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Writing Information (and Image) to a Jar File from another Java Program

Possibly a duplicate, though I doubt so since I have not seen anything so far completely answering my criteria in a way that I can complete my program
Background
What I need is to access another jar, from a seperate jar, read and write files to that jar. So far what I have done is change the jar to a zip and then I can delete files, but the problem I am having is with writing files back in, specifically image files (.txt works perfectly fine)
Question
How do I write image files to a zip (that was originally a jar) from another java program (in the end product another jar)
Note
I have looked around and most sources say this is not possible, but those questions dealt with this during the running of a program, my special case is that the other program is not running, but in file format. All I want to do is write and image in and convert it back to a jar and not have any problems with running that jar in the end.
Thank you!

Use FileSystems to access, write and replace the contents of the jar file:
try (FileSystem fs = FileSystems.newFileSystem(Paths.get("path/file.jar"), null)) {
Files.copy(Paths.get("path/to/image"), // path to an external image
fs.getPath("image.jpg"), // path inside a jar file
StandardCopyOption.REPLACE_EXISTING);
}

How do I move a file into a compiled .jar file using Java?

I'm making a file import system, and I can't move files into the compiled .jar file the application is in.
Here's what I'm trying to do:
Path FROM = Paths.get(filePath.getText());
Path TO = Paths.get("C:\\Users\\" + System.getProperty("user.name") +
"\\AppData\\Roaming\\.minecraft\\mods\\music_crafter-1.0\\src\\main\\resources\\assets\\music_crafter\\sounds\\block\\music_player");
//jar file
Files.move(FROM, TO.resolve(FROM.getFileName()), StandardCopyOption.REPLACE_EXISTING);

You need to handle the jar file internally. A Jar is not a directory, it is a compressed container file (pretty much a ZIP file with a different extension).
To do this, given that you are on Java 6, you have 2 options:
Unzip the contents to a temporary working directory (there are built
in APIs for this, or use a library such as Apache Commons Compress)
do your work (copying, deleting, etc) and then re-zip.
Make external command line calls to the Jar utilities that come with
Java
Of those, only (1) makes any real sense.
A third option would be available if you could up your Java to 7+ which would be:
3. Use a Zip File System Provider to to treat it as a file system in code
All that said, however:
As per comments on your question, you really might want to look at if this something you need to do at all? Why do you need to insert into existing jars? If this is 'external' data, it would be much better in a separate resource location/container, not the application jar.

Correct way to make an obb file

I'm trying to make an obb file and I've been unsuccessful with the JOBB tool see here: JOBB DirectoryFullException: de.waldheinz.fs.fat.DirectoryFullException: directory is full and when I follow the suggestion found in this answer: https://stackoverflow.com/a/21982186/1489990 and I try to access the contents with ZipFile as
expansionFile = APKExpansionSupport.getAPKExpansionZipFile(context, 46, 1);
InputStream fileStream = expansionFile.getInputStream("/storage/emulated/0/Android/obb/com.nick.app/main.46.com.nick.app.obb/img1large.jpg");
This does not work and I get a log message that says V/zipro: Not a Zip archive.When I made this OBB file I just selected all of my images and used WinRar to add them to a "ZIP" archive and renamed it to ".OBB". What I'm wondering is to use the ZipResourceFile tool are you supposed to create a zip archive of your content, and then add that archive to another archive and rename it to ".OBB"? What is the correct way to accomplish this?

Easy Method : get it to ES file explorer(app in android) you will see it as .zip then rename it as .obb and also this renaming method is not possible in windows

Split two files

I have combined two files in android, using this Linux command
cat file1.png file2.zip > file3.png
How can I split two files again?I just want the zip file to be retrieved separately.
Is there any specific command?I've tried these codes:
unzip file3.png
Replaced png with zip:
unzip file3.zip
but none of them work.
The only application with which I can open the combination, is winrar on windows
And also I tried several unzipping and unraring apps on android but none of them work except RAR app by rarlab
Is there any source for those apps I mentioned to unrar/unzip the file?

Strictly speaking : there's no way.
You might look for the PK 0x04 0x03 as a separator in the answer above, but you don't have any guaranty that this char sequence does not show up in the image data of the file1.png as well.
All together it's a funny question. If you want to split files like this on a regular basis rethink your strategy. If you need it to correct a one time mistake or something, you can split finding the seperator and be ok in over 99% of the cases.

What you need to know is that PNG files start with the hex value 0x89 followed by the text PNG. Zip files start with PK 0x04 0x03. You could write a utility which reads a file and outputs the bytes read in to a new file, using a new output file when a certain file signature is detected.
For a one-of solution, you can use vim, though you have to be careful to stop vim from adding a newline character to the end of the line.
Copy your input file for safety
cp file3.png f1
cp file3.png f2
vim -b f1
and in vim type
:set noeol
search for the start of the zip file
/PK
checking that the sequence found is PK^C^D. If not, look for the next match.
Delete the end of the line from PK with
d$
Move down a line, delete the remainder of the file and save
j
:.,$ d
ZZ
Similarly, delete the top of the file in f2 to get the zip file.
Note: don't name f2 as f2.zip because vim is smart enough to open this as as a zip file, which is not what you want here.

I'm not sure what you trying to accomplish by "hiding" a zip file into a PNG, but if you are trying to make a single file Winrar can open, then that's an odd way to do it.
You do not make a .zip (or any other type of archive) file when you cat a file to the either the start or end of a zip file. That simply appends two binary files together.
The reason winrar can open your "combined" binary file is that it most likely recognizes the file headers and can decipher you have 2 files.
I suggest you look into the usage of the zip command, for how to add files to an archive. I quick search shows, for example
zip -rv zipfile.zip newfile.txt
Will add newfile.txt to zipfile.zip.

JAR - Listing files into a folder

I would like to get a list of file contained in a directory which is in a jar package.
I have an "images" folder, within it I have an Images class that should load all images from that directory.
In the past i used the MyClass.class.getResourceAsStream("filename"); to read files, but how do I read a directory?
This is what I tried:
System.out.println(Images.class.getResource("").getPath());
System.out.println(new File(Images.class.getResource("").getPath()).listFiles());
I tried with Images.class.getResource because I have to work with File and there isn't a constructor that accepts an InputStream.
The code produces
file:/home/k55/Java/MyApp/dist/Package.jar!/MyApp/images/
null
So it is finding the folder which I want to list files from, but it is not able to list files.
I've read on other forums that in fact you can't use this method for folders in a jar archive, so how can I accomplish this?
Update: if possible, i would like to read files without having to use the ZipInputStream

You can't do that easily.
What you need to do:
Get the path of the jar file.
Images.class.getResource("/something/that/exists").getPath()
Strip "!/something/that/exists".
Use Zip File System to browse the Jar file.
It's a little bit of hacking.