Handling paths in portable JAR - java

I have written a java program which different classes where during the process it generates lots of files (say txt files) and then reads the files and operates on them. I have made the project as a JAR file where I can run it from command with no problem as JAR is in the same directory.
However, I want to run this JAR file on a remote server where the PATHs are not the same, so then it generates error, because in my project for instance it should read a file from /Programs/Folder/here whereas in the server this cannot happen as there is no /Programs/Folder... directory. Do I have to change all the paths in my program according to the new location I wanna put my jar or there is another way around it?
I would appreciate your help.
Best wishes

I would think the best way is to pass the directory name where the files in as an argumrnt to the java command line.
Read it from the parameter args in main

This simplest and best solution is to change all of your paths in files inside of your project to be relative paths. You can also simply pass the directory you would like the files stored in to String args[] as parameters

what i prefer is i use relative path and then read relative path using classloader.try it.

Related

Will my executable jar include files I added in the file structure?

I'm trying to comprehend how creating executable jars and file paths work, so please bear with me.
I have a small program, which is supposed to upload some Firmware to Servers. These Firmwares have a combined size of about 650MB. All in all, the Project Folder takes up about 2.2GB of space.
In the program itself, I have some Strings that refer to the name of the folder that contains the firmware.
"src\\com\\java\\myProgram\\data\\bios_fw\\"+biosFWPath+"\\"+biosFW
When testing in the IDE, this works without problem.
So here's my question(s): Will this still work if I build the program as executable jar? I have already noticed that the jar I created holds only about 450MB. Does it even contain the firmware then? And if it does, will the program be able to find the path specified?
Furthermore, what would be the proper way of "installing" these folders together with the jar on a Users PC?
Thanks!
As a .jar file is a java archive file, you can open it and look yourself as we don't know how you build your .jar.
But considering the future, I would remove the absolute path and use a relative path to a directory where you can insert the firmware in case the fw got updated, so you don't have to build your program each time.
The proper way, when you want to keep the FW in your project is using ClassLoader.

Set a Path for database file in java

I am developing a program that uses a sqlite file as a database. When I compile and test my program I haven't any problem:
Conection c = DriverManager.getConnection(
"jdbc:sqlite:/home/mehdi/my_database.sqite");
as you can see in above, the code shows a direct path that set to database file (this path is only in my system).
So it works fine, but my problem starts when I create an executable jar file of my program, if I create executable jar file and share it with other users, when they run the executable jar it doesn't work.
My first question: how do I set my database path for an executable jar file in my code?
My second question: is it possible for the database to be along side the executable jar? (and i can move my executable jar file with its database)
some_path/my_program_file.jar
some_path/my_database.sqite
The best option is externalizing the database path instead of hardcoding it, and letting the user choose where to put the actual sqlite file.
You could put the sqlite file in the same directory as the program archive, but please note that this does not really help in locating the file from your code, because relative paths are definitely resolved against the JVM (process) working directory, not the location of the JAR:
new File(".").getAbsolutePath()
There are ways to get the location of the JAR:
getClass().getProtectionDomain().getCodeSource().getLocation()
but they are deploy-dependent, so it's not a very robust strategy to rely on.
Most Java program are configured with external files put in well-known locations (like the current JVM directory or a .myapp/conf.properties in the user home), and individual properties can be overridden in the starting command line, as either system properties -Dkey=value or program arguments (there's a library to simplify this)
Java uses relative paths when you do not specify the absolute path. I'm not sure if it works for the DriverManager though. What this means is that if you call jdbc:sqlite:my_database.sqite it should look for my_database.sqite in the folder the jar is executed from. So if the user calls java -jar my_program_file.jar from the folder where the database is it will work. If the user calls the jar from another location it won't. To fix that, you have to grab the path where the jar is located, like so:
return new File(MyClass.class.getProtectionDomain().getCodeSource().getLocation().getPath());
This was showcased in another SO post.

Java Game refering to files

I have started getting into game programming.
My question is, that when I am working with files, either parsing data, writing to files, etc. Should I be using relative path names, or absolute pathnames, or something else which is better. I've heard about using jar files, but I am not sure
1. how that works
2. if it is a good way to do it.
So when developing a game that will be cross platform, what is the best method for managing files that the program will need to read from and write to.
there are several ways in which you can ship your code as a product. the most common are
packaging everything in one executable jar file.
having a set of folders where you place all necessary resources.
minecraft, for example, is written in java and distributed as a single executable jar file that contains all necessary class files and resources. to run the game (assuming you have java installed) all you need to do is double-click the jar file.
read this short tutorial about how to add a main class to a jar file.
either way, always treat classes and resources in your code as if they're in your classpath. for example, if you have a my.properties file on the root of the source tree then load it by using 'my.properties'. if you put it under a 'conf' folder then use 'conf/my.properties'.
i think it is the safest way not to get lost.
are you using maven?
The jar file is a zip of all your compiled *.class files and your resources. You can safely load your resources and even default data FROM a jar if you package your program, but you can NOT safely write data back to the jar. This detail is answered in depth already at
How can an app use files inside the JAR for read and write?
For information on how to package a jar see
http://docs.oracle.com/javase/tutorial/deployment/jar/

Create folder in jar file dir

Im looking for a solution to create a new folder in the folder, where the .jar file is located.
So if i have a test.jer file in C:\test\temp I would like to be able to create a new folder (newFolder) : C:\test\temp\newFolder.
Anyone know how to do this.
I have tried to use
System.getProperty("user.dir").
But this only gives me the correct, if i double click the jar file.
I need it to be run from a scheduler and then the location is changed.
In general you cannot easily figure out where your JAR file is at runtime. The execution environment of your VM can tell you the user directory (as you said), but that is the directory from which the VM was started. So if you double click your JAR, that's actually the directory where the JAR is. But if you call java on the commandline, for example
c:\some\directory> java -cp c:\temp\test.jar main.Class
then "user.dir" will contain "c:\some\directory".
This lack of information is intentional, because it allows the VM to transparently load classes from different locations.
Your best choice to do what you want would be to pass the directory of choice as a commandline argument. And in the worst case, you could use "user.dir" as default. But what the best solution really is depends on why you want to place the directory where the JAR is.

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.

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