I want to create a single jar file from multiple packages. I have created the jar using below command but when I'm importing it to a project as a dependency it is not working.
jar cfe output/jar/my-java.jar Main src/pkg1/pkg0/*.class src/pkg1/*.class src/pkg2/*.class
My project structure is something like below structure
src
pkg1
A.java
B.JAVA
pkg0
E.java
pkg2
C.java
D.java
My Example code is something like
import pkg1.A;
public class Main {
public static void main(String[] args) {
A.printMe("Hello World");
}
}
error that I'm getting is:
java pkg1 not exist
But in the editor(IntelliJ), it is not showing errors and also i'm able to import class but not package.
import pkg1: showing red means error in the editor
import pkg1.A: not showing red means no error in the editor
Note: I don't want to use maven.
An unzip -t something.jar shows the actual file structure of the jar file (zip). It is the same as the class structure of it (except that instead "/", a "." is the separator).
In your case, the problem will be that src will be on the top level, and not pkg1. Either import src.pkg1 (very dirty), or play a little bit more with the directories / jar flags.
just call your method it automatic get path if you correctly put your jar in your current project.`project 1
pkg com.test.demo
class test{
public static void m1(){
System.out.println("project 1 in method 1 );
}
}
in project 2 put jar of project 1
pkg com.test.demo
class Test1{
public static void main(String...){
System.out.println(test.m1())
}
}
I am learning how to use packages in Java, but I am running into trouble when trying to implement them. I have a simple class called Main which appears as follows:
public class Main
{
public static void main(String[]args)
{
System.out.println("Package Test...");
}
}
The directory of this class is: C:\Users\MyComputer\Desktop\Packages\Main.java
When I compile this class, I run into no trouble. However, when I add "package com.example.mypackage;" to the top of the .java file, compile the program, and try to run the program, I receive the following error: "Error: Could not find or load main class Main"
What can I do to solve this problem?
If the path of your class is C:\Users\MyComputer\Desktop\Packages\Main.java then your class is not in a package. In this instance "Packages" is your project folder, and it only contains the one java class.
If you want package com.example.mypackage; to work, then your path needs to be:
C:\Users\MyComputer\Desktop\Packages\com\example\mypackage\Main.java
Edited to restart question from scratch due to complaints. I am a newbie to this format and to intellij so please excuse...
I am building a project in intellij for class. This project imports jnetcap and uses it to process a captured pcap file. My issue is I have two class files I am trying to integrate. NetTraffic which is the user interface class, and ProcessPacket that actually reads in the packet and does the work.
I have tried to make a project and import ProcessPacket into NetPacket but have been unsuccessful so far. I am sure I am missing something simple in this process but I just can not find anything showing the proper way to do this.
I have gotten it working by making a package under the src directory and adding both files to that package. This doesn't require an import from the NetPacket class and seems to work but my worry is that I need to be able to run this from a linux command line. I have been working all semester so far with everything in one source file so it hasn't been an issue until now. I don't remember using packages in the past under eclipse to do this.
Can someone offer a step by step process on how to properly add these source files to my project so that I am able to import ProcessPacket into NetTraffic or will leaving like this in a package work fine?
The files in question reside in package named nettraffic in src directory.
NetTraffic.java
package nettraffic;
public class NetTraffic {
public static ProcessPacket pp;
public static void main (String args[]) {
pp = new ProcessPacket();
pp.PrintOut();
}
}
ProcessPacket.java
package nettraffic;
import org.jnetpcap.*;
public class ProcessPacket {
public ProcessPacket() {
}
public void PrintOut() {
System.out.println("Test");
}
}
Note there is no real functionality in these at this time. Just trying to get the class import syntax correct before continuing. Again while this seems to work as a package I want to have it done without using a package and importing ProcessPacket.java into NetTraffic.java.
public class NetTraffic {
ProcessPacket pp = new ProcessPacket();
pp.PrintOut();
}
You're calling the PrintOut() method outside of any constructor or method or similar block (static or non-static initializer blocks...), and this isn't legal. Put it in a constructor or method.
public class NetTraffic {
public NetTraffic() {
ProcessPacket pp = new ProcessPacket();
pp.PrintOut();
}
}
filename: mainoverloading.java
error: could not find or load main class mainoverloading
class simple{
public static void main(int a)
{
System.out.println(a);
}
public static void main(String args[])
{
System.out.println("Hi");
main(10);
}
}
Your class is named simple (not mainoverloading). Rename the class (or move the file "mainoverloading.java" to "simple.java").
When You compile the above given class using
javac mainoverloading.java
It is successfully compiled and a class file named simple.class is generated in your folder.
You can then run it by typing
java simple.
But this is actually not a good practice, as Elliott Frisch said rename your class into simple.java.
have a look For Windows- HelloWorld
For Linux - HelloWorld
Java allow us to use any name for file name, only when class is not public.
Its running nicely, because for eclipse main class it simple, its smart to identify that simple.class will be created.
If you are running from command line ,
class file created for your code is simple.class so JVM will be unable to find mainoverlading.class
Is there any way to compile a java program without having the java file name with its base class name.
If so, please explain..
To answer the question take a look at this example:
Create a file Sample.java
class A
{
public static void main(String args[])
{
String str[] = {""};
System.out.println("hi");
B.main(str);
}
}
class B
{
public static void main(String args[])
{
System.out.println("hello");
}
}
now you compile it as javac Sample.java and run as java A then output will be
hi
hello
or you run as java B then output will be hello
Notice that none of the classes are marked public therefore giving them default access. Files without any public classes have no file naming restrictions.
Your Java file name should always reflect the public class defined within that file. Otherwise, you will get a compiler error. For example, test.java:
public class Foo {}
Trying to compile this gives:
[steven#scstop:~]% javac test.java
test.java:1: class Foo is public, should be declared in a file named Foo.java
public class Foo {
^
1 error
So you must have your filename match your public class name, which seems to render your question moot. Either that or I don't understand what you're asking... spending some time explaining what you are actually trying to achieve would go a long way towards asking a more effective question :)
As long as you don't have a public class in your source file, you can name your source file to any name and can compile. But, if you have a public class in your source file, that file should have the name same as your class name. Otherwise, compiler will throw an error.
Example:
Filename: TestFileName.java
public class HelloWorld {
public static void main(String[] args) {
System.out.println("Hello,World\n");
}
}
Compiling: javac TestFileName.java
Error:
TestFileName.java:1: class HelloWorld is public, should be declared in a file named HelloWorld.java
public class HelloWorld
^
1 error
No, the public class name must match the file name. Inner, non public, class names may differ.
You must have a public class with the same name as the file name. This is a Very Good Thing. You CAN have secondary classes inside the same file as long as they are not public. They can still be "default" though, so they can still be used by other classes in the same package.
This should not be done for the most part. Java's naming patterns regarding classes and packages are one of the bigger advantages it has--makes a programmers life easier at no cost.
You can use the Java Compile API and compile any java source you wish, the source need not come from a file or could come from a file with an unrelated name. It depends on how obtuse you want to develop your program. ;)
yes, we compile a java file with a different name than the class, provided that there should not be any public class in that file.
If there is any public class in file then in that case you have to give that name as file name. But if your class does not contain any public class then you can give any name to you class.
Please refer below example to make it more clear:
file name : sample.java
class A
{
public static void main(String args[])
{
System.out.println("hi in Class A");
}
}
class B
{
public static void main(String args[])
{
System.out.println("hello in class B");
}
}
then compile it with(windows) : javac sample.java
then run it : java A
output : hi in Class A
then run it : java B
output : hello in class B
Please check and confirm.
It is not necessary to name your file same as the name of the class it has, until this class is public. Though it is a good practice to name the file same as the name of class.
The compiler will compile your file successfully and make a dot class file. Now at the run time you need to give class name to the JVM for that you have to keep the name of the class, which has main method, in your mind. If you keep both the file name and the class name same, it will become easy to remember the name of the compiled dot class file.
for example:
file Dummy.java
class Dummy
{
public static void main(String args[])
{
System.out.println("This is Dummy class running");
}
}
to run the above code we will use :
Javac Dummy.java // to compile
Java Dummy //to run
example:
file Dummy.java
class Diff
{
public static void main(String args[])
{
System.out.println("This is Diff class running");
}
}
to run the above code we will use :
Javac Dummy.java // to compile
Java Diff //to run
I guess what he means is the .java file is named differently than the actual class defined inside it.
I guess this is not possible.
No. You could write a shell script to rename the .java file before compiling it, but javac requires that filenames = class names.
(Also in windows, it's case insensitive, so ASDF.java can compile Asdf.class)
yes, you can choose any name for the file (.java). there is no matter to what are the name of classes in that file means that class names may be totaly different from the file name.
you should compile the program with file name and you should run the program with the class name in which the main method exist.
main methods may be multiple in different classes so you should run it with the class name in which the main method you want to run......
we can save the file tootle different name of class name because in java we compile the program but we run the method.
we have to compile our program with file name and run our class name
Yes,it is possible to compile a java source file with different file name but you need to make sure none of the classes defined inside are public...when you compile the source file the corresponding .class files for the classes inside the source file are created.
Yes,you can save your java source code file with any other name, not same as your main class name but when you comiple it than byte code file name will be same as your main class name. So for your ease of not to memorize to many names for java code run, You need to have your file name same as your main class than only your file name and byte code file will be with same name.
If class is not public you can save it using other name like if classname is Simple then save it Hard.java.
complie->Hard.java
run->Simple.java
Save your java file by .java only.
compile javac .java
run java yourclassname
For example if my program main class name is A then
save by .java only
compile by javac .java
run by java A
yes, we can compile a java file with a different name than the class, provided that there should not be any public class in that file.
If there is any public class in file then in that case you have to give that name as file name. But if your class does not contain any public class then you can give any name to you class.
Please refer below example to make it more clear:
file name : example.java
class A
{
public static void main(String args[])
{
System.out.println("You are in Class A");
}
}
class B
{
public static void main(String args[])
{
System.out.println("You are in class B");
}
}
then compile it with : javac example.java
then run it : java A
output : you are in Class A
then run it : java B
output : you are in class B
Please check and confirm.
You can write more than one main methods in java because java provides main method overloading in which main method can also be overloaded . Once you compile the file here example.java
Compiler create .class file which contains main method when you run the file with java A it will run the A.class file whih contains the main method of class A and that output will be display on you screen ,but when you run this file with java B ,It runs the B.class file which provides main method of B class
So your code is run successfully
Yes. This can be done.
The main reason for the class and file name to be same are to make the job of the complier easy to check which class it needs to run, in the whole list of the Java classes.
So it's a good practice to have filename and class name as same.
And you have compile and run a class with different name other than the filename, If you don't have any public methods in this class.
By convention, the name of the main class should match the name of the file that holds the program. You should also make sure that the capitalization of the filename matches the class name.
The convention that filenames correspond to class names may seem arbitrary. However, this convention makes it easier to maintain and organize your programs. Furthermore, in some cases, it is required.
According to the other answers the only viable solution is to somehow determine the classname from the source, then use it to rename the file to proper name and compile it as usual.
Another option is to alter the package and class name in the source to match file name:
sed -i -r "0,/package/s/^\s*package .*?;/package new.klass.pkg;/" %1
sed -i -r "0,/class/s/public\s+class .+?\{/public class NewClassName {/" %1
Via How to use sed to replace only the first occurrence in a file?
You can have your java file even without name ( simply ".java" ). Only thing is you should not have any public class in your file.