Below is my sample class file:
package org.foo.tutorial;
public class App
{
public static void main( String[] args )
{
System.out.println( "Hello World!" );
}
}
In order to execute the project (maven framework) we run:
>java -cp Something-1.0.SNAPSHOT.jar org.foo.tutorial.APP
The above command works fine and gives me the output 'HELLO WORLD'.
However, if I leave out the third argument in the above command (org.foo.tutorial.APP) I get the following error:
Error: Could not find or load main class target.MavenTutorialApp-1.0-SNAPSHOT.jar
My question is:
Why should the groupId and app name matter when I am supplying the entire 'jar' file ?
The error is a bit misleading. Your java command is incorrect since you don't specify a class. The Something-1.0.SNAPSHOT.jar is meant to be part of the -cp option but java is interpreting it as the class.
That's how java behaves
The java command starts a Java application. It does this by starting a
Java runtime environment, loading a specified class, and calling that
class's main method.
If your .jar file contains an entry point specified by a Main-Class header in the manifest, then you can simply run
java -jar Something-1.0.SNAPSHOT.jar
Let me try to answer your question.
Why should the groupId and app name matter when I am supplying the
entire 'jar' file ?
Lets divide the question into smaller part -
What is group id? - The id of the project's group.
What is artifactId? - The id of the artifact (project) and off course version is part of default artifact name.
While running a java program under jvm there are no such effect that how you built jar, for example it is produced by maven or gradle build process or even command line.
Things that matter is jar file and entry point or the class where main reside.
This may be out of scope of this question, but i felt relevant.
To run java program from jar file, you can explicitly mention like above. Also you can create executable jar file by adding manifest file, where it define the entry point of the executable -
Main-Class: org.foo.tutorial.APP
How to create executable jar?
Also maven maven-assembly-plugin helps to package executable jar.
help
Related
So I have a basic hello world set up in eclipse and I can compile it using cmd easily (I have set all the necessary paths), however when I then try to use the java command to execute the hello world, it always returns the same error:
Error: Could not find or load main class helloWorld
Caused by: java.lang.NoClassDefFoundError: net/codejava/helloWorld (wrong name: helloWorld)
This is the code used:
package net.codejava;
public class helloWorld {
public static void main(String[] args) {
System.out.println("Hello World");
}
}
I am cd in the right directory (I think, I cd into the src directory and then into the package file stored in src) and am using Windows 10 with java 18.0.1 and JRE build 18.0.1+10-24
Any help would be greatly appreciated, as this is highly frustrating, when the code runs effortlessly on the eclipse console. Thanks.
Your file has a 'package' of net.codejava and a name of helloWorld, meaning, the full name of this class is net.codejava.helloWorld.
The java command, at least in the way you're using it, requires that you pass the full name, thus, you must run java net.codejava.helloWorld. Just java helloWorld simply isn't going to work.
But that's not all.
Java needs to then find the class file that contains the code for class net.codejava.helloWorld. It does this by first turning that full name into a path of sorts: net/codejava/helloWorld.class, and it will then scan each entry in the classpath for that. You can put directories and jar files on the classpath.
Thus, you have a directory on your system; let's call this directory X. X contains a directory named net, which contains a directory named codejava, which contains a file named helloWorld.class. If there is no such X (i.e. your class file is not in a dir named codejava for example), you're going to have to fix that by making these directories.
Then, X (and not the codejava dir!) needs to be on the classpath. Usually (it depends on how you configured things), 'the current dir' is by default on the classpath.
Given that your code is in, say, /home/PythonSux/workspace/learningjava/net/codejava/helloWorld.class, that means the dir that needs to be on the classpath is /home/PythonSux/workspace/learningjava. After all, if you, from there, look for net/codejava/helloWorld.class, you find the right file.
Therefore, either cd to that directory, or run java -cp /home/PythonSux/workspace/learningjava net.codejava.helloWorld
Note that this isn't usually how you actually run java apps. You either run them from your IDE, or you ask your build tool to run it, or you package your java app into a jar file and run that, etcetera.
Hi Guys I have included the Webcam-Capture API in my project.
When I run it in Netbeans everything works fine. But when i compile everything to a runnable jar i get this message trying to run it by cmd line.
can anyone of you help me?
i already tried to unbound and rebound all jars and changing jdks but its not working
add -classpath flag in the command line ,pointing to the path where Webcam-Capture API exists in your file system, unless you want to create a single package executable.In your case It should be something like below
java -classpath YOURJAR.jar;folder_of_dependant_jar/*;. com.awesome.pagackage.Starter
Where YOURJAR.jar contains the com.awesome.pagackage.Starter.main(String args[])
You also mentioned that your jar is a runnable jar it also means that while exporting/building you can do one of the following way.( NOTE , this feature is in eclipse , but you would get the idea ).Each of the following options you see in the library handling does specific things.
The first option: Extracts the dependent jar into your target jar as java packaging.This means if your package is com.awesome.package and the dependent jar has package logic.package; , after the runnable jar is build you could find both these package exists in your jar file.
The second option: I think it is more on eclipse specific since eclipse adds few classes of its own , of runnable generation, so I am not explaining it here.
The third option : is the most interesting one. it creates folder stucture like below
ndon_lib\external.jar ( external jar file )
ndon.jar ( your jar file )
This time the manifest.mf file contains something like below.
Class-Path: . ndon_lib/external.jar
Main-Class: com.awesome.pagackage.Starter
You should set the classpath
java -cp "your.jar" "yourclass"
Java and Gradle beginner's question.
I made a project directory for java and gradle test:
The directory hierarchy :
HelloWorld.java:
package foo.bar;
public class HelloWorld {
public static void main(String[] args) {
System.out.println("Hello, world");
}
}
build.gradle:
apply plugin:'java'
Then,gradle build this project and generated what i need.
As you see above, my problem is why doesn't this execute correctly? Even through I cd to .class path.
======================================================================
While, if I remove package foo.bar; in HelloWorld.java, and repeat gradle commands and execute at he.bak directory then the error remained the same.
But when I cd to the directory where HelloWorld.java placed. everything goes OK！Why? something related with CLASSPATH environment variables or other causes?
////////////////////////////////////////////////////////////////////
UPDATE
////////////////////////////////////////////////////////////////////
Thought you guys' warm replies, I know that I should combine the CLASSPATH and the period-separated executable .class file to figure out what's going on when executing java class file.
I experiment my thought resulting in 2 point to this question:
The -cp option path parameter A/B plus the executable file c.d.e.class finally form the A/B/c.d.e.class full path where the class is actually located.
If I specify the package in source code file with package d,I must split the full path in the form of java -cp A/B/c/d e.class. split in other ways all will result in errors.
something I am not sure here is :
When I specify my package path in my source code file, It determined the only classpath when executing corresponding executable, right?
If it is the truth, How does a project with lots of package and sources files work?
What's the root principle?
When in build/classes/main try java foo.bar.HelloWorld instead of java HelloWorld
The reason you need to specify foo.bar.HelloWorld is because you specified package foo.bar;. This tells java that the class should be in foo/bar/HelloWorld and the fully qualified name for HelloWorld is foo.bar.HelloWorld. If you want to execute the class from a different working directory however, you can specify the classpath explicitly using the -cp option, e.g., java -cp c:\myproject\build\classes\main foo.bar.HelloWorld.
By the way, the classpath default is the current working directory (i.e., .) but java -cp c:\myproject\build\classes\main foo.bar.HelloWorld will NOT have the classpath set to the current working directory if it is explicitly set using the -cp option. If you want to include the current working directory but explicitly set it, or even add more directories, you can chain them using semicolons like this: java -cp .;c:\myproject\build\classes\main foo.bar.HelloWorld. So this will include both the current working directory and the directory I specified.
I want to use clearNLP (http://clearnlp.wikispaces.com/) for extracting semantic role labels of an input sentence. I followed the instructions here: http://clearnlp.wikispaces.com/installation (I downloaded the jar files, put them in a directory called ClearNLP and set the classpath) but when I run the command java com.clearnlp.run.Version, I face the error: Could Not find or Load Main.
I tried it twice: Once I set the classpath as an environment variable of windows and ran the command in CMD. But, when it didn't work, I tried to create a java project, set the libraries using NetBeans and run the program. But, it didn't work, too.
BTW, when I run echo %classpath% command, I see that the classpath is set correctly.
Can anybody help me?
Try Eclipse. I included the jars in a new project I created. I then created a simple class like so
package test;
import com.clearnlp.run.Version;
public class TestClearNLP {
public static void main String(args[]) {
Version.main(args);
}
}
When run, this creates output in the console of:
ClearNLP version 2.0.2
Webpage: clearnlp.com
Owner : Jinho D. Choi
Contact: support#clearnlp.com
The only weird situation I ran into was that Eclipse did not like the jar files beginning with a period. I removed those from my project and ran with the remaining libraries.
I have written two simple Java classes (one of them containing "main()", and the other called by "main()").
Class #1 (containing "main()"):
package daniel347x.outerjar;
import daniel347x.innerjar.Funky;
public class App
{
public static void main( String[] args )
{
Funky.foo();
}
}
Class #2 (called by "main()"):
package daniel347x.innerjar;
public class Funky
{
public static void foo()
{
System.out.println( "Funky!" );
}
}
The above classes appear in different project root folders, and use Maven as the build system (each project has its own POM). The pom.xml file for the main project includes the proper entry to add daniel347x.outerjar.App as the main class, and it properly includes the dependency on daniel347x.innerjar. Both projects build successfully into JAR files.
I use NetBeans to wrap these as Maven projects (in fact, I used NetBeans to create both projects). When I run the main project from within NetBeans, it runs successfully and I see Funky! as the output.
However, when I attempt to run the main class straight from the Windows command line (cmd.exe), passing the JAR file containing Funky on the command line's classpath, as such:
java -classpath "P:\_Dan\work\JavaProjects\JarFuckup\innerjar\target\innerjar-1.0-SNAPSHOT.jar" -jar "P:\_Dan\work\JavaProjects\JarFuckup\outerjar\target\outerjar-1.0-SNAPSHOT.jar"
... I receive the dreaded NoClassDefFoundError:
Exception in thread "main" java.lang.NoClassDefFoundError: daniel347x/innerjar/Funky
at daniel347x.outerjar.App.main(App.java:7)
I have carefully confirmed that, inside the innerjar JAR file noted above containing Funky, that the path structure is daniel347x\innerjar and that inside the innerjar folder is the Funky.class file - and this file looks correct within a HEX editor where I can see the ASCII strings representing the name of the class.
The fact that the class can't be found defies my understanding of Java, which I thought allows you to pass JAR files as a -classpath parameter and it will be able to find classes inside those JAR files.
This very basic point has me flummoxed - an answer that explains what I am doing wrong would be greatly appreciated.
The classpath is ignored when using the -jar option. A way to run your app would be java -classpath "P:\_Dan\work\JavaProjects\JarFuckup\innerjar\target\innerjar-1.0-SNAPSHOT.jar";"P:\_Dan\work\JavaProjects\JarFuckup\outerjar\target\outerjar-1.0-SNAPSHOT.jar" daniel347x.outerjar.App
Perhaps a better approach would be to add a manifest file to the Jar that specifies the class path of the dependent Jars by relative paths. Then..
java -jar "P:\_Dan\...\outerjar-1.0-SNAPSHOT.jar"
..should do it.
Double clicking the main Jar will also launch it. That is mostly useful for GUIs.