Turning a folder into an executable jar file in Java - java

I need some help with turning a folder into an executable jar file. I want to do this in my application not from a batch file.
Basically i have a folder on my desktop with class files already in it, all i want is to turn that folder into an executable jar file.
I have already tried this:
File file = new File(System.getProperty("user.home") + "/Desktop/myfolderwithclasses");
File file2 = new File(System.getProperty("user.home") + "/Desktop/myfolderwithclasses.jar");
file.renameTo(file2);
Unfortunately all this does is rename the folder to what i set but it doesn't change the file type.
Any help with this would be greatly appreciated.

Making an executable JAR isn't as simple as renaming a folder. You can use IDE functionality, as has been mentioned, but a more portable solution is to use something like Apache Ant, which is actually supported in a lot of the IDEs. You'll want to use the JAR Task. If you just have your own classes to package, and no complicated library structure, the examples at the bottom of the page should suffice. Here's a Hello World tutorial for Ant to help you set up your build file.

You can use eclipse and export runnable jar.

Related

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.

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i can't manage to get the folders contents to be jared (compressed into jar), i did about 3hrs research and found only outdated methods.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.

Export a java library in a .jar file

I'm still pretty new to java and I'm VERY new to exporting .jar files. I've got a little game that I want to send to some friends and I was told in another question that I should export it to an executable jar file. Well I finally got that working on my computer but when I send it to other people it doesn't work because they don't have the library.
I'm importing the objectdraw library and without that my program won't run at all!
So basically I need to find a way to export the object draw library as part of my .jar file so that they can use it too. Do I simply include it in the included files part of the jar command?
ex: jar cmf MANIFEST.mf Archery.jar * /System/Library/Java/Extensions/objectdraw.jar
or what? I'm working out of the command line right now.
The simplest way is to send the JAR library file too and add a Class-Path entry to the manifest. This entry would look like:
Class-Path: objectdraw.jar
You could also set the CLASSPATH environment variable manually.
Alternatively, you can unpack the library and add all (or just the required files) to your final jar. This doesn't always work though, because some libraries rely on the integrity of teir JAR file.
Finally, it is possible to include the dependency in the main JAR, but it would require a custom class loader.
Turns out the best way I've found to do this is to unpack the library and then put all the resulting files in with your final archive. This way it actually works on other computers.
jar xf library_wanted.jar; jar cvmf MANIFEST.mf end_result.jar *.class library_wanted/

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.

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.

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