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Modifying BufferedWriter argument in order to make JAR file

I am using BufferedReader and BufferedWriter in a progect in order to export it as Jar and use it,
My code using the streams is the following
br = new BufferedReader(new InputStreamReader(getClass().getResourceAsStream("/data/score.scr")));
bw= new BufferedWriter(new FileWriter("src/data/score.scr"));
I tried to use getClass().getResourceAsStream(); inside BufferedWriter but it doest work, the program works well but it seems ... that the file read is not the one written ... I save a score (BufferedWriter) when I get to the scorepanel (BufferedReader) it is not there
all this in order to export my project as a JAR so I need to modify the second line of my core

In general you cannot write inside a jar file but here is a more detailed answer How can my Java program store files inside of its .jar file?

You could try using your IDE to do this.
eclipse
or through command line
jar cf jar-file input-file(s)
From what it looks like your doing, you don't necessarily need to store data in the jar itself. Just write the data to the same directory or folder as your jar and have your user transfer the entire folder as part of your program.

Reading a file into compiled jar by relative path

How would you read a file into a program that's compiled into a jar next to it through its local directory? The type read would be a simple .txt file.

It depends on what the usage of the program is. Do you know how the jar is supposed to be executed? When you try to run it, does it spit out a "usage: somejar firstarg secondarg" type message?
Also, if its a jar that you've compiled and you know how it should be executed, then you may have forgot to set its main class or manifest.
Check this: http://www.mkyong.com/java/how-to-make-an-executable-jar-file/

If you want to read a file that exists within an external .jar file, you will need to unzip the .jar file first in your code and then retrieve the file. You can do this using Java's zip APIs. See this answer if this is the case: Easiest way to unpack a jar in java
If you want to read a file that is in the same .jar file that your code is executing, you can get the file as a resource. See this answer: Get a resource using getResource()
If the file is simply in the exact same directory as the executable .jar, create a new file like so:
File input = new File("myfile.txt");

Java - read file from directory for jar

I have an application that creates a temporary mp3-file and puts it in a directory like C:\
File tempfile = File.createTempFile("something", ".mp3", new File("C:\\));
I'm able to read it by just using that same tempfile again.
Everything works fine in the Eclipse IDE.
But when I export my project for as a Runnable jar, my files are still being made correctly (I can play them with some normal music player like iTunes) but I can't seem to read them anymore in my application.
I found out that I need to use something like getClass().getResource("/relative/path/in/jar.mp3") for using resource files that are in the jar. But this doesn't seem to work if I want to select a file from a certain location in my file system like C:\something.mp3
Can somebody help me on this one?

It seems you dont have file name of the temp files . When you was running your program in eclipse that instance was creating a processing files, but after you made a runable you are not able to read those file that instance in eclipse created, You runable file can create its own temp file and can process them,
To make temp files globe put there (path + name ) entries in some db or property file
For example of you will create a temp file from the blow code
File tempfile = File.createTempFile("out", ".txt", new File("D:\\"));
FileWriter fstream = new FileWriter(tempfile);//write in file
out = new BufferedWriter(fstream);
the out will not be out.txt file it will be
out6654748541383250156.txt // it mean a randum number will be append with file
and you code in runable jar is no able to find these temp files

getClass().getResource() only reads resources that are on your classpath. The path that is passed to getResource() is, in fact, a path relative to any paths on your current classpath. This sounds a bit confusing, so I'll give an example:
If your classpath includes a directory C:\development\resources, you would be able to load any file under this directory using getResource(). For example, there is a file C:\development\resources\mp3\song.mp3. You could load this file by calling
getClass().getResource("mp3/song.mp3");
Bottom line: if you want to read files using getResource(), you will need those files to be on your classpath.

For loading from both privileged JARs and the file system, I have had to use two different mechanisms:
getClass().getClassLoader().getResource(path), and if that returns null,
new File(path).toURI().toURL();
You could turn this into a ResourceResolver strategy that uses the classpath method and one or more file methods (perhaps using different base paths).

Extracting a file from the currently running JAR through code

Are there any built-in methods I can use to allow users to extract a file from the currently running JAR and save it on their disk?
Thanks in advance.

File file = new File("newname.ext");
if (!file.exists()) {
InputStream link = (getClass().getResourceAsStream("/path/resources/filename.ext"));
Files.copy(link, file.getAbsoluteFile().toPath());
}

Use getResourceAsStream (docs), you can do whatever you want with it after that.
For a two-liner you could use one of the Commons IO copy methods.

I am not sure whether you will get know from which jar your class is getting executed but you can try this to extract resources from jar :
How to write a Java program which can extract a JAR file and store its data in specified directory (location)?

Where does Java put resource files when I JAR my program?

Been looking for this for the past 2 hours and can't find anything (I've found solutions to the same problem but with images, not text files).
Pretty much, I made a program that reads a text file. The file is a list of names and IDs. Using Eclipse, I put the file in my src folder and in the program put the path file to it. Like this:
in = new BufferedReader(new FileReader(curDir+"\\bin\\items.txt"));
Where curDir is the user's current directory (found with System.getProperty("user.dir")).
Now, problem is, the program runs fine when I run it from Eclipse, but when I try to make it a runnable JAR and then run it, the program runs, but the info from the text file does not load. It look like Eclipse is not putting the text file with the JAR.
EDIT: Solved-ish the problem? So the JAR file needs to the in a folder with all the original files? I am so confused, what is a JAR file then?

A more robust way to get a file whether you are running from Eclipse or a JAR is to do
MyClass.getResource("items.txt")
where MyClass is a class in the same package (folder) as the resource you need.
If Eclipse is not putting the file in your JAR you can go to
Run -> Run Configurations -> -> Classpath tab -> Advanced -> Add Folders
Then add the folder containing your file to the classpath. Alternatively, export the Ant script and create a custom build script.

To the point, the FileReader can only read disk file system resources. But a JAR contains classpath resources only. You need to read it as a classpath resource. You need the ClassLoader for this.
Assuming that Foo is your class in the JAR which needs to read the resource and items.txt is put in the classpath root of the JAR, then you should read it as follows (note: leading slash needed!):
InputStream input = Foo.class.getResourceAsStream("/items.txt");
reader = new BufferedReader(new InputStreamReader(input, "UTF-8"));
// ...
Or if you want to be independent from the class or runtime context, then use the context class loader which operates relative to the classpath root (note: no leading slash needed!):
ClassLoader classLoader = Thread.currentThread().getContextClassLoader();
InputStream input = classLoader.getResourceAsStream("items.txt");
reader = new BufferedReader(new InputStreamReader(input, "UTF-8"));
// ...
(UTF-8 is of course the charset the file is encoded with, else you may see Mojibake)

Get the location of your jar file
Firstly create a folder(say myfolder) and put your files inside it
Consider the following function
public String path(String filename)
{
URL url1 = getClass().getResource("");
String ur=url1.toString();
ur=ur.substring(9);
String truepath[]=ur.split("myjar.jar!");
truepath[0]=truepath[0]+"myfolder/";
truepath[0]=truepath[0].replaceAll("%20"," ");
return truepath[0]+filename;
}//This method will work on Windows and Linux as well.
//You can alternatively use the following line to get the path of your jar file
//classname.class.getProtectionDomain().getCodeSource().getLocation().getPath();
Suppose your jar file is in D:\Test\dist
Then path() will return /D:/Test/dist/myfolder/filename
Now you can place 'myfolder' inside the folder where your jar file is residing
OR
If you want to access some read-only file inside your jar you should copy it to one
of your packages and can access it as
yourClassname.getResource("/packagename/filename.txt");