Given this .bat file :
#ECHO OFF
echo === Compiling and executing bat file ===
md C:\my_project\dir_prog\class_files
copy ProgAudioJ.java C:\my_project\dir_prog
javac -d C:\my_project\dir_prog\class_files ProgAudioJ.java
java -classpath C:\my_project dir_prog.class_files.ProgAudioJ
I'm wondering what's wrong with it.
It just doesn't work.
I mean it places the class files in the directory called class_files (so the compiling process is ok) but the program doesn't run...
Thanks for your help MAX
Do you have defined a main class within your project and do you have defined the
public static void main(String[] args) method for this class? Remember, that method gets
called when running stuff from shell...
EDIT: you might find this comprehensive overview usefull: Running a Java Program from Command Prompt # SkylightPublishing.com
HTH
I think the first problem is that you need to specify the class_files directory on the classpath.
The second problem is that you have to specify the fully qualified name of the class that you want to run. This depends on the package that you have defined in the Java source file.
So something like this should work: (I'm assuming your class is in the default package, i.e. no package)
java -classpath c:\my_project\dir_prog\class_files ProgAudioJ
Can you provide the contents of the source file?
If your code and command is correct, I would suspect a delay, related to the asynchronous execution of the javac or to the filesystem.
Try adding a pause command in order to check if problem is the rapidity of execution of the commands or if the problem is in the command.
Related
At the moment I am looking for another way to run my Java program from command line, other than adding it to a JAR file. My program has the following number of classes:
The name of the program file - MyProgram
Main class - Server1
second class - Client Handler
Package name - Items
3rd class - User1
4th class - User2
The main class and client handler alongside the package will have to run first in order for user 1 & user 2 to run, because they are client classes and are dependent on the main class.
javac *.java // compliles all java files in the dir
java MyClass // runs the particular file
If one class is dependent on another class that hasn't been compiled yet, the program won't run. So you should compile all files before trying to run the program dependent on other files.
If your files are packaged, then something like this
javac com.mypackage/.*java
java com.mypackage.MyClass
you must ensure that you add the location of your .class file to your classpath. So, if its in the current folder then add . to your classpath. Note that the windows classpath separator is a semi-colon ie ;
javac -cp . PackageName/*.java
java -cp . PackageName/ClassName_Having_main
Example. Suppose you have the following
Package Named: com.test
Class Name: Hello (Having main)
Java file is located inside "src/com/test/Hello.java"
then, from outside directory:
$ cd src
$ javac -cp . com/test/*.java
$ java -cp . com/test/Hello
Note that you can add -d to specify output directory of your class files whenever compiling
$ javac -d output_directory -cp . com/test/Hello
In windows the same thing will be working too, I already tried
Check out this from Oracle official site
Once you compile your code, you then run this from the top level:
java -cp . com.myprogram.MyProgram
That order thing you describe doesn't matter. They all get compiled together, and MyProgram will reference Server1, etc.
It may be more then you want to tackle right now but you might want to consider a build system like Maven. To start try out; How do I make my first Maven project?
You can use it to predefine the build order and if you want have it create a jar for you (or not).
Sounds like you will just need to open multiple command prompts and compile and run them in the order you need them to run. Let me know if I misunderstood question.
TO EXECUTE TWO JAVA PROGRAMS WHICH DEPENDS TO EACH OTHER.
(for example:two files Complex.java and Solution.java, where Soultion.java depends upon Complex.java.
So Complex.java should be compiled first and then the class file of Complex must be linked with Solution.java and then Solution.class must be executed for Output.)
REFER THE IMAGE WITH SYNTAX.
STEP 1:
COMPILE Complex.java
compiling Complex.java
syntax-
javac -d [path_where_class_File_build] [path_of_the_file\filename.java]
(Solution.java and Complex.java are Linked. ie-Solution.java calls Complex.java)
STEP 2:
COMPILE Solution.java
compiling Solution.java with linking Complex.class
with linking Complex.class(above created in step 1)
syntax-
javac -d [path_where_class_File_build] -cp [path_of_the_first_class_created] [path_of_the_file\filename.java]]
STEP 3:
EXECUTE THE Solution.class
java -cp [path_of_second_class_created] [class_Name]
(created in Step 3)
I am very new to linux environment.
I am trying to run an simple hello world java class in linux environment.
Hello .java
package com.util;
public class Hello {
/**
* #param args
*/
public static void main(String[] args) {
System.out.println("hi");
}
}
I have compiled java class in windows environment and uploaded the .class file to linux system into /home/scripts path.
my command is as follows,
java -cp /home/scripts com.util.Hello
when i am executing this command from this same /home/scripts where Hello.class is there i am getting,
Error: Could not find or load main class com.util.Hello and not able to proceed further.
can some one help me in this issue?
navigate to /home/scripts using terminal
javac com/util/Hello.java
then
cd /home/scripts
java -cp . com.util.Hello
Or,
java -cp "/home/scripts" com.util.Hello
At first you must generate your .class file :
javac ./hello.java
This command has generated hello.class file
And after you can run your class file ! :)
java hello
We first know javac command work well.
I also met this error,and i have resolved this.Let me share this.
First we need to find the parent path of your package in your java codes.
Then cd to that path using java package + fileName should work well at that moment.
I had the exact same issue on windows, and I solved it by adding path "." to both CLASSPATH and PATH, maybe you can try this on Linux as well.
Your .class file should not reside in /home/scripts/, but in /home/scripts/com/util/. Take a look at this document that explains the relation between classpath, packages and directories.
Before Specifying the path,ensure you follow these three things meticulously,
1. Close the command prompt window, before specifying the path.
2. When adding path, add bin and semi- colon at the end and
3. If JAVAC command has worked properly, try java -cp class name.
if you want to run program in current working directory where your class reside.
java gives three options.
first option
java -cp Tester
Second option for current working directory
java -cp . Tester
Third option export CLASSPATH variable
export CLASSPATH=$CLASSPATH:. (this is the best one if your directory changes) or
export CLASSPATH=$PWD
or
export CLASSPATH=
after that you must sorce the bashrc or bashprofile.
I have done a lot of researching on this concept but I can't seem to run a java program on the command prompt. Let's say we had a simple program like this:
public class Hello_World {
public static void main(String[] args) {
System.out.println("Hello World!");
}
}
On the command prompt I tried:
javac Hello_World.java
But I get:
'javac' is not recognized as an internal or external command, operable program or batch file
So I compiled it on BlueJ and then did this:
java Hello_World.java
But it said "cannot load or find main class Hello_World"!
I am currently using Windows 7, and made the programs on Notepad++ and BlueJ (to compile).
Any suggestions? Thanks!
This explains in detail what you have to do to set class path. Primarily you need to set your environment variables so that your shell finds the right directory containing javac to compile your program
javac' is not recognized ..
comes when you haven't point your java bin directory to your path environment variable. Because bin directory is the place where javac.exe exist.
To do it.
1) right click on mycomputer property
2) go to Advance system settings.
3) go to environment variable.
4) In system variable click on path
5) go to edit mode and provide your path to java bin directory.
in my case it is C:\Program Files\Java\jdk1.7.0_01\bin;
'javac' is not recognized as an internal...
means OS does not know where javac program is located. Either add it to PATH or run explicitly
my\path\to\file\javac Hello_World.java
Compiling will convert *.java to *.class
Hello_World.class file should be located according to it's package directive. Since you have no one, in your case it should be located in the same directory you will run java.
To run your class specify it's name not file name
java Hello_world
looking for the class is essential part of launching and occurs by rules.
Suppose I just created a package "example" and have two classes inside it, "Main" and "Helper".
With the simplest possible compilation (e.g., $javac Main.java Helper.java) I am already able to run it fine as long as I am in the directory containing the example package, by typing this in the command line:
$java example.Main
Questions:
Why would I want to set a CLASSPATH given I can already run the program? I am guessing to be able to type "$java example.Main" from any directory on my machine, but I am not sure.
What happens when I type "java -cp /path/to/your/java/class/file Main" on the command line? Right now I picture there's file containing all the different classpaths, and that command will just add another one to it. Is it the case?
Is there a difference between using "CLASSPATH=/path/to/your/java/class/file" and "java -cp /path/to/your/java/class/file Main" on the command line? How come the second one has the name of the class (i.e. Main) in the end?
Yea, pretty much. That of course assumes you have the path to java in your PATH variable
-cp or -classpath adds it's option (a string) in front of whatever is in your CLASSPATH
Yes, there is a difference. Using CLASSPATH is often more convenient as you tend to set your CLASSPATH once. From then on, java Main is enough to execute the main class. With java -cp /path/to/your/java/class/file Main you have to type the -cp /path/to/your/java/class/file every time.
That being said, both CLASSPATH and -cp or -classpath options usually contain entries pointing to directories containing java libraries used by your program, not the directory of your program itself.
I have just installed JDK on Windows Vista. After that I set proper values for the 4 environment variables: classpath, include, lib, path. After that I was able to compile my HelloWorld-program (I got a *.class file). But when I try to execute the compiled program (I type java HelloWorldApp) it does not work. The Java write a lot of stuff and in the end it is written that it "could not find the main class: HelloWorldApp". Can anybody, pleas, help me with this problem?
Just for clarity; you are saying that you have a class in the default package, that is you have not included a package specifier in the Java file, and your class is called HelloWorldApp. When you compiled this, you got a classfile HelloWorldApp.class in the current directory.
Assuming the above to be true then try:
java -cp . HelloWorldApp
For example, the following works on a unix box:
$ echo 'class HelloWorldApp { public static void main(String []argv) { System.out.println("Hello World!"); } }' > HelloWorldApp.java
$ javac HelloWorldApp.java
$ java -cp . HelloWorldApp
Hello World!
Of course, you should indent your code a little nicer than just shoving the whole thing onto one line ;-)
Edit: To answer the comment:
Normally, the default classpath is the runtime libraries and the current directory. However, if you have the CLASSPATH variable set, then this will override the default, and you need to explicitly set the classpath back to its "default value". To verify if the CLASSPATH environment variable is set, you can do (again assuming unix):
set | grep CLASSPATH
If it is set, that is why you need to manually include . on your classpath.
create a file called HelloWorld.java;
paste the code posted below inside HelloWorld.java:
compile it by executing the command: javac HelloWorld.java in the same folder as HelloWorld.java is in;
execute the code by doing: java -cp . HelloWorld in the same folder as HelloWorld.java is in.
public class HelloWorld {
public static void main(String[] args) {
System.out.println("HelloWorld works!");
}
}
How the classpath works, can be read here: http://en.wikipedia.org/wiki/Classpath_%28Java%29
Have you included . and .. in your path? Just for clarification . represents your current directory and .. represents your parent directory. You are telling that the java has to search the current directory and the parent directory to find the class. Add the same to your classpath too.
What happens if you use:
java -cp {path to directory with HelloWorldApp in it} HelloWorldApp
That path should be contained within your CLASSPATH environment variable. Is that exported to your command shell ? Do you need to start a new command shell to get the most recent version of CLASSPATH ?
Post your code. I believe the problem is that your main class is not defined properly. I did this the other day.
public static void main(String[] args){
//code
}
The class path concept and the logical difference between Java source code and compiled byte code is notoriously hard to get right.
I would strongly recommend you familiarize yourself with the Sun Java Tutorial. The relevant section is
http://java.sun.com/docs/books/tutorial/getStarted/cupojava/win32.html