();
javax.swing.JOptionPane.showMessageDialog(null, currentDir); //This line shows a graphical dialog with the current dir
When I run it through the terminal, it gives me the directory where the jar-file is located. But when I run it using the GUI file manager (that is, right click on the jar-file -> Open With -> OpenJDK Java 7 Runtime) - the working directory is my user home directory (/home/angstrem). How can I set the working directory to be the one, in which the jar-file is situated?

You can do this:
String jarPath = YourClass.class.getProtectionDomain().getCodeSource().getLocation().getPath();
replacing YourClass with an actual class defined in your jar.
You can then make your file paths be relative to jarPath, and your program will work regardless of its working directory.

You can't, and you shouldn't be able to. Consider multi-threading for example. You can only make your application current-path-insensitive.

distributing my java app, HOW do I manage directories

My java applications is almost complete and now I am giving it a finishing touch. Although certain code requires to read data from specific directories on the computer. On my machine I did it in the following way:
FileOutputStream fos=new FileOutputStream(C:/User/AnirudhVarma/Document/Appname/foldername/file,true);
But on the users machine how do I ensure that it creates it in the following directory?
C:/enduser/appname/folder/file

Hard coding folder structure in code is very bad idea.
One way to handle this is:
You need to have a properties file something like that where user need to enter path to required folder. Mention this as requirement in install docs.
In your code you need to make sure folder has required content before proceeding. If not valid folder, just show error message with proper information to setup the directory path.

System.getProperty("user.dir") will return C:/User/AnirudhVarma on you machine. From there, I'd suggest you need to consider the platform. Under windows, you should store this kind of information into the AppData folder.
So you would end up with something like System.getProperty("user.dir") + File.seperator + "AppData/Appname/foldername/file"
Under MacOS, I believe the convention is to use .AppName instead, something like System.getProperty("user.dir") + File.seperator + ".Appname/foldername/file"