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Reading and writing files within Jar

I have finally completed a program in Java and I have to upload it.
The problem is that I have to upload also the executable .jar file and not only the eclipse project.
The main functionality of my program consists by reading and writing .xml files (for example one file is used to read and add new users), and the files in the project folder are so located:
-Project Name
src
default package
main and all other classes
file1.xml
file2.xml
So the two .xml files are in the root of the project.
My question is: It is better to save the .xml files in the JAR and then writing and reading them from the executable program or it is better to store them in a folder outside the .JAR and reading and writing them as externally files?
It is a good practice to create a folder like that?:
-ProjectName
file1.xml
file2.xml
project.jar
I read in Stackoverflow a lot of people having my same issue and a lot of people doesnt know how to manage this problem properly.
Thank you in advance for the reply :)

Changing files in JAR-files can have all sorts of problems. That starts with simple things such as what should happen when you want to update your program to the newest version? Usually you'd just swap the jar, but then you loose everything you edited so far. You'd need a process to update inside the jar.
Other problems include that for changing the jar file you need to open it, possibly realign contents and rewrite the index which could conflict with the JVM that is reading the jar at the same time causing odd behaviour. On some systems (windows...) the Jar file might even be locked while the application is running and thus you cannot write to it at all.
I'd suggest that you add "default files" (in case that your files are initially not just empty) to you Jar file that represent the initial state. If the application is started you check if the XML files exist in the some normal writable directory and if they don't just copy the default files to that directory. This allows you to deploy still just a single jar file, but once started the appropriate files will be created.

You may read a XML file located inside the executable Jar but it is not possible to update (write) a XML file located inside that executable Jar file. So the best option would be:-
-ProjectName
file1.xml
file2.xml
project.jar

The jar should be kept read-only, the XML "files" inside the jar should be read using getResource[AsStream] (class path). You can use those resources as templates to create a copy in the user's (or application's) directory/sub-directory. For the user's directory:
System.getProperty("user.home")

jsr305_annotations Folder inside apk file

I've had reports of problems with the installation of the apk.
I tried to extract the apk as a zip file and I found the folder
jsr305_annotations in the root APK
with inside
folder: v0_r47 -> file: V0_r47.gwt.xml
File: Jsr305_annotations.gwt.xml
I would like information about those files and folders and a method to make sure that they are not included in my applications
thanks
Using
Eclipse
Version: 3.8.2
Build id: M20130131-0800
Android Development Toolkit
Version: 22.0.1.v201305230001--685705

I'd seen those files included in an app as a result of including the library GoogleAdMobAdsSdk in a project (as available from here https://developers.google.com/mobile-ads-sdk/)
If you open that JAR file you can see those annotations folders and XML files are present.
If you don't include that library directly, you may be including another JAR which has those files.
I'd suggest checking the JAR files you include in your project (by opening them in a file that can open APK/Zip files) to find which one(s) include the file and look for an alternative library.

The javascript that comes with it looks like it will lock you out of your device and start making phone calls, mail, SMS, etc while it plays audio and shows a video. It also creates a *.css file and calls on varible uri. I didnt inspect every file but from the *.js file it dont look like I would want to run it. I found this in my Downloads folder, file name, "yourdownloadisready.apk", Im a tech. I cant imagine if a novice saw this.

view jar file through java code

I want to create one java application that will iterate an jar file and will give classes of that jar file.
I searched a lot but didn't get sufficient information.
I want java code that will open the jar file and will able to iterate that jar.

See:
java.util.jar.JarFile.entries()
http://docs.oracle.com/javase/6/docs/api/java/util/jar/JarFile.html#entries%28%29

What exactly does a jar file contain?

As an intern, I use company code in my projects and they usually send me a jar file to work with. I add it to the build path in Eclipse and usually all is fine and dandy.
However, I got curious to know, what each class contained and when I try to open one of the classes in the jar file, it tells me that I need a source file.
What does this mean? I come from a C/C++ background so is a jar similar to an already compiled .o file and all I can see is the .h stuff? Or is there actual code in the jar file that I'm using that's encrypted so I can't read it?
Thanks for all the answers!
Edit: Thanks, guys, I knew it was a sort of like an archive but I was confused to why when I tried to open the .class files, I got a bunch of random characters. The output was similar when I tried to open a .o file in C so I just wanted to make sure.
Thanks!

A JAR file is actually just a ZIP file. It can contain anything - usually it contains compiled Java code (*.class), but sometimes also Java sourcecode (*.java).
However, Java can be decompiled - in case the developer obfuscated his code you won't get any useful class/function/variable names though.

However, I got curious to what each class contained and when I try to open one of the classes in the jar file, it tells me that I need a source file.
A jar file is basically a zip file containing .class files and potentially other resources (and metadata about the jar itself). It's hard to compare C to Java really, as Java byte code maintains a lot more metadata than most binary formats - but the class file is compiled code instead of source code.
If you either open the jar file with a zip utility or run jar xf foo.jar you can extract the files from it, and have a look at them. Note that you don't need a jar file to run Java code - classloaders can load class data directly from the file system, or from URLs, as well as from jar files.

The best way to understand what the jar file contains is by executing this :
Go to command line and execute jar tvf jarfilename.jar

A jar file is a zip file with some additional files containing metadata. (Despite the .jar extension, it is in zip format, and any utilities that deal with .zip files are also able to deal with .jar files.)
http://docs.oracle.com/javase/8/docs/technotes/guides/jar/index.html
Jar files can contain any kind of files, but they usually contain class files and supporting configuration files (properties), graphics and other data files needed by the application.
Class files contain compiled Java code, which is executable by the Java Virtual Machine.
http://en.wikipedia.org/wiki/Java_class_file

JAR stands for Java ARchive. It's a file format based on the popular ZIP file format and is used for aggregating many files into one. Although JAR can be used as a general archiving tool, the primary motivation for its development was so that Java applets and their requisite components (.class files, images and sounds) can be downloaded to a browser in a single HTTP transaction, rather than opening a new connection for each piece. This greatly improves the speed with which an applet can be loaded onto a web page and begin functioning. The JAR format also supports compression, which reduces the size of the file and improves download time still further. Additionally, individual entries in a JAR file may be digitally signed by the applet author to authenticate their origin.

Jar file contains compiled Java binary classes in the form of *.class which can be converted to readable .java class by decompiling it using some open source decompiler. The jar also has an optional META-INF/MANIFEST.MF which tells us how to use the jar file - specifies other jar files for loading with the jar.

Jar( Java Archive) contains group of .class files.
1.To create Jar File (Zip File)
if one .class (say, Demo.class) then use command jar -cvf NameOfJarFile.jar Demo.class (usually it’s not feasible for only one .class file)
if more than one .class (say, Demo.class , DemoOne.class) then use command jar -cvf NameOfJarFile.jar Demo.class DemoOne.class
if all .class is to be group (say, Demo.class , DemoOne.class etc) then use command jar -cvf NameOfJarFile.jar *.class
2.To extract Jar File (Unzip File)
jar -xvf NameOfJarFile.jar
3.To display table of content
jar -tvf NameOfJarFile.jar

A .jar file is akin to a .exe file.
In essence, they are both executable files.
A jar file is also a archive (JAR = Java ARchive). In a jar file, you will see folders and class files. Each .class file is similar to a .o you might get from C or C++, and is a compiled java archive.
If you wanted to see the code in a jar file, download a java decompiler (located here: http://java.decompiler.free.fr/?q=jdgui) and a .jar extractor (7zip works fine).

JD-GUI is a very handy tool for browsing and decompiling JARs

A .jar file contains compiled code (*.class files) and other data/resources related to that code. It enables you to bundle multiple files into a single archive file. It also contains metadata. Since it is a zip file it is capable of compressing the data that you put into it.
Couple of things i found useful.
http://www.skylit.com/javamethods/faqs/createjar.html
http://docs.oracle.com/javase/tutorial/deployment/jar/basicsindex.html
The book OSGi in practice defines JAR files as, "JARs are archive files based on the ZIP file format,
allowing many files to be aggregated into a single file. Typically the files
contained in the archive are a mixture of compiled Java class files and resource
files such as images and documents. Additionally the specification defines a
standard location within a JAR archive for metadata — the META-INF folder
— and several standard file names and formats within that directly, most
important of which is the MANIFEST.MF file."

Just check if the aopalliance.jar file has .java files instead of .class files. if so, just extract the jar file, import it in eclipse & create a jar though eclipse. It worked for me.

While learning about JAR, I came across this thread, but couldn't get enough information for people like me, who have .NET background, so I'm gonna add few points which can help persons like myself with .NET background.
First we need to define similar concept to JAR in .NET which is Assembly and assembly shares a lot in common with Java JAR files.
So, an assembly is the fundamental unit of code packaging in the .NET environment. Assemblies are self contained and typically contain the intermediate code from compiling classes, metadata about the classes, and any other files needed by the packaged code to perform its task. Since assemblies are the fundamental unit of code packaging, several actions related to interacting with types must be done at the assembly level. For instance, granting of security permissions, code deployment, and versioning are done at the assembly level.
Java JAR files perform a similar task in Java with most differences being in the implementation. Assemblies are usually stored as EXEs or DLLs while JAR files are stored in the ZIP file format.
Source of Information -> 5- Assemblies

Modifying a file inside a jar

I would like to modify a file inside my jar. Is it possible to do this without extracting and re jarring, from within my application?
File i want to modify are configuration files, mostly xml based.
The reason i am interested in not un jarring is that the application is wrapped with launch4j if i unjar it i can't create the .exe file again.

You can use the u option for jar
From the Java Tutorials:
jar uf jar-file input-file(s)
"Any files already in the archive having the same pathname as a file being added will be overwritten."
See Updating a JAR File.
Much better than making the whole jar all over again. Invoking this from within your program sounds possible too. Try Running Command Line in Java

You can use Vim:
vim my.jar
Vim is able to edit compressed text files, given you have unzip in your environment.

Java jar files are the same format as zip files - so if you have a zip file utility that would let you modify an archive, you have your foot in the door. Second problem is, if you want to recompile a class or something, you probably will just have to re-build the jar; but a text file or something (xml, for instance) should be easily enough modified.

As many have said, you can't change a file in a JAR without recanning the JAR. It's even worse with Launch4J, you have to rebuild the EXE once you change the JAR. So don't go this route.
It's generally bad idea to put configuration files in the JAR. Here is my suggestion. Search for your configuration file in some pre-determined locations (like home directory, \Program Files\ etc). If you find a configuration file, use it. Otherwise, use the one in the JAR as fallback. If you do this, you just need to write the configuration file in the pre-determined location and the program will pick it up.
Another benefit of this approach is that the modified configuration file doesn't get overwritten if you upgrade your software.

Not sure if this help, but you can edit without extracting:
Open the jar file from vi editor
Select the file you want to edit from the list
Press enter to open the file do the changers and save it
pretty simple
Check the blog post for more details
http://vinurip.blogspot.com/2015/04/how-to-edit-contents-of-jar-file-on-mac.html

I have similar issue where I need to modify/update a xml file inside a jar file.
The jar file is created by a Spring-boot application and the location of the file is BOOT-INF/classes/properties
I was referring this document and trying to replace/update the file with this command:
jar uf myapp.jar BOOT-INF/classes/properties/test.xml
But with this, it wont change the file at the given location. I tried all the options also but wont work.
Note: The command I am executing from the location where jar file is present.
The solution I found is:
From the current location of jar file, I created folders BOOT-INF/classes/properties
Copy the test.xml file into the location BOOT-INF/classes/properties.
Run the same command again. jar uf myapp.jar BOOT-INF/classes/properties/test.xml
The xml file has been changed in the jar file.
Basically you need to create a folder structure like where the file is located into the jar file. Copy the file at that location and then execute the command.
The problem with the documentation is that, it does not have enough examples as well as explanation around common scenarios.

This may be more work than you're looking to deal with in the short term, but I suspect in the long term it would be very beneficial for you to look into using Ant (or Maven, or even Bazel) instead of building jar's manually. That way you can just click on the ant file (if you use Eclipse) and rebuild the jar.
Alternatively, you may want to actually not have these config files in the jar at all - if you're expecting to need to replace these files regularly, or if it's supposed to be distributed to multiple parties, the config file should not be part of the jar at all.

To expand on what dfa said, the reason is because the jar file is set up like a zip file. If you want to modify the file, you must read out all of the entries, modify the one you want to change, and then write the entries back into the jar file. I have had to do this before, and that was the only way I could find to do it.
EDIT
Note that this is using the internal to Java jar file editors, which are file streams. I am sure there is a way to do it, you could read the entire jar into memory, modify everything, then write back out to a file stream. That is what I believe utilities like 7-Zip and others are doing, as I believe the ToC of a zip header has to be defined at write time. However, I could be wrong.

Yes you can, using SQLite you can read from or write to a database from within the jar file, so that you won't have to extract and then re jar it, follow my post http://shoaibinamdar.in/blog/?p=313
using the syntax "jdbc:sqlite::resource:" you would be able to read and write to a database from within the jar file

Check out TrueZip.
It does exactly what you want (to edit files inline inside a jar file), through a virtual file system API. It also supports nested archives (jar inside a jar) as well.

Extract jar file for ex. with winrar and use CAVAJ:
Cavaj Java Decompiler is a graphical freeware utility that reconstructs Java source code from CLASS files.
here is video tutorial if you need:
https://www.youtube.com/watch?v=ByLUeem7680

The simplest way I've found to do this in Windows is with WinRAR:
Right-click on the file and choose "Open with WinRAR" from the context menu.
Navigate to the file to be edited and double-click on it to open it in the default editor.
After making the changes, save and exit the editor.
A dialogue will then appear asking if you wish to update the file in the archive - choose "Yes" and the JAR will be updated.

most of the answers above saying you can't do it for class file.
Even if you want to update class file you can do that also.
All you need to do is that drag and drop the class file from your workspace in the jar.
In case you want to verify your changes in class file , you can do it using a decompiler like jd-gui.

As long as this file isn't .class, i.e. resource file or manifest file - you can.