Compiling java program using javac from command prompt - java

I have new laptop on which I have installed jdk1.8.0_91 and jre1.8.0_91.
Both are in the "C:\Program Files\Java" folder.
I have NOT set any classpath or any environment variables.
I wrote a a HelloWorld.java program and saved it in "C:\my Data" folder.
I then went to Command Prompt using cmd.
Then I changed the current directory to "C:\Program Files\Java\jdk1.8.0_91\bin" ..since here is the javac.exe
and then tried to compile my HelloWorld program and its giving the following error -
C:\Program Files\Java\jdk1.8.0_91\bin>javac -sourcepath C:\my Data\HelloWorld.java
javac: invalid flag: Data\HelloWorld.com
Usage: javac <options> <source files>
use -help for a list of possible options
I am not sure whether I am correctly using the "sourcepath" or not...
How should I tell the compiler where my source file is ?(and I want to resolve this without setting any classpath or any environment variables)

Use this instead...
javac -sourcepath "C:\my Data" "C:\my Data\HelloWorld.java"
The sourcepath parameter allows you to specify the DIRECTORY where source files will be found. As per the javac command line output:
-sourcepath Specify where to find input source files
The parameter after that specifies the actual Java files to compile. You will need " around the parameters, given that your paths have spaces in them. Avoid spaces in your paths where ever possible to avoid this issue.

-sourcepath is a PATH, you are giving a file name that's not a java file, that's not valid. From the docs:
-sourcepath sourcepath
Specify the source code path to search for class or interface
definitions. As with the user class path, source path entries are
separated by colons (:) and can be directories, JAR archives, or ZIP
archives. If packages are used, the local path name within the
directory or archive must reflect the package name.
[EDIT: OP changes the file name to .java in the question, as the other answer noted, it needed quotes.]

You need to place the source path in quotes so that the command line processes it as a single argument. The source path must also be the directory in your case, not the file:
javac -sourcepath "C:\my Data"

Path C:\my Data\HelloWorld.java has space in it hence the error.
Please enclose path in "" (double quotes)

Related

Compiling packages in java at windows command line

I have trying to compile java files at the windows command line using commands such as:
java myProg once I have used javac to create class files.
Problems arise when I use packages with a number of source files.
Often but not always I get main not found errors even though a main exists.
I am not quite sure what some of the directives mean and that is why it seems hit or miss.
Question
what does -cp mean exactly? java -cp src\myDirectory.myfile
sometimes I see:
./ infront of source eg .\src\myDirectory.myfile
on other sites I have found
% javac -cp .;stdlib.jar MyProgram.java
% java -cp .;stdlib.jar MyProgram
while compiling a jar library with java source files
what doesthe ".;" mean?
basically how do I compile three java source java files in one package at the windows command line and what does -cp and .; mean?
-cp means class path if I'm not mistaken.
try reading the following java docs
-classpath path
Specifies the path javac uses to look up classes needed to run javac or being referenced by other classes you are compiling. Overrides the default or the CLASSPATH environment variable if it is set. Directories are separated by semi-colons. It is often useful for the directory containing the source files to be on the class path. You should always include the system classes at the end of the path. For example:
javac -classpath .;C:\users\dac\classes;C:\tools\java\classes ...
https://www.cis.upenn.edu/~bcpierce/courses/629/jdkdocs/tooldocs/win32/javac.html
Answering your question directly, -cp means classpath or path.
Details on commandline arguments used while compiling and running a Java application can be found here: javac - Java programming language compiler
Extracting the description of -cp from that page:
-cp path or -classpath path:
Specify where to find user class files, and (optionally) annotation processors and source files. This class path overrides the user class path in the CLASSPATH environment variable. If neither CLASSPATH, -cp nor -classpath is specified, the user class path consists of the current directory. See Setting the Class Path for more details.
. means the current directory.
To compile multiple files in a directory use the following:
javac *.java // compliles all java files in the dir
java MyClass // runs the particular file
There are also a bunch of other related questions that should help you resolve this:
How to run a java program from the command line
How do I run java program with multiple classes from cmd?
Problems running a java program from the command line interface
Can't run multiple-class program from command line using packages

What are the common errors you see when you run 'java -cp ...' or 'java -classpath'? How do you set a directory of jars in classpath? [duplicate]

Is there a way to include all the jar files within a directory in the classpath?
I'm trying java -classpath lib/*.jar:. my.package.Program and it is not able to find class files that are certainly in those jars. Do I need to add each jar file to the classpath separately?
Using Java 6 or later, the classpath option supports wildcards. Note the following:
Use straight quotes (")
Use *, not *.jar
Windows
java -cp "Test.jar;lib/*" my.package.MainClass
Unix
java -cp "Test.jar:lib/*" my.package.MainClass
This is similar to Windows, but uses : instead of ;. If you cannot use wildcards, bash allows the following syntax (where lib is the directory containing all the Java archive files):
java -cp "$(printf %s: lib/*.jar)"
(Note that using a classpath is incompatible with the -jar option. See also: Execute jar file with multiple classpath libraries from command prompt)
Understanding Wildcards
From the Classpath document:
Class path entries can contain the basename wildcard character *, which is considered equivalent to specifying a list of all the files
in the directory with the extension .jar or .JAR. For example, the
class path entry foo/* specifies all JAR files in the directory named
foo. A classpath entry consisting simply of * expands to a list of all
the jar files in the current directory.
A class path entry that contains * will not match class files. To
match both classes and JAR files in a single directory foo, use either
foo;foo/* or foo/*;foo. The order chosen determines whether the
classes and resources in foo are loaded before JAR files in foo, or
vice versa.
Subdirectories are not searched recursively. For example, foo/* looks
for JAR files only in foo, not in foo/bar, foo/baz, etc.
The order in which the JAR files in a directory are enumerated in the
expanded class path is not specified and may vary from platform to
platform and even from moment to moment on the same machine. A
well-constructed application should not depend upon any particular
order. If a specific order is required then the JAR files can be
enumerated explicitly in the class path.
Expansion of wildcards is done early, prior to the invocation of a
program's main method, rather than late, during the class-loading
process itself. Each element of the input class path containing a
wildcard is replaced by the (possibly empty) sequence of elements
generated by enumerating the JAR files in the named directory. For
example, if the directory foo contains a.jar, b.jar, and c.jar, then
the class path foo/* is expanded into foo/a.jar;foo/b.jar;foo/c.jar,
and that string would be the value of the system property
java.class.path.
The CLASSPATH environment variable is not treated any differently from
the -classpath (or -cp) command-line option. That is, wildcards are
honored in all these cases. However, class path wildcards are not
honored in the Class-Path jar-manifest header.
Note: due to a known bug in java 8, the windows examples must use a backslash preceding entries with a trailing asterisk: https://bugs.openjdk.java.net/browse/JDK-8131329
Under Windows this works:
java -cp "Test.jar;lib/*" my.package.MainClass
and this does not work:
java -cp "Test.jar;lib/*.jar" my.package.MainClass
Notice the *.jar, so the * wildcard should be used alone.
On Linux, the following works:
java -cp "Test.jar:lib/*" my.package.MainClass
The separators are colons instead of semicolons.
We get around this problem by deploying a main jar file myapp.jar which contains a manifest (Manifest.mf) file specifying a classpath with the other required jars, which are then deployed alongside it. In this case, you only need to declare java -jar myapp.jar when running the code.
So if you deploy the main jar into some directory, and then put the dependent jars into a lib folder beneath that, the manifest looks like:
Manifest-Version: 1.0
Implementation-Title: myapp
Implementation-Version: 1.0.1
Class-Path: lib/dep1.jar lib/dep2.jar
NB: this is platform-independent - we can use the same jars to launch on a UNIX server or on a Windows PC.
My solution on Ubuntu 10.04 using java-sun 1.6.0_24 having all jars in "lib" directory:
java -cp .:lib/* my.main.Class
If this fails, the following command should work (prints out all *.jars in lib directory to the classpath param)
java -cp $(for i in lib/*.jar ; do echo -n $i: ; done). my.main.Class
Short answer: java -classpath lib/*:. my.package.Program
Oracle provides documentation on using wildcards in classpaths here for Java 6 and here for Java 7, under the section heading Understanding class path wildcards. (As I write this, the two pages contain the same information.) Here's a summary of the highlights:
In general, to include all of the JARs in a given directory, you can use the wildcard * (not *.jar).
The wildcard only matches JARs, not class files; to get all classes in a directory, just end the classpath entry at the directory name.
The above two options can be combined to include all JAR and class files in a directory, and the usual classpath precedence rules apply. E.g. -cp /classes;/jars/*
The wildcard will not search for JARs in subdirectories.
The above bullet points are true if you use the CLASSPATH system property or the -cp or -classpath command line flags. However, if you use the Class-Path JAR manifest header (as you might do with an ant build file), wildcards will not be honored.
Yes, my first link is the same one provided in the top-scoring answer (which I have no hope of overtaking), but that answer doesn't provide much explanation beyond the link. Since that sort of behavior is discouraged on Stack Overflow these days, I thought I'd expand on it.
Windows:
java -cp file.jar;dir/* my.app.ClassName
Linux:
java -cp file.jar:dir/* my.app.ClassName
Remind:
- Windows path separator is ;
- Linux path separator is :
- In Windows if cp argument does not contains white space, the "quotes" is optional
For me this works in windows .
java -cp "/lib/*;" sample
For linux
java -cp "/lib/*:" sample
I am using Java 6
You can try java -Djava.ext.dirs=jarDirectory
http://docs.oracle.com/javase/6/docs/technotes/guides/extensions/spec.html
Directory for external jars when running java
Correct:
java -classpath "lib/*:." my.package.Program
Incorrect:
java -classpath "lib/a*.jar:." my.package.Program
java -classpath "lib/a*:." my.package.Program
java -classpath "lib/*.jar:." my.package.Program
java -classpath lib/*:. my.package.Program
If you are using Java 6, then you can use wildcards in the classpath.
Now it is possible to use wildcards in classpath definition:
javac -cp libs/* -verbose -encoding UTF-8 src/mypackage/*.java -d build/classes
Ref: http://www.rekk.de/bloggy/2008/add-all-jars-in-a-directory-to-classpath-with-java-se-6-using-wildcards/
If you really need to specify all the .jar files dynamically you could use shell scripts, or Apache Ant. There's a commons project called Commons Launcher which basically lets you specify your startup script as an ant build file (if you see what I mean).
Then, you can specify something like:
<path id="base.class.path">
<pathelement path="${resources.dir}"/>
<fileset dir="${extensions.dir}" includes="*.jar" />
<fileset dir="${lib.dir}" includes="*.jar"/>
</path>
In your launch build file, which will launch your application with the correct classpath.
Please note that wildcard expansion is broken for Java 7 on Windows.
Check out this StackOverflow issue for more information.
The workaround is to put a semicolon right after the wildcard. java -cp "somewhere/*;"
To whom it may concern,
I found this strange behaviour on Windows under an MSYS/MinGW shell.
Works:
$ javac -cp '.;c:\Programs\COMSOL44\plugins\*' Reclaim.java
Doesn't work:
$ javac -cp 'c:\Programs\COMSOL44\plugins\*' Reclaim.java
javac: invalid flag: c:\Programs\COMSOL44\plugins\com.comsol.aco_1.0.0.jar
Usage: javac <options> <source files>
use -help for a list of possible options
I am quite sure that the wildcard is not expanded by the shell, because e.g.
$ echo './*'
./*
(Tried it with another program too, rather than the built-in echo, with the same result.)
I believe that it's javac which is trying to expand it, and it behaves differently whether there is a semicolon in the argument or not. First, it may be trying to expand all arguments that look like paths. And only then it would parse them, with -cp taking only the following token. (Note that com.comsol.aco_1.0.0.jar is the second JAR in that directory.) That's all a guess.
This is
$ javac -version
javac 1.7.0
All the above solutions work great if you develop and run the Java application outside any IDE like Eclipse or Netbeans.
If you are on Windows 7 and used Eclipse IDE for Development in Java, you might run into issues if using Command Prompt to run the class files built inside Eclipse.
E.g. Your source code in Eclipse is having the following package hierarchy:
edu.sjsu.myapp.Main.java
You have json.jar as an external dependency for the Main.java
When you try running Main.java from within Eclipse, it will run without any issues.
But when you try running this using Command Prompt after compiling Main.java in Eclipse, it will shoot some weird errors saying "ClassNotDef Error blah blah".
I assume you are in the working directory of your source code !!
Use the following syntax to run it from command prompt:
javac -cp ".;json.jar" Main.java
java -cp ".;json.jar" edu.sjsu.myapp.Main
[Don't miss the . above]
This is because you have placed the Main.java inside the package edu.sjsu.myapp and java.exe will look for the exact pattern.
Hope it helps !!
macOS, current folder
For Java 13 on macOS Mojaveā€¦
If all your .jar files are in the same folder, use cd to make that your current working directory. Verify with pwd.
For the -classpath you must first list the JAR file for your app. Using a colon character : as a delimiter, append an asterisk * to get all other JAR files within the same folder. Lastly, pass the full package name of the class with your main method.
For example, for an app in a JAR file named my_app.jar with a main method in a class named App in a package named com.example, alongside some needed jars in the same folder:
java -classpath my_app.jar:* com.example.App
For windows quotes are required and ; should be used as separator. e.g.:
java -cp "target\\*;target\\dependency\\*" my.package.Main
Short Form: If your main is within a jar, you'll probably need an additional '-jar pathTo/yourJar/YourJarsName.jar ' explicitly declared to get it working (even though 'YourJarsName.jar' was on the classpath)
(or, expressed to answer the original question that was asked 5 years ago: you don't need to redeclare each jar explicitly, but does seem, even with java6 you need to redeclare your own jar ...)
Long Form:
(I've made this explicit to the point that I hope even interlopers to java can make use of this)
Like many here I'm using eclipse to export jars: (File->Export-->'Runnable JAR File'). There are three options on 'Library handling' eclipse (Juno) offers:
opt1: "Extract required libraries into generated JAR"
opt2: "Package required libraries into generated JAR"
opt3: "Copy required libraries into a sub-folder next to the generated JAR"
Typically I'd use opt2 (and opt1 was definitely breaking), however native code in one of the jars I'm using I discovered breaks with the handy "jarinjar" trick that eclipse leverages when you choose that option. Even after realizing I needed opt3, and then finding this StackOverflow entry, it still took me some time to figure it out how to launch my main outside of eclipse, so here's what worked for me, as it's useful for others...
If you named your jar: "fooBarTheJarFile.jar"
and all is set to export to the dir: "/theFully/qualifiedPath/toYourChosenDir".
(meaning the 'Export destination' field will read: '/theFully/qualifiedPath/toYourChosenDir/fooBarTheJarFile.jar' )
After you hit finish, you'll find eclipse then puts all the libraries into a folder named 'fooBarTheJarFile_lib' within that export directory, giving you something like:
/theFully/qualifiedPath/toYourChosenDir/fooBarTheJarFile.jar
/theFully/qualifiedPath/toYourChosenDir/fooBarTheJarFile_lib/SomeOtherJar01.jar
/theFully/qualifiedPath/toYourChosenDir/fooBarTheJarFile_lib/SomeOtherJar02.jar
/theFully/qualifiedPath/toYourChosenDir/fooBarTheJarFile_lib/SomeOtherJar03.jar
/theFully/qualifiedPath/toYourChosenDir/fooBarTheJarFile_lib/SomeOtherJar04.jar
You can then launch from anywhere on your system with:
java -classpath "/theFully/qualifiedPath/toYourChosenDir/fooBarTheJarFile_lib/*" -jar /theFully/qualifiedPath/toYourChosenDir/fooBarTheJarFile.jar package.path_to.the_class_with.your_main.TheClassWithYourMain
(For Java Newbies: 'package.path_to.the_class_with.your_main' is the declared package-path that you'll find at the top of the 'TheClassWithYourMain.java' file that contains the 'main(String[] args){...}' that you wish to run from outside java)
The pitfall to notice: is that having 'fooBarTheJarFile.jar' within the list of jars on your declared classpath is not enough. You need to explicitly declare '-jar', and redeclare the location of that jar.
e.g. this breaks:
java -classpath "/theFully/qualifiedPath/toYourChosenDir/fooBarTheJarFile.jar;/theFully/qualifiedPath/toYourChosenDir/fooBarTheJarFile_lib/*" somepackages.inside.yourJar.leadingToTheMain.TheClassWithYourMain
restated with relative paths:
cd /theFully/qualifiedPath/toYourChosenDir/;
BREAKS: java -cp "fooBarTheJarFile_lib/*" package.path_to.the_class_with.your_main.TheClassWithYourMain
BREAKS: java -cp ".;fooBarTheJarFile_lib/*" package.path_to.the_class_with.your_main.TheClassWithYourMain
BREAKS: java -cp ".;fooBarTheJarFile_lib/*" -jar package.path_to.the_class_with.your_main.TheClassWithYourMain
WORKS: java -cp ".;fooBarTheJarFile_lib/*" -jar fooBarTheJarFile.jar package.path_to.the_class_with.your_main.TheClassWithYourMain
(using java version "1.6.0_27"; via OpenJDK 64-Bit Server VM on ubuntu 12.04)
You need to add them all separately. Alternatively, if you really need to just specify a directory, you can unjar everything into one dir and add that to your classpath. I don't recommend this approach however as you risk bizarre problems in classpath versioning and unmanagability.
The only way I know how is to do it individually, for example:
setenv CLASSPATH /User/username/newfolder/jarfile.jar:jarfile2.jar:jarfile3.jar:.
Hope that helps!
class from wepapp:
> mvn clean install
> java -cp "webapp/target/webapp-1.17.0-SNAPSHOT/WEB-INF/lib/tool-jar-1.17.0-SNAPSHOT.jar;webapp/target/webapp-1.17.0-SNAPSHOT/WEB-INF/lib/*" com.xx.xx.util.EncryptorUtils param1 param2
Think of a jar file as the root of a directory structure. Yes, you need to add them all separately.
Not a direct solution to being able to set /* to -cp but I hope you could use the following script to ease the situation a bit for dynamic class-paths and lib directories.
libDir2Scan4jars="../test";cp=""; for j in `ls ${libDir2Scan4jars}/*.jar`; do if [ "$j" != "" ]; then cp=$cp:$j; fi; done; echo $cp| cut -c2-${#cp} > .tmpCP.tmp; export tmpCLASSPATH=`cat .tmpCP.tmp`; if [ "$tmpCLASSPATH" != "" ]; then echo .; echo "classpath set, you can now use ~> java -cp \$tmpCLASSPATH"; echo .; else echo .; echo "Error please check libDir2Scan4jars path"; echo .; fi;
Scripted for Linux, could have a similar one for windows too. If proper directory is provided as input to the "libDir2Scan4jars"; the script will scan all the jars and create a classpath string and export it to a env variable "tmpCLASSPATH".
Set the classpath in a way suitable multiple jars and current directory's class files.
CLASSPATH=${ORACLE_HOME}/jdbc/lib/ojdbc6.jar:${ORACLE_HOME}/jdbc/lib/ojdbc14.jar:${ORACLE_HOME}/jdbc/lib/nls_charset12.jar;
CLASSPATH=$CLASSPATH:/export/home/gs806e/tops/jconn2.jar:.;
export CLASSPATH
I have multiple jars in a folder. The below command worked for me in JDK1.8 to include all jars present in the folder. Please note that to include in quotes if you have a space in the classpath
Windows
Compiling: javac -classpath "C:\My Jars\sdk\lib\*" c:\programs\MyProgram.java
Running: java -classpath "C:\My Jars\sdk\lib\*;c:\programs" MyProgram
Linux
Compiling: javac -classpath "/home/guestuser/My Jars/sdk/lib/*" MyProgram.java
Running: java -classpath "/home/guestuser/My Jars/sdk/lib/*:/home/guestuser/programs" MyProgram
Order of arguments to java command is also important:
c:\projects\CloudMirror>java Javaside -cp "jna-5.6.0.jar;.\"
Error: Unable to initialize main class Javaside
Caused by: java.lang.NoClassDefFoundError: com/sun/jna/Callback
versus
c:\projects\CloudMirror>java -cp "jna-5.6.0.jar;.\" Javaside
Exception in thread "main" java.lang.UnsatisfiedLinkError: Unable

Error compiling servlet: file not found

I'm trying to compile a servlet called BeerSelect.java and I'm getting this error:
javac: file not found: BeerSelect.java
I compiled using: javac -classpath "C:\Program Files\Apache Software Foundation\Apache Tomcat 7.0.34\lib\servlet-api.jar"; -d classes \BeerSelect.java
I used this command to compile with my current directory set to where the servlet is stored, my class path is set correctly.
I checked many related questions on this site and cannot get the answer
You shouldn't set your classpath to point to your JDK bin directory -- instead it should be the PATH environment variable, which serves a different purpose to classpath. (The classpath defines a list of jars and directories containing compiled Java .class code; the PATH variable defines a list of paths where the shell needs to look and locate programs to execute when they are not found in the current directory -- so if you type for instance zip -- it would look in all the directories defined in PATH and figure out that zip program is located under /usr/bin) Secondly if you want to compile sources from both directory you need to specify:
all the paths where the sources are (both home/pathToFolderA/src and home/pathToFolderB/gen-java)
the path where the compiled .class files to be generated
specify in the classpath any library you might use in your source files
To sum it up, it would be something like this to compile:
javac -d /home/pathToFolderWithResultsOfCompilation -classpath /path/to/some.jar:/path/to/another.jar home/pathToFolderA/src/*.java home/pathToFolderB/gen-java/*.java
and to run your compiled programs:
java -classpath /path/to/some.jar:/path/to/another.jar:/home/pathToFolderWithResultsOfCompilation full.name.of.your.Java
The problem with the command i was using:
javac -classpath "C:\Program Files\Apache Software Foundation\Apache Tomcat 7.0.34\lib\servlet-api.jar"; -d classes \BeerSelect.java
is with the path name to store the results of compilation (.class file) and putting a backslash on the BeerSelect servlet name (i highlighted the errors).
I refined the command to look like this:
javac -d C:\Users\ModernWarFare\Desktop\MyProject\beerV1\classes BeerSelect.java
the highlited path is where i'm going to store the .class file and i didn't specify the path to the BeerSelect.java file because it is in the current directory
note that i ommited the path to the servlet-api.jar file because i am using tomcat 7, it already have all the jar files i need stored on C:\Program Files\Apache Software Foundation\Apache Tomcat 7.0.34\lib directory.
Sorry for responding late, i have limited time online

java classpath in unix

I can run java in cygwin+windows using the following settings (the sw/jar directory has several jar files, and I pick the relevant one from the java command line):
CLASSPATH=.;C:\sw\java_6u35\lib\\*;C:\sw\jar\\*
java org.antlr.Tool Calc.g
But I am having the following problems when running in linux:
(1) I can't set a directory name in a classpath, the following line reports an error:
setenv CLASSPATH .:/sw/jdk1.6.0_35/lib/\*:/sw/jar/*
(2) when I run explictly with -jar option, I still get an error:
java -jar /sw/jar/antlr-3.4.jar org.antlr.Tool Calc.g
error(7): cannot find or open file: org.antlr.Tool
However, the class does exist. When I do jar tf /sw/jar/antlr-3.4.jar, I get:
...
org/antlr/Tool.class
So my question is: (a) how do I specify in unix that my jar-directory is xxx that contains several jar files, and (2) how do I pick the relevant jar from this dir at runtime?
To specify multiple jars in a directory, directly in the java command, use this
java -cp "/sw/jar/*" org.antlr.Tool Calc.g
This will include all the jars
If you want to set the classpath in Unix/Linux systems, use this
export CLASSPATH=/sw/jar/a.jar:/sw/jar/b.jar
in unix use this to set the classpath:
export CLASSPATH=myClassPath
about not finding your jar, you're using a leading slash (/), that means that you path is absolute (not relative to your home folder) is this what you want?
if you want the path to be relative to your folder try:
java -jar ~/mypathToMyJar

Compiling java file from different directory

I am very new to JAVA. I have written simple program (in Linux -VIM editor), compiled and executed it, everything is fine.
Now,I have moved that file to a different directory and am trying to compile(javac Myfile.java) it, but it throws an error message as javac-not found.
Can somebody explain what is the problem?
Your original question was not totally clear (since it did not contain the complete error message).
From your comment:
$ javac Example1.java
javac: file not found: Example1.java
Usage: javac <options> <source files> use -help for a list of possible options
So, javac did not find your file example java.
Normally, you should not have to set the CLASSPATH (use export CLASSPATH= in bash), and javac would search the source in the current directory. Is your Example1.java in the current directory? (Type ls and look at the output.)
If not, you should give the path to this file to javac as a parameter ... but it really is better so simply move to the right directory with cd.
If you are using packages, position your shell to the directory on top of the package directory hierarchy, and call the compiler with the relative filename from there.
Edit, since I see the next questions coming:
The compiler will put the resulting class files in the output directory tree given by the -d parameter (or the current directory, if not given), by their package structure, so make sure you search them there later (when invoking the program).
If the compiler needs other classes to compiler the files indicated in the command line, it searches class files in the classpath (given by the -classpath or -cp option, or by the CLASSPATH environment variable, or the current directory) and source files in the sourcepath (given by the -sourcepath option or the classpath if no sourcepath is set). If for a needed both exist and the source file is newer, it is recompiled too. (They are searched according to the package-structure, too.)
So in this case you should make sure to pass the -sourcepath option so the compiler can find your other source files.
set the classpath and path properly and check whether its working fine.
USAGE:
SET CLASSPATH=%<CLASSPATH>%
SET PATH=%<PATH_WHERE_JDKS_BIN_LOCATED>%
The path environment variable must point to the bin directory in the jdk installation...
USAGE:
Variable : JAVA_HOME
Value : C:\Program Files\Java\jdk1.5.0\bin;.
Variable: PATH
Value : C:\Program Files\Java\jdk1.5.0\lib
System Variables :
Variable : PATH (This will be there already)
Value : %JAVA_HOME%\bin;
Since the file is not in current directory do the below at the prompt
$cd home/kiddosr/Kiddo/Java_Programs/ and press enter
home/kiddosr/Kiddo/Java_Programs at this point of time type javac Example1.java

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