Here My first file code
package com.shubham.packages.a;
import static com.shubham.packages.b.Message.Hello;
public class Greeting {
public static void main(String[] args) {
System.out.println("Hello World");
Hello();
}
}
Here my second file code
package com.shubham.packages.b;
public class Message {
public static void main(String[] args) {
}
public static void Hello() {
System.out.println("This is Awesome.");
}
}
Here the error I got when I run the program.
When you compile your code javac needs to know where to look for all of your source/class files. You could go to "java_tutorial" folder and run.
javac com/shubham/packages/a/Greeting.java com/shubham/packages/b/Message.java
That should compile both of your java files into class files at the same location. Then you should be able to run.
java com.shubham.packages.a.Greeting
I might be using the wrong slashes for windows
You can explicitly name all of the necessary java files, so this should cause both Greeting.java and Message.java to be compiled in place.
When you run java, the CWD is on the classpath by default so that means the package com.shubham.packages.a and ...b should be on the classpath in their correct location.
A slightly better way to do this is to create a folder called "build" or whatever you like.
javac -d build com/shubham/packages/a/Greeting.java com/shubham/packages/b/Message.java
That will output the class files to the build folder. Then when you run it.
java -cp build com.shubham.packages.a.Greeting
package concurrencyTest;
public class concurrencyTest implements Runnable
{
#Override
public void run()
{ System.out.println("Hello from a thread!"); }
public static void main(String[] args)
{
concurrencyTest c = new concurrencyTest();
Thread t = new Thread(c);
t.start();
}
}
Hi, I'm just trying to get my java concurrency test to run. But I'm getting this error :
Error: Main method not found in class concurrencytest.ConcurrencyTest, please
define the main method as:
public static void main(String[] args)
or a JavaFX application class must extend javafx.application.Application
I'm guessing that somewhere, in the myriad of project directories and subfiles that a java program needs to run, the name of the project or class has been wrongly referenced in lowercase letters. I've manually been thru all the files I can find that span from the root directory and renamed any instance of lowercase 'concurrencytest'. But still it seems the compiler finds a reference to lowercase concurrencytest and so refuses to compile. Any idea where this reference may be?
My root directory, source directory, and java code file are all called 'concurrencyTest'
edit
amended the original code for this question to include 'static' in the main method definition. Doing this was necessary but has not fixed the problem.
You wrote the main definition wrong. You should change it as below:
public static void main(String[] args) {
// ...
}
I found a reference to 'concurrencytest' in the meta file ...\concurrencyTest\nbproject\project.properties
So I changed it to 'concurrencyTest'
Also when NetBeans can't find the expected main method in the class it expects, it opens a 'RunProject' window where it lists main methods from classes it has found. For me it found a main in 'concurrencyTest' so I selected it.
One of these actions solved the problem, but not sure which one.
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
I wrote a java program to test RESTful web services by using Netbeans7.0.1 and it works fine there. Now I wrote the build.xml file to compile the code and when I try to run the generated .class file I always got this exception:
Exception in thread "main" java.lang.NoClassDefFoundError: ClientREST (wrong name: clientrest/ClientREST)
at java.lang.ClassLoader.defineClass1(Native Method)
at java.lang.ClassLoader.defineClassCond(ClassLoader.java:632)
at java.lang.ClassLoader.defineClass(ClassLoader.java:616)
at java.security.SecureClassLoader.defineClass(SecureClassLoader.java:141)
at java.net.URLClassLoader.defineClass(URLClassLoader.java:283)
at java.net.URLClassLoader.access$000(URLClassLoader.java:58)
at java.net.URLClassLoader$1.run(URLClassLoader.java:197)
at java.security.AccessController.doPrivileged(Native Method)
at java.net.URLClassLoader.findClass(URLClassLoader.java:190)
at java.lang.ClassLoader.loadClass(ClassLoader.java:307)
at sun.misc.Launcher$AppClassLoader.loadClass(Launcher.java:301)
at java.lang.ClassLoader.loadClass(ClassLoader.java:248)
Could not find the main class: ClientREST. Program will exit.
The name and path are correct, so any thoughts why I'm getting this exception?
Exception in thread "main" java.lang.NoClassDefFoundError: ClientREST
So, you ran it as java ClientREST. It's expecting a ClientREST.class without any package.
(wrong name: clientrest/ClientREST)
Hey, the class is trying to tell you that it has a package clientrest;. You need to run it from the package root on. Go one folder up so that you're in the folder which in turn contains the clientrest folder representing the package and then execute java clientrest.ClientREST.
You should not go inside the clientrest package folder and execute java ClientREST.
I encountered this error using command line java:
java -cp stuff/src/mypackage Test
where Test.java resides in the package mypackage.
Instead, you need to set the classpath -cp to the base folder, in this case, src, then prepend the package to the file name.
So it will end up looking like this:
java -cp stuff/src mypackage.Test
To further note on Garry's reply: The class path is the base directory where the class itself resides. So if the class file is here -
/home/person/javastuff/classes/package1/subpackage/javaThing.class
You would need to reference the class path as follows:
/home/person/javastuff/classes
So to run from the command line, the full command would be -
java -cp /home/person/javastuff/classes package1/subpackage/javaThing
i.e. the template for the above is
java_executable -cp classpath the_class_itself_within_the_class_path
That's how I finally got mine to work without having the class path in the environment
Probably the location you are generating your classes in doesnt exists on the class path. While running use the jvm arg -verbose while running and check the log whether the class is being loaded or not.
The output will also give you clue as to where the clasess are being loaded from, make sure that your class files are present in that location.
Try the below syntax:
Suppose java File resides here: fm/src/com/gsd/FileName.java
So you can run using the below syntax:
(Make current directory to 'fm')
java src.com.gsd.FileName
Suppose you have class A
and a class B
public class A{
public static void main(String[] args){
....
.....
//creating classB object
new classB();
}
}
class B{
}
this issue can be resolved by moving class B inside of class A and using static keyword
public class A{
public static void main(String[] args){
....
.....
//creating class B
new classB();
static class B{
}
}
Here is my class structure
package org.handson.basics;
public class WithoutMain {
public static void main() {
System.out.println("With main()...");
}
}
To compile this program, I had to use absolute path. So from src/main/java I ran:
javac org/handson/basics/WithoutMain.java
Initially I tried with the below command from basics folder and it didn't work
basics % java WithoutMain
Error: Could not find or load main class WithoutMain
Caused by: java.lang.NoClassDefFoundError: org/handson/basics/WithoutMain (wrong name: WithoutMain)
Later I went back to src\main\java folder and ran the class with relevant package structure, which worked as expected.
java % java org.handson.basics.WithoutMain
With main()...
I also have encountered this error on Windows when using Class.forName() where the class name I use is correct except for case.
My guess is that Java is able to find the file at the path (because Windows paths are case-insensitive) but the parsed class's name does not match the name given to Class.forName().
Fixing the case in the class name argument fixed the error.
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.