There was a program that I used that made runnable .jar files.. All the ones I'm finding now are ones that make .exe files.. I remember it also has the option to make the file a .sh script as well. Anyone knows its name? I've been searching for hours with no avail :/
The command line
java -jar file.jar
Will run your jar file if it has a Main-Class defined as explained here.
You can use that command in a shell script.
You can create a runnable jar using NetBeans IDE or Eclipse IDE by just providing the main class to run. Rest of the things it will take automatically. That class must be having a main() method in it. Then you can run that jar file using java -jar yourjarfile.jar
Do you mean actually coding java and then compiling to .jar? If you do try
eclipse code editor
I used eclipse to make minecraft mods. It will work if you want to make .jar programs.
If you want to have a jar that you can execute using the usual syntax ./app.jar (instead of java -jar), here is a post explaining the process: how to create executable jars.
Basically, JAR is a variant of ZIP, which allows random bytes to be pre/appended to the JAR without corrrupting it. This means it is possible to prepend a launcher script at the beginning of the jar to make it "executable".
Here is a simple example of the process:
# Append a basic launcher script to the jar
cat \
<(echo '#!/bin/sh')\
<(echo 'exec java -jar $0 "$#"')\
<(echo 'exit 0')\
original.jar > executable.jar
# Make the new jar executable
chmod +x executable.jar
With this, you can now run ./executable.jar instead of java -jar original.jar. This works on all unix like systems including Linux, MacOS, Cygwin, and Windows Linux subsystem.
I am making a build script as a batch file (don't ask me why or suggest alternatives. You won't be helping). I have a variable called CLASSPATH that I use with the java compiler. CLASSPATH contains paths to numerous directories and jar files. In addition to those, I'd like to add every jar file in the .[some-long-path]\lib\ directory
It looks something like this:
SET /p dummy=%CLASSPATH%>classpath.tmp~<nul
SET WAR_LIB_PATH=war\WEB-INF\lib
DIR %WAR_LIB_PATH% /B | findstr /L ".jar" > jars.tmp~
REM Have to put it into an external file
FOR /f %%j in (jars.tmp~) do (
SET /p dummy=;%WAR_LIB_PATH%\%%j>>classpath.tmp~<nul
)
SET /P CLASSPATH=<classpath.tmp~
ECHO %CLASSPATH%
This ALMOST works. There are just two problems:
Somehow a space appears between entries, which ruins the classpath.
It abruptly ends after 1024 characters.
Can someone help me with this?
If you use java6, it's enough to write dir/* to include all jars in the dir
http://download.oracle.com/javase/6/docs/technotes/tools/windows/javac.html
For example, if directory foo contains
a.jar and b.JAR, then the class path
element foo/* is expanded to
A.jar;b.JAR
If you are running javac, then try using the -classpath command-line argument instead of the environment variable, since those variables are size-limited on different operating systems.
And purely as a side note, if you are running a program from a JAR (ex java -jar app.jar), you can add metadata do the JAR file that accomplishes the same thing.
I am writing a Java program and I need to set a temporary classpath that includes my package. The package is on my Ubuntu desktop, which I am importing as /home/gaurav/Desktop. Do you have
any idea how to set the Java CLASSPATH temporarily?
You set the Java classpath on Ubuntu the same way as you do on any Linux / UNIX platform (or indeed on Windows ... modulo some minor syntactic differences). There are two ways:
$ java -cp <classpath> some.ClassName arg1 arg2 ...
or
$ export CLASSPATH=<classpath>
$ java some.ClassName arg1 arg2 ...
where <classpath> is a sequence of pathnames with ':' separators.
For more details refer to the manual entry for the 'java' command; e.g. here and here.
If you don't understand export CLASSPATH=... read the Ubuntu manual entry for bash, paying attention to what it says about setting variables, environment variables, and the export built-in shell command. (Hint: $ man bash.)
This is temporary. To make it permanent, add the line to the relevant shell init script; see man bash for details.
how do i get details for my path which path i have set
The classpath is a list of pathnames of directories and JAR files that you want the JVM to search in order to find the classes it needs to run your application. You'll need to figure that out yourself ... or (re-)read the documentation of whatever it is you are trying to run.
If you want to run a java program from the desktop, you have three choices.
The easy choice is to write a small shell script and put that on the desktop. The smallest example might be:
#!/bin/sh
java -cp YOUR_CLASSPATH YOUR_CLASS_NAME "$*"
Next comes the use of 'jarjar' or 'shade' to make one big jar containing all your dependencies, and then run that with java -jar. (As a sub-option, if it's really just for you, you can make a jar with a META-INF/MANIFEST.MF containing a class path with absolute pathnames.)
The more complex choice is to learn to use JNLP to build a launchable item.
IIRC you can control the classpath with either an environment variable or a command-line option to java.
Is there a way to include all the jar files within a directory in the classpath?
I'm trying java -classpath lib/*.jar:. my.package.Program and it is not able to find class files that are certainly in those jars. Do I need to add each jar file to the classpath separately?
Using Java 6 or later, the classpath option supports wildcards. Note the following:
Use straight quotes (")
Use *, not *.jar
Windows
java -cp "Test.jar;lib/*" my.package.MainClass
Unix
java -cp "Test.jar:lib/*" my.package.MainClass
This is similar to Windows, but uses : instead of ;. If you cannot use wildcards, bash allows the following syntax (where lib is the directory containing all the Java archive files):
java -cp "$(printf %s: lib/*.jar)"
(Note that using a classpath is incompatible with the -jar option. See also: Execute jar file with multiple classpath libraries from command prompt)
Understanding Wildcards
From the Classpath document:
Class path entries can contain the basename wildcard character *, which is considered equivalent to specifying a list of all the files
in the directory with the extension .jar or .JAR. For example, the
class path entry foo/* specifies all JAR files in the directory named
foo. A classpath entry consisting simply of * expands to a list of all
the jar files in the current directory.
A class path entry that contains * will not match class files. To
match both classes and JAR files in a single directory foo, use either
foo;foo/* or foo/*;foo. The order chosen determines whether the
classes and resources in foo are loaded before JAR files in foo, or
vice versa.
Subdirectories are not searched recursively. For example, foo/* looks
for JAR files only in foo, not in foo/bar, foo/baz, etc.
The order in which the JAR files in a directory are enumerated in the
expanded class path is not specified and may vary from platform to
platform and even from moment to moment on the same machine. A
well-constructed application should not depend upon any particular
order. If a specific order is required then the JAR files can be
enumerated explicitly in the class path.
Expansion of wildcards is done early, prior to the invocation of a
program's main method, rather than late, during the class-loading
process itself. Each element of the input class path containing a
wildcard is replaced by the (possibly empty) sequence of elements
generated by enumerating the JAR files in the named directory. For
example, if the directory foo contains a.jar, b.jar, and c.jar, then
the class path foo/* is expanded into foo/a.jar;foo/b.jar;foo/c.jar,
and that string would be the value of the system property
java.class.path.
The CLASSPATH environment variable is not treated any differently from
the -classpath (or -cp) command-line option. That is, wildcards are
honored in all these cases. However, class path wildcards are not
honored in the Class-Path jar-manifest header.
Note: due to a known bug in java 8, the windows examples must use a backslash preceding entries with a trailing asterisk: https://bugs.openjdk.java.net/browse/JDK-8131329
Under Windows this works:
java -cp "Test.jar;lib/*" my.package.MainClass
and this does not work:
java -cp "Test.jar;lib/*.jar" my.package.MainClass
Notice the *.jar, so the * wildcard should be used alone.
On Linux, the following works:
java -cp "Test.jar:lib/*" my.package.MainClass
The separators are colons instead of semicolons.
We get around this problem by deploying a main jar file myapp.jar which contains a manifest (Manifest.mf) file specifying a classpath with the other required jars, which are then deployed alongside it. In this case, you only need to declare java -jar myapp.jar when running the code.
So if you deploy the main jar into some directory, and then put the dependent jars into a lib folder beneath that, the manifest looks like:
Manifest-Version: 1.0
Implementation-Title: myapp
Implementation-Version: 1.0.1
Class-Path: lib/dep1.jar lib/dep2.jar
NB: this is platform-independent - we can use the same jars to launch on a UNIX server or on a Windows PC.
My solution on Ubuntu 10.04 using java-sun 1.6.0_24 having all jars in "lib" directory:
java -cp .:lib/* my.main.Class
If this fails, the following command should work (prints out all *.jars in lib directory to the classpath param)
java -cp $(for i in lib/*.jar ; do echo -n $i: ; done). my.main.Class
Short answer: java -classpath lib/*:. my.package.Program
Oracle provides documentation on using wildcards in classpaths here for Java 6 and here for Java 7, under the section heading Understanding class path wildcards. (As I write this, the two pages contain the same information.) Here's a summary of the highlights:
In general, to include all of the JARs in a given directory, you can use the wildcard * (not *.jar).
The wildcard only matches JARs, not class files; to get all classes in a directory, just end the classpath entry at the directory name.
The above two options can be combined to include all JAR and class files in a directory, and the usual classpath precedence rules apply. E.g. -cp /classes;/jars/*
The wildcard will not search for JARs in subdirectories.
The above bullet points are true if you use the CLASSPATH system property or the -cp or -classpath command line flags. However, if you use the Class-Path JAR manifest header (as you might do with an ant build file), wildcards will not be honored.
Yes, my first link is the same one provided in the top-scoring answer (which I have no hope of overtaking), but that answer doesn't provide much explanation beyond the link. Since that sort of behavior is discouraged on Stack Overflow these days, I thought I'd expand on it.
Windows:
java -cp file.jar;dir/* my.app.ClassName
Linux:
java -cp file.jar:dir/* my.app.ClassName
Remind:
- Windows path separator is ;
- Linux path separator is :
- In Windows if cp argument does not contains white space, the "quotes" is optional
For me this works in windows .
java -cp "/lib/*;" sample
For linux
java -cp "/lib/*:" sample
I am using Java 6
You can try java -Djava.ext.dirs=jarDirectory
http://docs.oracle.com/javase/6/docs/technotes/guides/extensions/spec.html
Directory for external jars when running java
Correct:
java -classpath "lib/*:." my.package.Program
Incorrect:
java -classpath "lib/a*.jar:." my.package.Program
java -classpath "lib/a*:." my.package.Program
java -classpath "lib/*.jar:." my.package.Program
java -classpath lib/*:. my.package.Program
If you are using Java 6, then you can use wildcards in the classpath.
Now it is possible to use wildcards in classpath definition:
javac -cp libs/* -verbose -encoding UTF-8 src/mypackage/*.java -d build/classes
Ref: http://www.rekk.de/bloggy/2008/add-all-jars-in-a-directory-to-classpath-with-java-se-6-using-wildcards/
If you really need to specify all the .jar files dynamically you could use shell scripts, or Apache Ant. There's a commons project called Commons Launcher which basically lets you specify your startup script as an ant build file (if you see what I mean).
Then, you can specify something like:
<path id="base.class.path">
<pathelement path="${resources.dir}"/>
<fileset dir="${extensions.dir}" includes="*.jar" />
<fileset dir="${lib.dir}" includes="*.jar"/>
</path>
In your launch build file, which will launch your application with the correct classpath.
Please note that wildcard expansion is broken for Java 7 on Windows.
Check out this StackOverflow issue for more information.
The workaround is to put a semicolon right after the wildcard. java -cp "somewhere/*;"
To whom it may concern,
I found this strange behaviour on Windows under an MSYS/MinGW shell.
Works:
$ javac -cp '.;c:\Programs\COMSOL44\plugins\*' Reclaim.java
Doesn't work:
$ javac -cp 'c:\Programs\COMSOL44\plugins\*' Reclaim.java
javac: invalid flag: c:\Programs\COMSOL44\plugins\com.comsol.aco_1.0.0.jar
Usage: javac <options> <source files>
use -help for a list of possible options
I am quite sure that the wildcard is not expanded by the shell, because e.g.
$ echo './*'
./*
(Tried it with another program too, rather than the built-in echo, with the same result.)
I believe that it's javac which is trying to expand it, and it behaves differently whether there is a semicolon in the argument or not. First, it may be trying to expand all arguments that look like paths. And only then it would parse them, with -cp taking only the following token. (Note that com.comsol.aco_1.0.0.jar is the second JAR in that directory.) That's all a guess.
This is
$ javac -version
javac 1.7.0
All the above solutions work great if you develop and run the Java application outside any IDE like Eclipse or Netbeans.
If you are on Windows 7 and used Eclipse IDE for Development in Java, you might run into issues if using Command Prompt to run the class files built inside Eclipse.
E.g. Your source code in Eclipse is having the following package hierarchy:
edu.sjsu.myapp.Main.java
You have json.jar as an external dependency for the Main.java
When you try running Main.java from within Eclipse, it will run without any issues.
But when you try running this using Command Prompt after compiling Main.java in Eclipse, it will shoot some weird errors saying "ClassNotDef Error blah blah".
I assume you are in the working directory of your source code !!
Use the following syntax to run it from command prompt:
javac -cp ".;json.jar" Main.java
java -cp ".;json.jar" edu.sjsu.myapp.Main
[Don't miss the . above]
This is because you have placed the Main.java inside the package edu.sjsu.myapp and java.exe will look for the exact pattern.
Hope it helps !!
macOS, current folder
For Java 13 on macOS Mojave…
If all your .jar files are in the same folder, use cd to make that your current working directory. Verify with pwd.
For the -classpath you must first list the JAR file for your app. Using a colon character : as a delimiter, append an asterisk * to get all other JAR files within the same folder. Lastly, pass the full package name of the class with your main method.
For example, for an app in a JAR file named my_app.jar with a main method in a class named App in a package named com.example, alongside some needed jars in the same folder:
java -classpath my_app.jar:* com.example.App
For windows quotes are required and ; should be used as separator. e.g.:
java -cp "target\\*;target\\dependency\\*" my.package.Main
Short Form: If your main is within a jar, you'll probably need an additional '-jar pathTo/yourJar/YourJarsName.jar ' explicitly declared to get it working (even though 'YourJarsName.jar' was on the classpath)
(or, expressed to answer the original question that was asked 5 years ago: you don't need to redeclare each jar explicitly, but does seem, even with java6 you need to redeclare your own jar ...)
Long Form:
(I've made this explicit to the point that I hope even interlopers to java can make use of this)
Like many here I'm using eclipse to export jars: (File->Export-->'Runnable JAR File'). There are three options on 'Library handling' eclipse (Juno) offers:
opt1: "Extract required libraries into generated JAR"
opt2: "Package required libraries into generated JAR"
opt3: "Copy required libraries into a sub-folder next to the generated JAR"
Typically I'd use opt2 (and opt1 was definitely breaking), however native code in one of the jars I'm using I discovered breaks with the handy "jarinjar" trick that eclipse leverages when you choose that option. Even after realizing I needed opt3, and then finding this StackOverflow entry, it still took me some time to figure it out how to launch my main outside of eclipse, so here's what worked for me, as it's useful for others...
If you named your jar: "fooBarTheJarFile.jar"
and all is set to export to the dir: "/theFully/qualifiedPath/toYourChosenDir".
(meaning the 'Export destination' field will read: '/theFully/qualifiedPath/toYourChosenDir/fooBarTheJarFile.jar' )
After you hit finish, you'll find eclipse then puts all the libraries into a folder named 'fooBarTheJarFile_lib' within that export directory, giving you something like:
/theFully/qualifiedPath/toYourChosenDir/fooBarTheJarFile.jar
/theFully/qualifiedPath/toYourChosenDir/fooBarTheJarFile_lib/SomeOtherJar01.jar
/theFully/qualifiedPath/toYourChosenDir/fooBarTheJarFile_lib/SomeOtherJar02.jar
/theFully/qualifiedPath/toYourChosenDir/fooBarTheJarFile_lib/SomeOtherJar03.jar
/theFully/qualifiedPath/toYourChosenDir/fooBarTheJarFile_lib/SomeOtherJar04.jar
You can then launch from anywhere on your system with:
java -classpath "/theFully/qualifiedPath/toYourChosenDir/fooBarTheJarFile_lib/*" -jar /theFully/qualifiedPath/toYourChosenDir/fooBarTheJarFile.jar package.path_to.the_class_with.your_main.TheClassWithYourMain
(For Java Newbies: 'package.path_to.the_class_with.your_main' is the declared package-path that you'll find at the top of the 'TheClassWithYourMain.java' file that contains the 'main(String[] args){...}' that you wish to run from outside java)
The pitfall to notice: is that having 'fooBarTheJarFile.jar' within the list of jars on your declared classpath is not enough. You need to explicitly declare '-jar', and redeclare the location of that jar.
e.g. this breaks:
java -classpath "/theFully/qualifiedPath/toYourChosenDir/fooBarTheJarFile.jar;/theFully/qualifiedPath/toYourChosenDir/fooBarTheJarFile_lib/*" somepackages.inside.yourJar.leadingToTheMain.TheClassWithYourMain
restated with relative paths:
cd /theFully/qualifiedPath/toYourChosenDir/;
BREAKS: java -cp "fooBarTheJarFile_lib/*" package.path_to.the_class_with.your_main.TheClassWithYourMain
BREAKS: java -cp ".;fooBarTheJarFile_lib/*" package.path_to.the_class_with.your_main.TheClassWithYourMain
BREAKS: java -cp ".;fooBarTheJarFile_lib/*" -jar package.path_to.the_class_with.your_main.TheClassWithYourMain
WORKS: java -cp ".;fooBarTheJarFile_lib/*" -jar fooBarTheJarFile.jar package.path_to.the_class_with.your_main.TheClassWithYourMain
(using java version "1.6.0_27"; via OpenJDK 64-Bit Server VM on ubuntu 12.04)
You need to add them all separately. Alternatively, if you really need to just specify a directory, you can unjar everything into one dir and add that to your classpath. I don't recommend this approach however as you risk bizarre problems in classpath versioning and unmanagability.
The only way I know how is to do it individually, for example:
setenv CLASSPATH /User/username/newfolder/jarfile.jar:jarfile2.jar:jarfile3.jar:.
Hope that helps!
class from wepapp:
> mvn clean install
> java -cp "webapp/target/webapp-1.17.0-SNAPSHOT/WEB-INF/lib/tool-jar-1.17.0-SNAPSHOT.jar;webapp/target/webapp-1.17.0-SNAPSHOT/WEB-INF/lib/*" com.xx.xx.util.EncryptorUtils param1 param2
Think of a jar file as the root of a directory structure. Yes, you need to add them all separately.
Not a direct solution to being able to set /* to -cp but I hope you could use the following script to ease the situation a bit for dynamic class-paths and lib directories.
libDir2Scan4jars="../test";cp=""; for j in `ls ${libDir2Scan4jars}/*.jar`; do if [ "$j" != "" ]; then cp=$cp:$j; fi; done; echo $cp| cut -c2-${#cp} > .tmpCP.tmp; export tmpCLASSPATH=`cat .tmpCP.tmp`; if [ "$tmpCLASSPATH" != "" ]; then echo .; echo "classpath set, you can now use ~> java -cp \$tmpCLASSPATH"; echo .; else echo .; echo "Error please check libDir2Scan4jars path"; echo .; fi;
Scripted for Linux, could have a similar one for windows too. If proper directory is provided as input to the "libDir2Scan4jars"; the script will scan all the jars and create a classpath string and export it to a env variable "tmpCLASSPATH".
Set the classpath in a way suitable multiple jars and current directory's class files.
CLASSPATH=${ORACLE_HOME}/jdbc/lib/ojdbc6.jar:${ORACLE_HOME}/jdbc/lib/ojdbc14.jar:${ORACLE_HOME}/jdbc/lib/nls_charset12.jar;
CLASSPATH=$CLASSPATH:/export/home/gs806e/tops/jconn2.jar:.;
export CLASSPATH
I have multiple jars in a folder. The below command worked for me in JDK1.8 to include all jars present in the folder. Please note that to include in quotes if you have a space in the classpath
Windows
Compiling: javac -classpath "C:\My Jars\sdk\lib\*" c:\programs\MyProgram.java
Running: java -classpath "C:\My Jars\sdk\lib\*;c:\programs" MyProgram
Linux
Compiling: javac -classpath "/home/guestuser/My Jars/sdk/lib/*" MyProgram.java
Running: java -classpath "/home/guestuser/My Jars/sdk/lib/*:/home/guestuser/programs" MyProgram
Order of arguments to java command is also important:
c:\projects\CloudMirror>java Javaside -cp "jna-5.6.0.jar;.\"
Error: Unable to initialize main class Javaside
Caused by: java.lang.NoClassDefFoundError: com/sun/jna/Callback
versus
c:\projects\CloudMirror>java -cp "jna-5.6.0.jar;.\" Javaside
Exception in thread "main" java.lang.UnsatisfiedLinkError: Unable