I have a web program where I want the user to be able to import a .war file and I can extract certain files out of the .war file. I have found two class libraries: java.util.zip.* and java.util.jar.*. From what I understand, a WAR file is a special JAR file which is a special ZIP file. So would it be better to use java.util.jar? If ZIP and JAR files are pretty much the same why is there a need for two different libraries?
WAR file is just a JAR file, to extract it, just issue following jar command –
jar -xvf yourWARfileName.war
If the jar command is not found, which sometimes happens in the Windows command prompt, then specify full path i.e. in my case it is,
c:\java\jdk-1.7.0\bin\jar -xvf my-file.war
If you look at the JarFile API you'll see that it's a subclass of the ZipFile class.
The jar-specific classes mostly just add jar-specific functionality, like direct support for manifest file attributes and so on.
It's OOP "in action"; since jar files are zip files, the jar classes can use zip functionality and provide additional utility.
Just rename the .war into .jar and unzip it using Winrar (or any other archive manager).
If you using Linux or Ubuntu than you can directly extract data from .war file.
A war file is just a jar file, to extract it, just issue following command using the jar program:
jar -xvf yourWARfileName.war
You can use a turn-around and just deploy the application into tomcat server: just copy/paste under the webapps folder.
Once tomcat is started, it will create a folder with the app name and you can access the contents directly
For mac users: in terminal command :
unzip yourWARfileName.war
Like you said, a jar is a zip file (not a special type, but just a plain old zip), so either library could be made to work. The reasoning is that the average person, seeing a *.zip extension, tends to unzip it. Since the app server wants it unzipped, a simple rename keeps people from unzipping it simply out of habit. Likewise, *.war file also should remain uncompressed.
java.util.jar basically just adds additional functionality to java.util.zip with very little extra overhead. Let the java.util.jar be a helper in posting, etc... and use it.
Jar class/package is for specific Jar file mechanisms where there is a manifest that is used by the Jar files in some cases.
The Zip file class/package handles any compressed files that include Jar files, which is a type of compressed file.
The Jar classes thus extend the Zip package classes.
This is the way to unarchive war
mkdir mywarfile
cp -r mywarfile.war mywarfile/
cd mywarfile/
jar -xvf mywarfile.war
ls
rm -rf mywarfile.war
Related
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.
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
I took an old jar file and edited it. Now i have a folder that has all the files and I want to recompile them back to a jar file. How do i do this?
Jar files aren't "compiled" in the normal sense of the word. But you just create one with the jar command:
jar cvf myfile.jar file1 file2 etc
You can use globbing as normal:
jar cvf ../myfile.jar *
Run jar -? or read the docs for more information.
Note that for an executable jar file, you'll need to specify the manifest file as well.
How did you edit it?
Anyway, jar files follow the same format as regular zip files. If you want to update the content of the jar ( probably with a new file ) you could do:
jar -uf yourJar.jar path/to/updated/Class.class
That would update your jar with the file path/to/updated/Class.class If there's already a class with that name in that path, it would be replaced. If is not it would be created.
After this your jar is up to date.
Jar files usually contain compiled java files (class-files) and resources.
If you are allowed to make changes on this jar, you could disassemble the class files using JAD and after recompilation assemble them again with the command jar
How did you edit a jar file, hex editor?
A jar file is just a zip file with a few extra files. You Can use any zip utility to rejar an unjarred jar file.
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.
I have updated my ant build.xml file to include a new file and a new folder. After creating the .jar I check if they exist in the jar by 'unzip\extract', and they are there.
But when executing the .jar neither the folder or the file gets extracted.
Am I missing a step?
Look into getResourceAsStream. It'll keep you from having to extract the files from the jar file. Unless that's your goal.
Your application should be able to use the file directly from within the jar, no need for extracting it. Or do you mean something else?
Are you doing something specific to extract the jar file? I ask because normally jar files are not extracted when executing them.
If you run "java -jar myJar.jar" or "java -cp myJar.jar com.example.MyMainClass" the jar files that is referenced will not be extracted. Java will load your classes and resources directly from the jar file without extracting it.
If you wrap your application up using One-JAR, you can specify an attribute in the Manifest file to extract files that you want (See the One-Jar-Expand manifest attribute).
As a bonus, you will also be able to wrap any dependent libraries along with your code, creating a single distributable jar.