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Java - How do I copy a java aplication jar file?

Is it possible to copy a jar file from which I'm running my application?
How do I find a source to this jar file?
Each user can save my application's jar file in a different folder.

I can answer one part of your question: how do you find the location of your JAR at runtime:
File jarDir = new File(ClassLoader.getSystemClassLoader().getResource(".").getPath());
locationOfJar = jarDir.getAbsolutePath();
This will give you the path to the folder the Jar is currently in as a String.

What IDE or Compiler are you using? If you are using Eclipse you should find where your saves go and check your package, there you'll find the .jar and the .java forms of your application.

JAR doesn't include root files

I have simple Java Application and trying to create JAR to distribute using eclipse.
But when I look inside JAR it doesn't contain the test.txt file. I created JAR as Export>Runnable JAR File

You need to put that file into one of your source folders (such as src). Only those get copied into the jar file (by default).

See the following image:
Choose the file which you want to add in jar file.

Netbeans created Jar does not work, but inside IDE program works

The Netbeans created Jar does not work, but inside the IDE program it works perfectly.
I believe that the main class is set, so I'm not sure what the problem is, I think it might have something to do with the txt files I'm using, in the IDE they are in C:\Users\J\Documents\NetBeansProjects\PointOfSale\src\pointofsale (the text files are with my java files). After building the dist/ jar though the text files are inside the jar with no folders or anything (Jar is in C:\Users\J\Documents\NetBeansProjects\PointOfSale\dist). I this this might be the problem, if its helpful, I access the files using
File file = new File(System.getProperty("user.dir")+"\\src\\pointofsale\\list.txt");

You need to use Class.getResourceAsStream() to load the file. It searches from inside the classpath (and therefore from inside the jar). Now you can't load the list.txt because it doesn't exist in the directory you're specifying, it's inside your jar.
Something along the lines of
getClass().getResourceAsStream("list.txt"); // Or "/list.txt"
Will give you an InputStream you can use to load the file contents.

How to turn folder into jar - Code

Hi im busy on a application that decompiles a jar the pastes files into the folder of the decompiled jar, it then compresses the folder into a jar.
Decompiling and copying works, but i can't manage to get the folders contents to be jared (compressed into jar), i did about 3hrs research and found only outdated methods. please help.
-Regards
marko5049
EDIT MORE INFO:
I apologize i mean i cant get my application to turn a folder into a jar file, my application is an modification installer for a jar file. and it extracts the jars files, then adds the modification and then, is supposed to then turn the folder back into a jar file so that the modification is installed. The jar file is not executable.

This worked for me for a MAC OSX:
Open Terminal at the folder with the jar file and run the following commands
unzip mylib.jar -d jarfolder
//You can then change whatever you need and finally run the command below
jar cvf mylib.jar -C jarfolder/ .

Given that you want to create the JAR through code; you can use JarOutputStream for that. There is an example at this link that contains code to create a JAR file given a File[] containing all the input files.
As for creating the list of files given a starting input path, see Recursively list files in Java.
You could either build a list of files then just use code like in the above example, or you could recursively scan files and add them to the JAR as you go.
If you are using Java 7 and you know your users are too you can also use Files.walkFileTree() with a FileVisitor that adds entries to the JAR as it visits files.
Original answer before OP clarified:
Is there something wrong with:
jar cf my-application.jar folder1 folder2 folder3 etc
The JDK comes with a jar utility to create JAR files.
You can read an official tutorial on it here: Creating JAR Files. It is very straightforward.
If you want to create a runnable JAR, you can create a manifest file that has the main class and other options in it. The linked tutorial describes that process.

The short answer is, ZIP the folder, then rename it to a JAR file.

for windows just make the folder as winrar file.,
to do this right click the folder and click "7 -zip" then
choose "add to foldername.zip".
now a rar file is created with the same folder name.
Then open the cmd in current folder directory
type "mv foldername.zip foldername.jar"
Now you got the executable jar file with your corresponding folder.

The easiest way to make it .. put your folder to C:\Program Files\Java\jdk1.8.0_221\bin and then reach till the same path from CMD then run this
jar cvf Name_your_jar.jar folder

Following command worked for me in Windows 10 and jdk-8u212
jar cf my-application.jar folder1 folder2 folder3 etc

You can put your files in a zip folder. Then convert the zip file into Jar format.A .jar extension file is a Java Archive format file. It is used to store a large amount of files into one single file. You can try a free online file converter without downloading a new software on your computer. There are various online file converters available on Google. I would recommend Convert zip to jar
I hope it helps.

I just found this question and its answers are more useful for your problem:
how to zip a folder itself using java
2 tips :
1、a jar is exactly a zip. So, you just need to zip your folder, and rename it to jar
2、be careful that you should zip your whole folder without changing the relative path of the files, but not just extract all the .class files and zip them together. Because when you run the jar, the class package should be consistent with its path.

I suggest trying to create a regular .ZIP file in Windows.
You need to get 7-zip in order to view the .JAR file you are creating. You should just paste contents into the .ZIP, then rename the file type from .ZIP to .JAR, this worked for me and I hope this works for you.
.JARs are basically .ZIPs created by the Oracle Java client, so you need special file viewing software such as 7-zip or WinRAR to view it for some reason.
You can also revert .JARs to .ZIPs by renaming the file type. You might have to mod your computer with RegeEdit or something to have access to renaming your file types.
I hope this helps.

Updating .class file in jar

I want to update a .class file in a jar with a new one. What is the easiest way to do it, especially in the Eclipse IDE?

This tutorial details how to update a jar file
jar -uf jar-file <optional_folder_structure>/input-file(s)
where 'u' means update.

Do you want to do it automatically or manually? If manually, a JAR file is really just a ZIP file, so you should be able to open it with any ZIP reader. (You may need to change the extension first.) If you want to update the JAR file automatically via Eclipse, you may want to look into Ant support in Eclipse and look at the zip task.

Use jar -xvf to extract the files to a directory.
Make your changes and replace the classes.
Use jar -cvf to create a new jar file.

Simply drag and drop your new class file to the JAR using 7-Zip or Winzip. You can even modify a JAR file that is included in a WAR file using the parent folder icon, and click Ok when 7zip detects that the inside file has been modified

Jar is an archive, you can replace a file in it by yourself in your favourite file manager (Total Commander for example).

A JAR file is just a .zip in disguise. The zipped folder contains .class files.
If you're on macOS:
Rename the file to possess the '.zip' extension. e.g. myJar.jar -> myJar.zip.
Decompress the '.zip' (double click on it). A new folder called 'myJar' will appear
Find and replace the .class file with your new .class file.
Select all the contents of the folder 'myJar' and choose 'Compress x items'. DO NOT ZIP THE FOLDER ITSELF, ONLY ITS CONTENTS
Miscellaneous - Compiling a single .class file, with reference to a original jar, on macOS
Make a file myClass.java, containing your code.
Open terminal from Spotlight.
javac -classpath originalJar.jar myClass.java This will create your compiled class called myClass.class.
From here, follow the steps above. You can also use Eclipse to compile it, simply reference the original jar by right clicking on the project, 'Build Path' -> 'Add External Archives'. From here you should be able to compile it as a jar, and use the zip technique above to retrieve the class from the jar.

Editing properties/my_app.properties file inside jar:
"zip -u /var/opt/my-jar-with-dependencies.jar properties/my_app.properties". Basically "zip -u <source> <dest>", where dest is relative to the jar extract folder.

High-level steps:
Setup the environment
Use JD-GUI to peek into the JAR file
Unpack the JAR file
Modify the .class file with a Java Bytecode Editor
Update the modified classes into existing JAR file
Verify it with JD-GUI
Refer below link for detailed steps and methods to do it,
https://www.talksinfo.com/how-to-edit-class-file-from-a-jar/

1) you can extract the file into a folder called
jarname.jar
and then replace the file in the folder, handy if you are updating the class a lot while debugging
2) you can extract the jar replace the file then the jar it up again
3) Open the jar with 7 zip and drag and drop your new class in to copy over the old one

You can find source code of any .jar file online, import the same project in your IDE with basic setups. Make necessary changes in .java file and compile it for .class files.
Once compilation is done You need to extract the jar file, replace the old .class file with new one.
And use below command for reconstruct .jar file
Jar cf test.jar *
Note : I have done so many time this changes in our project, hope you will find it useful.

An alternative is not to replace the .class file in the jar file. Instead put it into a new jar file and ensure that it appears earlier on your classpath than the original jar file.
Not sure I would recommend this for production software but for development it is quick and easy.