I have a simple MyHelloWorld.java file within a directory myhelloworld. I set the classpath to appropriate direcotries by exporting CLASSPATH='[some needed class]' in my ~/.bashrc file. and I still need to use javac MyHelloWorld.jaca -cp 'the above classpath' to compile the .java file.
So here's my first question, why do I still need to explicitly set classpath by having a -cp option given the CLASSPATH is already specified?
After the file was compiled, I saw no file like MyHelloWorld, but the java commmand can autocomplete 'java MyHelloWorld' when in fact there's no file named MyHelloWorld exists within the directory? (If I typed "java MyH", the command can be automatically completed).
When I tried to execute 'java MyHelloWorld', it prompted me there's NOClassDefFoundError, which was an indicator of missing library. However when I tried to do 'java -cp MyHello', The command wouldn't complete itself, suggesting it couldn't find anything executable at all
So what are the reasons for above confusing signs. Can anyone take from here and explain to me how java deal with classpath and package, etc. Thanks!
Follow the java tuturial and especialy this section.
The section about JARs is for you too.
I'd like to compile a very basic servlet from command prompt, but it is always unsuccessful and the compiler tells me the following:
error: package javax.servlet does not exist.
I googled for the solution and I found that I need to include the servlet.jar libraries into my PATH.
I believe I did.
I strongly believe that the location for those libraries in my computer is:
C:\apache-tomcat-7.0.23\lib\servlet-api.jar\
and the end (the relevant part) of my PATH is the following:
%JAVA_HOME%\bin;C:\apache-tomcat-7.0.23\lib\servlet-api.jar\
For me, it looks ok, but it is obviously not. Can anyone tell me what could be the problem?
classpath not path ... and you don't need it as an evironment variable.
You can set the classpath for javac with option -cp or -classpath (several other ways are also available).
javac uses the environment variable CLASSPATH to look for classes, that can be useful and can also be a source for hard-to-track-down-problems.
An example to compile a java file that uses a library(that is classes from outside the standard JavaSE) would be:
javac -classpath C:\apache-tomcat-7.0.23\lib\servlet-api.jar MyTestServlet.java
if your environmental variable CLASSPATH contains libraries you need you might want to do:
javac -classpath %CLASSPATH%;C:\apache-tomcat-7.0.23\lib\servlet-api.jar MyTestServlet.java
(please be aware that I don't have access to a windows machine, and therefore haven't tested the idiosyncratic parts of the syntax above)
(also note that in this example "C:\apache-tomcat-7.0.23\lib\servlet-api.jar" is a jar file and not a directory which it might be from your question on your machine)
For command line compiling on windows OS it is a good idea to have the environmental variable JAVA_HOME set correctly and the bin directory of the JDK in the PATH.
I do however suggest getting to write-compile-execute-deploy via/in an IDE for servlet development before figuring out how to do it with just the JDK from a command line.
Java Servlets are not stand-alone executable classes but needs to be deployed into for example tomcat to be tested/used.
First copy the servlet-api.jar file from following path
C:\apache-tomcat-7.0.23\lib\servlet-api.jar;
and paste it in the bin folder of java software by following the path
C:\java\jdk1.6\bin;
Hope now you can successfully compile your servlet program.
1.You can copy your javax.servlet.jar in the jdk1.6\lib folder.
2.You can go to Control Panel\System\Advanced System Properties\Environment Variables
Enter the location of the jar in the CLASSPATH variable as follows:
Then compile and run the servlet.
Pretty basic problem here. So I have a Java package that I have created that has three classes (one has the main method). I am trying to use a few Apache Jars, and have added these to my build path in Eclipse. However Eclipse wont let me build and run it properly, so I am trying the command line. I have added the env var CLASSPATH and pointed it to my lib directory which hold the Apache Jars. However, when I try to use javac I get a bunch of errors:
package org.apache.xmlrpc does not exist
import org.apache.xmlrpc.client.XmlRpcClient;
I was reading the man page for javac and it said that:
If neither CLASSPATH, -cp nor -classpath is specified, the user class path consists of the current directory.
So I tried copying the Jars to the same location as my three source files, but no change.
Can someone please tell me what I'm doing wrong?
Thanks.
Classpath variable (or command line option of javac) must contain all jars explicitly. It cannot go through jar files stored in specified directory.
You can compile this by specifying the option -cp on the command line:
javac -cp foo.jar:bar.jar foo/bar/Baz.java
You then run it with the same option:
java -cp foo.jar:bar.jar foo.bar.Baz
It sounds like you've just set the classpath to the directory containing the jar files. You need to set it to the individual jar files, or use java.ext.dirs to set an "extension" directory containing jar files. I'd recommend using the specific jar files. Something like:
// Assuming Windows...
CLASSPATH = c:\libs\foo.jar;c:\libs\bar.jar
I'd also personally recommend specifying the classpath on the command line instead of using an environment variable - the latter will work, but it ends up being a bit more fiddly if you want to compile different projects against different libraries.
However, I'd actually recommend getting Eclipse working first, rather than retreating to the command line. It should be fine - if you could give us more information about what's failing in Eclipse, we may be able to help you with that instead.
The jar files in the current directory are not automatically included; that only refers to .class files in normal package/directory hierarchy. Jar files must be added either explicitly, or via a wildcard like javac -cp ./* (Assuming JDK6+)
(Some OSes may require an escape of the * to avoid globbing; OSX does not.)
I agree with previous answers, but I would also recommend to use proper java build tool - like ant (perceived easier to use, but not necessary) or maven ( perceived more difficult to use, but really worth learning )
I have a file which imports org.w3c.dom.Document. Compiling and running is fine, but I don't understand how it knows where to find this package and I'm just curious how it works. I used the locate command to try and find org.w3c.dom but I get nothing. Where are these packages located? It seems to me that the right place to look would the CLASSPATH environment variable since my search results seem to be suggesting that. Is this correct? In any case, I don't know how to find out what my CLASSPATH variable is. It doesn't seem to be an environment variable that my shell knows about.
That would be part of the core libraries (rt.jar), so it'd be wherever you installed the java JRE; specifically under $JAVA_HOME/jre/lib
You can look inside the .jar files using the jar command. To see the class you mention, you can do:
jar tvf rt.jar
This lists all the classes in that jar.
Note that this location is automatically searched by the JVM - it's not needed nor included in the CLASS_PATH environment variable. (You could add it, but it would simply be redundant)
Edit for clarity:
The JVM includes <Where_you_installed_jdk>/jre/lib and <Where_you_installed_jdk>/jre/lib/ext by default. Anything else has to be explicitly added by you via either passing it to java directly via the -cp option or adding it to the CLASS_PATH environment variable.
The relavent documentation can be found at: http://download.oracle.com/javase/6/docs/technotes/tools/findingclasses.html
The JVM finds classes using classpath settings where alll paths to required packages are set. The classpath could be set with a number of ways. The first mentioned by you is CLASSPATH environment variable. It is optional and can be unset. The second way is an explicit option "-cp" for "java" executable.
Also some JRE runtime jars are added to classpath by default implicitly so you don't need to search and add standard packages by yourself (particulary the one you mentioned in your question).
try compiling messconvener.java like this from its own directory
javac -d ..\..\. -cp ..\..\. messconvener.java
-d - creates directory structure for your package
-cp - provides class path for user file, where it can find user defined classes
I was just reading this line:
The first thing the format() method does is load a Velocity template from the classpath named output.vm
Please explain what was meant by classpath in this context, and how I should set the classpath.
When programming in Java, you make other classes available to the class you are writing by putting something like this at the top of your source file:
import org.javaguy.coolframework.MyClass;
Or sometimes you 'bulk import' stuff by saying:
import org.javaguy.coolframework.*;
So later in your program when you say:
MyClass mine = new MyClass();
The Java Virtual Machine will know where to find your compiled class.
It would be impractical to have the VM look through every folder on your machine, so you have to provide the VM a list of places to look. This is done by putting folder and jar files on your classpath.
Before we talk about how the classpath is set, let's talk about .class files, packages, and .jar files.
First, let's suppose that MyClass is something you built as part of your project, and it is in a directory in your project called output. The .class file would be at output/org/javaguy/coolframework/MyClass.class (along with every other file in that package). In order to get to that file, your path would simply need to contain the folder 'output', not the whole package structure, since your import statement provides all that information to the VM.
Now let's suppose that you bundle CoolFramework up into a .jar file, and put that CoolFramework.jar into a lib directory in your project. You would now need to put lib/CoolFramework.jar into your classpath. The VM will look inside the jar file for the org/javaguy/coolframework part, and find your class.
So, classpaths contain:
JAR files, and
Paths to the top of package hierarchies.
How do you set your classpath?
The first way everyone seems to learn is with environment variables. On a unix machine, you can say something like:
export CLASSPATH=/home/myaccount/myproject/lib/CoolFramework.jar:/home/myaccount/myproject/output/
On a Windows machine you have to go to your environment settings and either add or modify the value that is already there.
The second way is to use the -cp parameter when starting Java, like this:
java -cp "/home/myaccount/myproject/lib/CoolFramework.jar:/home/myaccount/myproject/output/" MyMainClass
A variant of this is the third way which is often done with a .sh or .bat file that calculates the classpath and passes it to Java via the -cp parameter.
There is a "gotcha" with all of the above. On most systems (Linux, Mac OS, UNIX, etc) the colon character (':') is the classpath separator. In windowsm the separator is the semicolon (';')
So what's the best way to do it?
Setting stuff globally via environment variables is bad, generally for the same kinds of reasons that global variables are bad. You change the CLASSPATH environment variable so one program works, and you end up breaking another program.
The -cp is the way to go. I generally make sure my CLASSPATH environment variable is an empty string where I develop, whenever possible, so that I avoid global classpath issues (some tools aren't happy when the global classpath is empty though - I know of two common, mega-thousand dollar licensed J2EE and Java servers that have this kind of issue with their command-line tools).
Think of it as Java's answer to the PATH environment variable - OSes search for EXEs on the PATH, Java searches for classes and packages on the classpath.
The classpath is one of the fundamental concepts in the Java world and it's often misunderstood or not understood at all by java programmes, especially beginners.
Simply put, the classpath is just a set of paths where the java compiler and the JVM must find needed classes to compile or execute other classes.
Let's start with an example, suppose we have a Main.java file thats under C:\Users\HP\Desktop\org\example,
package org.example;
public class Main {
public static void main(String[] args) {
System.out.println("Hello world");
}
}
And Now, suppose we are under C:\ directory and we want to compile our class, Its easy right, just run:
javac .\Users\HP\Desktop\org\example\Main.java
Now for the hard question, we are in the same folder C:\ and we want to run the compiled class.
Despite of what you might think of to be the answer, the right one is:
java -cp .\Users\HP\Desktop org.example.Main
I'll explain why, first of all, the name of the class that we want ro tun is org.exmaple.Main not Main, or Main.class or .\users\hp\desktop\org\example\Main.class ! This is how things works with classes declared under packages.
Now, we provided the name of the class to the JVM (java command in this case), But how it (JVM) will know where to find the .class file for the Main class? Thats where the classpath comes into picture. Using -cp flag (shortcut for -classpath), we tell the JVM that our Main.class file will be located at C:\users\hp\Desktop.. In fact, not really, we tell it to just go to the Desktop directory, and, because of the name of the class org.example.Main, the JVM is smart and it will go from Desktop to org directory, and from org to example directory, searching for Main.class file, and it will find it and it will kill it, I mean, it will run it :D .
Now lets suppose that inside the Main class we want to work with another class named org.apache.commons.lang3.StringUtils and the latter is located in a jar file named commons-lang3-3.10.jar thats inside C:\Users\HP\Downloads. So Main.java will look like this now:
package org.example;
import org.apache.commons.lang3.StringUtils;
public class Main {
public static void main(String[] args) {
System.out.println("Hello world");
System.out.println(StringUtils.equals("java", "java")); //true
}
}
How to compile the Main.java if we are always inside C:\ ? The answer is:
javac -cp .\Users\HP\Downloads\commons-lang3-3.10.jar .\Users\HP\Desktop\org\example\Main.java
.\Users\HP\Desktop\org\example\Main.java is because our .java file is there in the filesystem.
-cp .\Users\HP\Downloads\commons-lang3-3.10.jar is because the java compiler (javac in this case) need to know the location of the class org.apache.commons.lang3.StringUtils, so we provided the path of the jar file, and the compiler will then go inside the jar file and try to find a file StringUtils.class inside a directory org\apache\commons\lang3.
And if we want to run the Main.class file, we will execute:
java -cp ".\Users\HP\Desktop\;.\Users\HP\Downloads\commons-lang3-3.10.jar" org.example.Main
org.example.Main is the name of the class.
".\Users\HP\Desktop\;.\Users\HP\Downloads\commons-lang3-3.10.jar" are the paths (separated by ; in Windows) to the Main and StringUtils classes.
The classpath is the path where the Java Virtual Machine look for user-defined classes, packages and resources in Java programs.
In this context, the format() method load a template file from this path.
The classpath in this context is exactly what it is in the general context: anywhere the VM knows it can find classes to be loaded, and resources as well (such as output.vm in your case).
I'd understand Velocity expects to find a file named output.vm anywhere in "no package". This can be a JAR, regular folder, ... The root of any of the locations in the application's classpath.
Setting the CLASSPATH System Variable
To display the current CLASSPATH variable, use these commands in Windows and UNIX (Bourne shell):
In Windows: C:\> set CLASSPATH
In UNIX: % echo $CLASSPATH
To delete the current contents of the CLASSPATH variable, use these commands:
In Windows: C:\> set CLASSPATH=
In UNIX: % unset CLASSPATH; export CLASSPATH
To set the CLASSPATH variable, use these commands (for example):
In Windows: C:\> set CLASSPATH=C:\users\george\java\classes
In UNIX: % CLASSPATH=/home/george/java/classes; export CLASSPATH
Classpath is an environment variable of system. The setting of this variable is used to provide the root of any package hierarchy to java compiler.
CLASSPATH is an environment variable (i.e., global variables of the operating system available to all the processes) needed for the Java compiler and runtime to locate the Java packages used in a Java program. (Why not call PACKAGEPATH?) This is similar to another environment variable PATH, which is used by the CMD shell to find the executable programs.
CLASSPATH can be set in one of the following ways:
CLASSPATH can be set permanently in the environment: In Windows, choose control panel ⇒ System ⇒ Advanced ⇒ Environment Variables ⇒ choose "System Variables" (for all the users) or "User Variables" (only the currently login user) ⇒ choose "Edit" (if CLASSPATH already exists) or "New" ⇒ Enter "CLASSPATH" as the variable name ⇒ Enter the required directories and JAR files (separated by semicolons) as the value (e.g., ".;c:\javaproject\classes;d:\tomcat\lib\servlet-api.jar"). Take note that you need to include the current working directory (denoted by '.') in the CLASSPATH.
To check the current setting of the CLASSPATH, issue the following command:
> SET CLASSPATH
CLASSPATH can be set temporarily for that particular CMD shell session by issuing the following command:
> SET CLASSPATH=.;c:\javaproject\classes;d:\tomcat\lib\servlet-api.jar
Instead of using the CLASSPATH environment variable, you can also use the command-line option -classpath or -cp of the javac and java commands, for example,
> java –classpath c:\javaproject\classes com.abc.project1.subproject2.MyClass3
For linux users, and to sum up and add to what others have said here, you should know the following:
$CLASSPATH is what Java uses to look through multiple directories to find all the different classes it needs for your script (unless you explicitly tell it otherwise with the -cp override). Using -cp requires that you keep track of all the directories manually and copy-paste that line every time you run the program (not preferable IMO).
The colon (":") character separates the different directories. There is only one $CLASSPATH and it has all the directories in it. So, when you run "export CLASSPATH=...." you want to include the current value "$CLASSPATH" in order to append to it. For example:
export CLASSPATH=.
export CLASSPATH=$CLASSPATH:/usr/share/java/mysql-connector-java-5.1.12.jar
In the first line above, you start CLASSPATH out with just a simple 'dot' which is the path to your current working directory. With that, whenever you run java it will look in the current working directory (the one you're in) for classes. In the second line above, $CLASSPATH grabs the value that you previously entered (.) and appends the path to a mysql dirver. Now, java will look for the driver AND for your classes.
echo $CLASSPATH
is super handy, and what it returns should read like a colon-separated list of all the directories, and .jar files, you want java looking in for the classes it needs.
Tomcat does not use CLASSPATH. Read what to do about that here: https://tomcat.apache.org/tomcat-8.0-doc/class-loader-howto.html
Static member of a class can be called directly without creating object instance.
Since the main method is static Java virtual Machine can call it without creating any instance of a class which contains the main method, which is start point of program.