RE: RPG - Java Call

["Larry" <Larry_Ducie@xxxxxxxxxxx>]

Hi Sudha,
 
<snip>
The code is similar to Browser() called BrowseControl() and I removed
all the package declarations & jar to call from command line (I still
don't understand why it would find the class from jar when the classpath
refers to jar).  I get browser not found or cannot be opened error now.
</snip>
 
I think we may need to move this thread over to the JAVA400 list, as we've
gone way off-topic.
 
But, before we go...
 
Understanding paths, classpaths, and how classes are located seems to cause
more confusion than anything else java-related. It must be a case of; "Once
you get classpaths, then the rest of java is a breeze!". Probably akin to;
"You don't understand unix until you understand the find command!".
 
Anyway, here's a very basic overview...
 
1) To load a class you must specify it's location with a fully qualified
path - that is, the path must start from the root directory.
When you create a class in the default package (no package declaration at
the top of your source) you MUST have the directory in which the class is
located placed within your classpath. This is the only way the class can be
located.
 
Now, to ensure you don't have two classes from two vendors with the same
name, the package declaration was created. This allows the vandors to
"qualify" their class with something akin to a namespace. To ensure the two
vendors do not accidentally use the same package name, convention states
that you use your domain name (reversed) at the beginning of your package
name. As domain names MUST be unique, the packages are unique.
 
For example, if you worked on a utilily package within your MIS department -
and you work for a company with the domain name http://ReallyBigCompany.com
then you would create a package with a name similar to
com.reallybigcompany.mis.util. This is guaranteed unique!
 
The thing to note is that, to include a class within the package you must do
two things: You must include the package statement in the source and you
must also place the class within a directory tree that maps to the package
name. You then set your classpath to point to the directory that contains
your "com" directory.
 
So, if you have a class within the directory
C:\myclasses\com\reallybigcompany\mis\util you would set your classpath to
C:\myclasses. The package declaration will implicitly declare the rest of
the file path.
 
To make life easier to distribute packages you can "wrap" the constituent
classes up together in a java archive file (jar file). This file will
contain your package within the "com\reallybigcompany\mis\util" directory
tree. They are zipped up just like a standard zip file. All you need to do
to access these classes is to set your classpath to point to the jar file.
The java runtime will navigate through the jar file, locate the class, unzip
and load it. This is part of java support - just like java, javac, javadoc.
I believe it's part of the SDK definition.
 
Finally, when you declare a java member (method or variable) without any
access modifier (public, private, protected) then you are implicitly
declaring it to be package protected. This means that only other classes
within the same package can access them. By removing all of the package
information from your BrowseControl class you may run into some problems -
as your class no longer has package access to some of the package class
members.
 
Cheers
 
Larry Ducie