I am trying to execute a java application, which is packaged as jar archive.
java -cp .\lib\json.jar -jar ".\myarchive.jar"
I get an error saying that the classes inside my json.jar archive cannot be found.
Exception in thread "main" java.lang.NoClassDefFoundError: json/serializers/JsonTypeResolversInstance
Caused by: java.lang.ClassNotFoundException: json.serializers.JsonTypeResolversInstance
at java.net.URLClassLoader.findClass(URLClassLoader.java:387)
at java.lang.ClassLoader.loadClass(ClassLoader.java:418)
at sun.misc.Launcher$AppClassLoader.loadClass(Launcher.java:352)
at java.lang.ClassLoader.loadClass(ClassLoader.java:351)
... 2 more
The jar file contains the class, so i think this error should not happen. When executing the code using my IDE it runs without errors.
I have tried to fix this in many ways, without success.
Using the -cp and -jar options at the same time does not work. When you use the -jar option, then the -cp option is ignored.
There are two ways to run code in a JAR file in Java.
First way: Use the -cp option and specify the name of the class that contains the main method on the command line. For example:
java -cp .\lib\json.jar;.\myarchive.jar com.mypackage.MyMainClass
When you do it like this, you specify the classpath on the command line (using -cp). The classpath must contain all JAR files and/or directories that contain all the classes that the application needs. You must also specify the name of the class to run on the command line.
Also, when you do it like this, the manifest file that might be present in the JAR file is ignored.
Second way: Use the -jar option. For example:
java -jar .\myarchive.jar
When you do it like this, then Java will look at the manifest file in the JAR file. The classpath and the name of the class to run will be taken from the manifest file, and the -cp option on the command line will be ignored.
For details see the documentation of the java command.
I'm a dev student
I would love to use Picocli in my project, unfortunately I doesn't understand how to compile using Picocli
I trie to follow the instruction given here https://picocli.info/ or here https://picocli.info/quick-guide.html but the step to compile aren't detailed. I'm not using Gradle nor Maven but they aren't really listed as required.
This is how it tried to compile the Checksum example given in the picocli.info webpage :
jar cf checksum.jar Checksum.java ; jar cf picocli-4.6.1.jar CommandLine.java && echo "hello" > hello
Then I simply copy paste this gived command : https://picocli.info/#_running_the_application
java -cp "picocli-4.6.1.jar:checksum.jar" CheckSum --algorithm SHA-1 hello
And get the following result :
Error: Could not find or load main class CheckSum
Caused by: java.lang.ClassNotFoundException: CheckSum
I tried to compile everything myself and then add the .jar like this :
java CheckSum -jar picocli-4.6.1.jar
But then the error output looks like this:
Exception in thread "main" java.lang.NoClassDefFoundError: picocli/CommandLine
at CheckSum.main(Checksum.java:33)
Caused by: java.lang.ClassNotFoundException: picocli.CommandLine
at java.base/jdk.internal.loader.BuiltinClassLoader.loadClass(BuiltinClassLoader.java:581)
at java.base/jdk.internal.loader.ClassLoaders$AppClassLoader.loadClass(ClassLoaders.java:178)
at java.base/java.lang.ClassLoader.loadClass(ClassLoader.java:521)
... 1 more
Witch I don't understand since I added the dependency.
What am I missing ?
Thanks in advance
The problem is that the command jar cf checksum.jar Checksum.java only creates a jar file (jar files are very similar to zip files) that contains the Checksum.java source file.
What you want to do instead is compile the source code first. After that, we can put the resulting Checksum.class file (note the .class extension instead of the .java extension) in the checksum.jar. The Java SDK includes the javac tool that can be used to compile the source code. Detailed steps follow below.
First, open a terminal window and navigate to a directory that contains both the Checksum.java source file and the picocli-4.6.1.jar library.
Now, the command to compile (on Windows) is:
javac -cp .;picocli-4.6.1.jar Checksum.java
Linux uses : as path separator instead of ;, so on Linux, the command to compile is:
javac -cp .:picocli-4.6.1.jar Checksum.java
The -cp option allows you to specify the classpath, which should contain the directories and jar/zip files containing any other class files that your project uses/depends on. Since Checksum.java uses the picocli classes, we put the picocli jar in the classpath. Also add the current directory . to the classpath when the current directory contains any classes. I just add . habitually now.
Now, if you list the files in the current directory, you should see that a file Checksum.class has been created in this directory.
Our Checksum class has a main method, so we can now run the program with the java tool:
On Windows:
java -cp .;picocli-4.6.1.jar Checksum
On Linux:
java -cp .:picocli-4.6.1.jar Checksum
Note that when running the program with java you specify the class name Checksum, not the file name Checksum.class.
You can pass arguments to the Checksum program by passing them on the command line immediately following the class name:
java -cp .:picocli-4.6.1.jar Checksum --algorithm=SHA-1 /path/to/hello
When your project grows, you may want to keep the source code and the compiled class files in separate directories. The javac compile utility has a -d option where you can specify the destination for the compiled class files. For example:
javac -cp picocli-4.6.1.jar:otherlib.jar -d /destination/path /path/to/source/*.java
This should generate .class files for the specified source files in the specified destination directory (/destination/path in the example above).
When you have many class files, you may want to bundle them in a single jar file. You can use the jar command for that. I often use the options -v (verbose) -c (create) -f (jar file name) when creating a jar for the compiled class files. For example:
jar -cvf MyJar.jar /destination/path/*.class /destination/path2/*.class
Enjoy!
I'm trying to run tests from the command line (PowerShell in Windows 10).
Before asking this question I looked in several sources and read a lot of topics, like
How to run JUnit test cases from the command line
https://github.com/junit-team/junit4/wiki/Getting-started
But when I'm trying to run tests from PowerShell like in JUnit wiki
cd E:\Dev\AutoTest\Example
java -cp .;/libs/junit-4.12.jar;/libs/hamcrest-core-1.3.jar org.junit.runner.JUnitCore tests.Booking.BookingTests
I get the output
CommandNotFoundException
If I run the same, but via old command line (cmd.exe):
cd E:\Dev\AutoTest\Example
java -cp .;E:\Dev\AutoTest\Example\libs\junit-4.12.jar;E:\Dev\AutoTest\Example\libs\hamcrest-core-1.3.jar org.junit.runner.JUnitCore tests.Booking.BookingTests
I get the output
java.lang.IllegalArgumentException: Could not find class [tests.Booking.BookingTests]
Caused by: java.lang.ClassNotFoundException: tests.Booking.BookingTests
In IDEA the project structure look like this:
On the hard drive the structure look like this:
• "out" folder cointains *.class files
• "src" folder contains *.java files
The question:
How to run JUnit test cases from the command line in PowerShell with my structure?
In PowerShell the semicolon is a command separator (allowing you to put two statements on one line), so you're running java -cp . (which should output the command help) and then /libs/junit-4.12.jar (which is not recognized as a command). The example in the JUnit wiki clearly isn't made for PowerShell, but for CMD, which uses the ampersand (&) for chaining commands, so the issue doesn't occur there.
Also, you made the paths in the classpath absolute (/libs/junit-4.12.jar), but your libs directory is in your project folder. That is why java complains that it can't find the class org.junit.runner.JUnitCore. When you're running JUnit from the root of your project directory you need to make the paths relative to that location.
And since you compiled your code to a different folder (.\out\production\Afl_AutoTest) you must add that folder to your classpath as well, otherwise JUnit won't be able to find the compiled classes (because they're outside the classpath).
Put your classpath in quotes, add the output directory, and remove the leading slashes from the library paths, and the command should work in CMD and PowerShell alike:
java -cp ".;out/production/Afl_AutoTest;libs/junit-4.12.jar;libs/hamcrest-core-1.3.jar" org.junit.runner.JUnitCore tests.Booking.BookingTests
Better yet, define all arguments as an array and splat it:
$classpath = '.',
'out/production/Afl_AutoTest',
'libs/junit-4.12.jar',
'libs/hamcrest-core-1.3.jar'
$params = '-cp', ($classpath -join ';'),
'org.junit.runner.JUnitCore',
'tests.Booking.BookingTests'
java #params
The latter only works in PowerShell, though.
Final comand must include all "libs" folder files:
• java-client-4.1.2.jar
• junit-4.12.jar
• selenium-server-standalone-3.4.0.jar
• hamcrest-core-1.3.jar
java -cp ".;out/production/Afl_AutoTest;libs/java-client-4.1.2.jar;libs/junit-4.12.jar;libs/selenium-server-standalone-3.4.0.jar;libs/hamcrest-core-1.3.jar" org.junit.runner.JUnitCore tests.Booking.BookingTests
if not, the output will be
Exception in thread "main" java.lang.NoClassDefFoundError: org/openqa/selenium/Capabilities
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