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NPE while try to get Resource in jar

I have tried many variants but I cant find correct.
I have something like
Inside my jar, which created by Maven I can see that
That is my folder with classes. And, by the way, If I start my program in IDEA, not from Console, there is not any exception with paths
Here, I am in debug mode start my jar trying to see, where is the problem.
If I do 'file.exists()' it would be false but file inside. I think, that problem because of '.jar!\' in the path, but I don`t know how to remove that.
Anyway I've tried absolute and relative path, I've tried
Thread.getCurrentThread.getContextLoader.getResource()
GUI.class.getResource()
GUI.class.getClassLoader.getResource()
Nothing help

You can't use File to open resources inside a jar file. File can only be used with normal files and directories.
Note: using File works fine within in the IDE, since all files are not packaged in a jar file yet. But the program will break after you package it.
Once you locate the resource eg. URL res = GUI.class.getResource("/rxtx64/myres.dll") , you can open that resource as a stream InputStream is = res.openStream(); .
See also related answers Utils to read resource text file to String (Java) and How to read a text-file resource into Java unit test?

Unable to reach resources folder when compiling from .jar to .exe using Launch4j

I made a small Java program for academic purposes, its main focus is to read some .txt files and present the information to the user. These files are present in the resources folder, under the src folder.
The program runs as intended when launched from Eclipse.
Using the Launch4j app I was able to successfully create an exe which runs fine and does what's intended, up until I try to read the .txt files I have in the resources folder, which appears not to be able to reach.
I'm guessing that when I launch the exe the run time path would change to where the exe was created, so I created the program in a desktop folder and specified this path in the program, but that doesn't seem to solve the situation.
As an alternative, I moved the .txt files out of the program and once again created the exe in a desktop folder with said .txt files, linked the program to this path and once again it didn't work.
The command used to get the .txt files is:
Files.readAllLines(Paths.get(doc)).get(line)
And doc is simply the path to the intended .txt file.
It's worth noting that I have no previous experience in Java and throughout the development of the program I tried my best to use commands I'd fully understand and to keep it as simple as possible. I hope the solution can be along these lines! I'm very confident this must be a rookie mistake, but I can't seem to find the solution to this specific problem anywhere.

The paths to files in Eclipse are different than the paths to files in an .exe or JAR file.
I will let this other user explain it because I am lazy :p
Rather than trying to address the resource as a File just ask the
ClassLoader to return an InputStream for the resource instead via
getResourceAsStream:
InputStream in = getClass().getResourceAsStream("/file.txt");
BufferedReader reader = new BufferedReader(new InputStreamReader(in));
As long as the file.txt resource is available on the classpath then
this approach will work the same way regardless of whether the
file.txt resource is in a classes/ directory or inside a jar.
The URI is not hierarchical occurs because the URI for a resource
within a jar file is going to look something like this:
file:/example.jar!/file.txt. You cannot read the entries within a jar
(a zip file) like it was a plain old File.
This is explained well by the answers to:
How do I read a resource file from a Java jar file?
Java Jar file: use resource errors: URI is not hierarchical
The original post is here, all credit to its author.
Fixing your URL should let you read from that file when you are using the .exe.
EDITED FOR CORRECTION. Thanks #VGR (see comments) for correcting my mistake.

JAR file that reads .cfg file and produces a text file

I developed a small program that takes as input some configuration parameters from a .cfg file and produces an output .txt file based on the values taken from the .cfg file.
While the program runs perfectly in eclipse, I receive a NullPointerException error when I create a JAR file of this program and try to run it. From my understanding I have to make the JAR access its internal files or try to receive the needed information (in this case the .cfg file) externally, e.g. create a resource folder next to the JAR file.
I have searched many related questions asked here but I got even more confused whether there is an optimal way to produce a JAR file that can access input files and produces output files. Should I modify my code to achieve this or there is another way?
For the record, I use FileReader and FileWriter to access and produce the files.

If your .cfg file is outside of the JAR file, it should work just as in Eclipse.
If you want to access it from inside of the JAR archive, then you should use a class loader to load it, instead of FileReader...

Problems with loading CLP file from JAR

I am using CLIPSJNI.
What I have is:
Environment clips = new Environment();
clips.load("main.clp");
where main.clp is put in the same level as src and bin folder.
This runs fine in Eclipse. However when I export to JAR. It cannot work.
I understand that there are some problems with the path when we export to JAR.
So I've seen people suggesting using this.getClass().getResourceStream() but this is not the case. Because what I need is the name of the file, not its content.
Any suggestions on how to fix this?

The issue is that the load is being done within the native library on the C side which is being passed a file name as an argument. The C code has no concept of a JAR file or how to extract files embedded within one. I think what you would need to do is always place your .clp files within the JAR file and then have a routine which extracts the data from the JAR file and saves it to a file. You can then load it using the load method and delete the file once done.

Creating a standalone package in java and take some settings from the configuration file

Anyone plz let us know what to do when we have some configuration file which is basically xml.I want to for example give the path to save the image(for my java program) in a folder from some config file (xml in my case).In that case where should the config file be kept.Rt now every thing is converted to jar file when i create a java standalone package.But i want to give some setting from xml file.What to do in that case.How is it possible.This article only provides to create a single jar file for java project but talks nothing about the configuration settings that u can provide from some external source.
Regards
Sagar

I'm not sure I fully understand your question, but if it is where to put the XML file with configuration information, you can place your xml file in the same directory as your jar file, and then pass the XML file name and path into the Jar on the command line when calling the Jar. If you're running it in Windows, this is often done using a shortcut. Then you can get the full path string for the Jar from the main method's String[] arg array that accepts the command parameters.

Sagar,
The fact your java program is a standalone package (.jar file) has no bearing on where your configuration file is stored. Your java package is a program and that program can read any file from the file system that it so desires; it does not have to be part of the code inside the IDE i.e. you don't have to write it when you write the program. What you do need is some way, when you start the program, to find and read said configuration file.
Depending on how you expect the program to be configured, you might put that file in a number of locations. For example, /etc/yourimageprogram/config.xml or c:\program files\yourimageprogram\config.xml or perhaps c:\users\Sagar\Application Settings\yourimageprogram\config.xml. Which you choose of those options really depends on what the use case is and that I can't help with.
However, there are some main points to reading any file:
Does it exist?
Are we allowed to open it for reading?
Are we allowed to open it for writing? Might want to know if we want to update the config?
In Java, typically, you would test this with:
File configfile = new File("C:\test.xml");
if ( configfile.exists() && configfile.canRead() )
{
// read the file
}
else
{
// decide what to do if no config exists.
// might be first run of app.
}
The next stage is to parse the file. There are a number of parsers available for XML including sax and org.w3c.dom. What you need to do is to use these to extract the information you require and store that in a class. Probably a singleton class as you're unlikely to have multiple configuration instances per instance of the program.
I suggest you read about XML Parsers and File Handling under Java. Also look at the File object. See all your options for file io in java. These should give you some indication of how to proceed.