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How to load a java class outside the classpath?

I have a program where I want the user to be able to choose a .java class file from the file system, and then have that class loaded into the program.
I'm using a JFileChooser to allow the user to select a file. Then, I tried converting that file to a URL, and using a URLClassLoader to load the class (as suggested by these answers).
The problem is that, when I want to use the loadClass() method, I don't know the "full class name" of the class (e.g. java.lang.String). So, I don't know how to make this method work. Is there a way to get this class name? Or is there another way to do this?
Here is a sample of my code:
// Open the file chooser
JFileChooser fileChooser = new JFileChooser();
fileChooser.showOpenDialog(null);
File obtainedFile = fileChooser.getSelectedFile();
// Create the class loader from the file
URL classPath = obtainedFile.toURI().toURL();
URLClassLoader loader = new URLClassLoader(new URL[] {classPath});
// Get the class from the loader
Class<?> theClassIWant = loader.loadClass("the file name"); // What do I put here??

Load a single class file is generally completely useless. Said class file isn't alone; it has more class files that are relevant. Even if you think 'nah, there is just one source file, do not worry about this', note that a single java file can easily generate multiple class files.
Thus, two options:
Don't load class files. Load jar files.
Use the usual mechanisms (META-INF/services or META-INF/MANIFEST.MF) to put some sort of class name in there so you know what to load. Then create a new classloader with the provided jar, load the manifest, figure out the main class, load that, and run it.
Attempt to determine the 'root' for the loaded class file and include that on the classpath.
This is quite difficult - the problem is, to 'load' a class file you need to tell the loader what the fully qualified name is of that class before it is loaded. But how do you know the fully qualified name? You can surmise the class name from the file (not quite always true, but usually), but the package is a more difficult issue.
You can open the class file yourself as a binary stream and write a basic class file format parser to get the fully qualified class name. Easy for an experienced java programmer. Quite tricky for someone new to java (which I gather you are, if you think this is a good idea).
You can also use existing tools to do this, such as bytebuddy or asm.
Finally, you can try a spaghetti-at-the-wall method: Keep travelling up the directory until it works. You know it isn't working if exceptions occur.
For example, to load C:\MyDir\Whatever\com\foo\MyApp.class, You first try creating a new classloader (see the API of URLClassLoader which is part of core java) using as root dir C:\MyDir\Whatever\com\foo, and then you ask it to load class MyApp.
If that works, great (but usually trying to load package-less classes is simply a non-starter, you're not supposed to do that, the CL API probably doesn't support it, intentionally, there is no fixing that).
If it doesn't, instead try C:\MyDir\Whatever\com, and load class foo.MyApp. If that doesn't work, try C:\MyDir\Whatever and load class com.foo.MyApp, and so on.
The considerable advantage is, if there is another class sitting right next to MyApp.class, and MyApp needs it, this will work fine.
You'll need to write a while loop (traversing the path structure using Paths.get and p.getParent()), catch the right exception, manipulate the path into the class name (using .replace and +), and, of course, create a class loader (URLClassLoader), load classes with it (invoke loadClass), and if you intend on running it, something like thatClass.getConstructor().newInstance() and then thatClass.getMethod("someMethod", String.class, /* all the other args here */).invoke(theInstanceYouJustMade, "param1", /*all other params */) to actually 'run' it, more to be found in the java.lang.reflect package.

Class reference is taken from different jar

There are some classes used in a java program referred from abc package from xyz.jar. The package is imported in the java file.
Also the same class is in other lmn.jar.
So if I delete the jar file from the project, i should be getting the error.
But the class compiles and takes the class from the other lmn.jar.
Eg.
weblogic.jdbc.oci.Blob is a class in weblogic.jar
But if i delete weblogic.jar, it takes it from java.sql.Blob.
I don't want this to happen, the program should display error.

In such a case you can use fully qualified name of class, which will include package name. For example, instead of:
Blob blob = new Blob();
You can write:
weblogic.jdbc.oci.Blob blob = new weblogic.jdbc.oci.Blob();

When importing a class you are also defining the package. So when you import "weblogic.jdbc.oci.Blob" and remove this class from classpath it will not automatically import it from a different package unless you change the import statement as well.
Some IDEs might automatically try to resolve classes and add the missing import statements. Maybe check that.

For one of the case try to use the entire path, ie
For example if you need Blob from weblogic.jar
then try to call weblogic.jdbc.oci.Blob bl = ...

Access classes from package

I'm developing an android test app and i'm going to access all internal class of android.view package. android.view is a package that is present in jar file. I tried by loading package name but it doesn't display the classes if any one tried
this already, please help.
Here's what I tried so far:
public static void main() throws ClassNotFoundException{
Class o =Class.forName("android.view");
Class[] C=o.getDeclaredClasses();
for(int i=0;i<C.length;i++) {
Classname = C[i].getName();
ClassesDisplayActivity.your_array_list3.add(Classname);
Log.i("Ramu","classname "+ C[i].getName());
}
}
}

It is not possible to determine at runtime all of the classes that are in a package using a standard class loader.
You might have some luck with this library though:
https://code.google.com/p/reflections/

Package is not a class. You cannot call Class.forName() for package and access classes that belong to class using getDelcaredClasses().
I do not know what do you really need, so I'd recommend you to explain this in separate question. probably you will receive better solutions.
However if you really need this you have to do the following:
Get your classpath by calling System.getProperty(java.class.path)
split this property to its elements by colon
iterate over the list and read each resource. If resource is jar you can use ZipInputStream, if it is a directory use File class.
filter list of resources you got at #3.
Fortunately you can use 3rd party library named Reflections that helps you to do all this without writing code.

How do I get all the class names and method names of a project?

I have downloaded a huge project written in Java. I wish to know the Classes and Methods of every class that are available in the project (for further analysis). How can I recover this information. Can I try javadoc in eclipse?

I guess you may ask about changing SVN properties.
Follow this step if that so.
press Alt + Shift + Q
Select Show view (view : Outline)
then under that u can see all details

I have wrote a custom doclet to list the classname and its methods:
public class ListClassAndMethods {
public static boolean start(RootDoc root) {
ClassDoc[] classes = root.classes();
for(ClassDoc clazz : classes){
System.out.println("Class Name: "+clazz);
System.out.println("--------------------------");
for(MethodDoc methodz :clazz.methods()){
System.out.println(methodz.name());
}
}
return true;
}
}
you need to run create a jar of this class and refer it while creating
a javadoc using Eclipse IDE

I would extract all the class source files (.java) with find (if you're on a *nix implementation) and create an empty NetBeans project with just one package and all the classess inside it. Netbeans will correct the package declaration and you can easily use autogenerate javadoc to get a navigable web archive listing all the classes and public/protected methods.
Of course the code may not run anymore but you'll get what you want in minutes.

How to get current class name including package name in Java?

I'm working on a project and one requirement is if the 2nd argument for the main method starts with “/” (for linux) it should consider it as an absolute path (not a problem), but if it doesn't start with “/”, it should get the current working path of the class and append to it the given argument.
I can get the class name in several ways: System.getProperty("java.class.path"), new File(".") and getCanonicalPath(), and so on...
The problem is, this only gives me the directory in which the packages are stored - i.e. if I have a class stored in ".../project/this/is/package/name", it would only give me "/project/" and ignores the package name where the actual .class files lives.
Any suggestions?
EDIT:
Here's the explanation, taken from the exercise description
sourcedir can be either absolute (starting with “/”) or relative to where we run the program from
sourcedir is a given argument for the main method. how can I find that path?

Use this.getClass().getCanonicalName() to get the full class name.
Note that a package / class name ("a.b.C") is different from the path of the .class files (a/b/C.class), and that using the package name / class name to derive a path is typically bad practice. Sets of class files / packages can be in multiple different class paths, which can be directories or jar files.

There is a class, Class, that can do this:
Class c = Class.forName("MyClass"); // if you want to specify a class
Class c = this.getClass(); // if you want to use the current class
System.out.println("Package: "+c.getPackage()+"\nClass: "+c.getSimpleName()+"\nFull Identifier: "+c.getName());
If c represented the class MyClass in the package mypackage, the above code would print:
Package: mypackage
Class: MyClass
Full Identifier: mypackage.MyClass
You can take this information and modify it for whatever you need, or go check the API for more information.

The fully-qualified name is opbtained as follows:
String fqn = YourClass.class.getName();
But you need to read a classpath resource. So use
InputStream in = YourClass.getResourceAsStream("resource.txt");

use this.getClass().getName() to get packageName.className and use this.getClass().getSimpleName() to get only class name