Tukukkk

A classpath is a list of locations to load classes from.

These 'locations' can either be directories, or jar files.

For directories, the JVM will follow an expected pattern for loading a class. If I have the directory C:/myproject/classes in my classpath, and I attempt to load a class com.mycompany.Foo, it will look under the classes directory for a directory called com, then under that a directory called mycompany, and finally it will look for a file called Foo.class in that directory.

In the second instance, for jar files, it will search the jar file for that class. A jar file is in reality just a zipped collection of directories like the above. If you unzip a jar file, you'll get a bunch of directories and class files following the pattern above.

So the JVM traverses a classpath from start to finish looking for the definition of the class when it attempts to load the class definition. For example, in the classpath :

C:/myproject/classes;C:/myproject/lib/stuff.jar;C:/myproject/lib/otherstuff.jar

The JVM will attempt to look in the directory classes first, then in stuff.jar and finally in otherstuff.jar.

When you get a ClassNotFoundException, it means the JVM has traversed the entire classpath and not found the class you've attempted to reference. The solution, as so often in the Java world, is to check your classpath.

You define a classpath on the command line by saying java -cp and then your classpath. In an IDE such as Eclipse, you'll have a menu option to specify your classpath.

This is the best solution I found so far.

Suppose we have a package called org.mypackage containing the classes:

• HelloWorld (main class)


• SupportClass


• UtilClass

and the files defining this package are stored physically under the directory D:myprogram (on Windows) or /home/user/myprogram (on Linux).

The file structure will look like this:

When we invoke Java, we specify the name of the application to run: org.mypackage.HelloWorld. However we must also tell Java where to look for the files and directories defining our package. So to launch the program, we have to use the following command:

NOTE: You have to execute the above java command no matter what your current location is. But this is not the case for javac. For
compiling you can even directly go into the directory where you have
your .java files and directly execute javac ClassName.java.

If you know the path of the class or the jar containing the class then add it to your classpath while running it. You can use the classpath as mentioned here:

on Windows

java -classpath .;yourjar.jar YourMainClass

on UNIX/Linux

java -classpath .:yourjar.jar YourMainClass

Try these if you use maven. I use maven for my project and when I do mvn clean install and try to run a program it throws the exception. So, I clean the project and run it again and it works for me.

I use eclipse IDE.

For Class Not Found Exception when running Junit test, try running mvn clean test once. It will compile all the test classes.

To add the location of a class to your classpath via command line simply add -cp or -classpath and the location of the class while running it. I.E.

java -cp "c:/location/of/file" YourProgram

Or if you're running an IDE such as eclipse you can right click on the project -> build path -> configure build path
and add the external JAR containing your class to the build path then it should work fine.

Use ';' as the separator. If your environment variables are set correctly, you should see your settings. If your PATH and CLASSPATH is correct, windows should recognize those commands. You do NOT need to restart your computer when installing Java.

Add the full path of jar file to the CLASSPATH.
In linux use: export CLASSPATH=".:/full/path/to/file.jar:$CLASSPATH". Other way worked (without editing the CLASSPATH) was unzipping the jar in the current project folder.

Ways didn't work for me:

1) Using -cp option with full path of jar file.

2) Using -cpwith only the name of jar when located in the current folder

3) Copying the jar to the current project folder

4) Copying the jar to standard location of java jars (/usr/share/java)

This solution is reported for class com.mysql.jdbc.Driver in mysql-connector-java.5-*.jar, working on linux with OpenJDK version 1.7

Go up to the top and remove the import statement if there is one, and re import the class. But if that isn't the case do a clean then build. Are you using Netbeans or Eclipse?

I ran into this as well and tried all of the other solutions. I did not have the .class file in my HTML folder, I only had the .java file. Once I added the .class file the program worked fine.

1. It could happen if your classpath is not correct




2. Let us posit a serializable class and deserializable class under same projectname. You run the serializable class, creating a serializable object in specific folder. Now you need the desearialized data. In the meantime, if you change the name of the project it will not work. You have to run the serializable class first and then deserialize the file.

If you are using maven
try to maven update all projects and force for snapshots.
It will clean as well and rebuilt all classpath..
It solved my problem..

I just did

1.Invalidate caches and restart

2.Rebuilt my project which solved the problem

I was trying to run .jar from C# code using Process class. The java code ran successfully from eclipse but it doesn't from C# visual studio and even clicking directly on the jar file, it always stopped with ClassNotFoundException: exception. Solution for my, was export the java program as "Runnable JAR file" instead of "JAR File". Hope it can help someone.

This can happen on Windows after a java update where the old version of the java SDK is missing and a new one is present. I would check if your IDE is using the installed java SDK version (IntelliJ: CTRL + SHIFT + ALT + S)

If you have added multiple (Third-Party)**libraries and Extends **Application class

Then it might occur.

For that, you have to set multiDexEnabled true and replace your extended Application class with MultiDexApplication.

It will be solved.