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.
I have an external jar file mysql-connector-java-5.1.26 under /Users/remy/Downloads/mysql-connector-java-5.1.26/mysql-connector-java-5.1.26.jar . I want to create a jar file from hello.java which should include the external jar file as specified as well, where hello.java is dependent on the external jar file.
Users
-->remy
-->Downloads
-->mysql-connector-java-5.1.26
-->mysql-connector-java-5.1.26.jar
Users
-->remy
-->Documents
-->Hello.java
How to create the myManifest.txt and include the external jar file so as to compile and run them together?
There are two ways.
Create a MANIFEST.MF that refers to the other jar in its classpath
Whenever you run the program, make sure you include both jars in your classpath.
That said, I think you should go back and read a tutorial because you've made a few mistakes in your question.
"hello.java" is not a jar file as you've said. This is a java source file. It will be compiled into a class file. You can take this class file and put it in a jar file.
You can't name your manifest file "myManifest.txt" AFAIK. It has to be named something specific and be located in a specific place. Read a tutorial to see the details on that.
Please any one give me the suggestion for this. I'm having the Xerces-J-source.zip file and I need to convert to Xerces-J-source.jar file.
Generally if your ZIP contains the class files (And not the sources):
Just rename the file, a jar is a zip file. But if you're aiming for a class to be launched with command java -jar myProject.jar you should create a MANIFEST file containing the main-class and libraries to use in the classpath.
So in your case since you got the sources (Java file), you'll have to compile classes and create the JAR. Eclipse has an Export as JAR feature if you use it.
That doesn't make any sense. If you need Xerces-J the better is download the jars from the website.
If you want the get the jar form the source, you need to compile it, but there is no need when you can download it.
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.
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.