Develop Reference

Linux equivalent of including the classpath during compilation

I'm following a guide that only includes compilation instructions on windows. How would one run this build.bat file on Linux?
The batch file looks like this:
#echo off
#echo Compiling...
javac -classpath ..\..\lib\OneWireAPI.jar;%classpath% -d . .\src\*.java
And when I run the javac command on Linux, it fails:
javac -classpath ../../lib/OneWireAPI.jar;%classpath% -d . ./src/ReadTemp.java
The output is:
javac: no source files
What is the correct way to do this?

On Linux, you have to use : (colon) in place of ; (semicolon) as the path separator in Java options.
Also, if you have a classpath variable, in most common Linux shells it is referenced by $classpath rather than by %classpath%
javac -classpath ../../lib/OneWireAPI.jar:$classpath -d . ./src/ReadTemp.java

You have two items that did not get translated correctly from Windows CMD to Unix:
Path separator ; should be :.
Environment variables should be changed from %classpath% to $CLASSPATH format. Note that pretty much everything is case-sensitive in Linux, including environment variable names, and the Java path is traditionally all-caps.
Try
javac -classpath ../../lib/OneWireAPI.jar:$CLASSPATH -d . ./src/ReadTemp.java

Import multiple libraries in makefile within Javac command

The problem
This is a problem I just faced using makefile in java, on Windows.
I wanted to set up my classpath with multiple path (libraries, etc.). The new command work by hands, but not from the makefile which throws me this error :
javac : no source files
Example
Let's say I have this makefile :
JFLAGS = -g
JARFLAGS = -cvfm
CLASSPATH = ./bin
LIBS = C:/java/lib/mylib.jar
SOURCEPATH = ./src/client
compileAll:
javac $(JFLAGS) -d $(CLASSPATH) -cp $(CLASSPATH)\;$(LIBS) $(SOURCEPATH )/*.java
jar $(JARFLAGS) app.jar bin/client/MANIFEST.MF bin/client/*.class
So the command line to compile the project is :
javac -g -d ./bin -cp ./bin;C:/java/libs/lib.jar ./src/client/*.java
It works well.
The class files goes to ./bin directory. It imports classes from ./bin and the lib.jar library. And it compliles all the source files from the ./src/client directory.
This command works perfectly by hands, but no from the makefile which doesn't compile anything.

Thanks to my text editor which colored the ';' character, I understood that I just needed to escape (disable) the ';' character by using a '\' :
javac -g -d ./bin -cp ./bin\;C:/java/libs/lib.jar ./src/client/*.java
Now, it works well from makefile !

How to create package folders with javac command

I am compiling java file with command prompt with some package name come.test.
But while running with java command it gives no class definition error. If I compile with IDE it is running because folders are created like com\test
How can I get those folders if I compile with javac command.

For such questions, when an online search at oracle does yield immediate results, try
javac -help
javac -source 1.8
-target 1.8
-encoding UTF-8
javac -d target/classes --> generated classes
-s src/main/java --> generated source files
src/main/java/x/y/*.java --> java files to be compiled
The directory target/classes should exist. Package folders are created under classes (x/y).

Batch file to execute java program

I have multiple .java files inside a folder (e.g temp/code/project)
I want a batch file that will compile and run these java files.
The batch file should create class files inside the same structure where java files are located starting with the classes folder (i.e classes/temp/code/project)
I don't know how to write a batch file. Can any body help me with this? Thanks in advance.

Theoretically you should write the following:
#echo off
::compile classes
javac -cp YOUR_CLASSPATH com/yourcompany/YourClass1.java
javac -cp YOUR_CLASSPATH com/yourcompany/YourClass2.java
javac -cp YOUR_CLASSPATH com/yourcompany/YourClass3.java
javac -cp YOUR_CLASSPATH com/yourcompany/YourClassLauncher.java
:: create jar
jar cvfM Manifest.txt myjar.jar *.class
echo.
echo Hit any key to launch project.
pause
java -jar myjar.jar
pause
If you have one class that depends on all other classes in your project it is enough to run javac with this class only: compiler will compile everything.
# Manifest.txt
Manifest-Version: 1.0
Class-Path: .;MyUtils.jar
Created-By: 1.6.0
Main-Class: com.yourcompany.YourClassLauncher
This is only example and a good exercise. In real life people use special build tools like good old Ant, Maven or newer Graidle or Buildr. I'd recommend you to take one of them.

Something along the lines of
cd C:\temp\code\project
javac -classpath . -d C:\classes\temp\code\project\ *.java
You may not need to include the -classpath tag (I always do so that I don't have to worry about maintaining my CLASSPATH variable. The -d tag specifies a particular directory to place generated class files.

javac no source files found

I have the .java file on the current working directory but javac reports:
javac: no source files
Usage: javac <options> <source files>
use -help for a list of possible options
I'm working on ubuntu.

From your comment above, it looks like you tried:
javac -cp .;lib.jar a.java on your Ubuntu system. The CLASSPATH separator is : on Unix systems and ; on Windows.
Ubuntu considered the command up to the ;, java -cp . and thus gave the message.
javac -cp .:lib.jar a.java should compile fine.

For anyone who is using powersehll on windows use CLASSPATH separator : instead of ;

I tried a similar thing and found that you need to mention the absolute path when you are using the
-cp and -d option with javac like this
javac -cp 'ur location of jars & files'; -d 'location to add your classes to' 'absolute path of file'
eg:
javac -cp C:\home\lib\mywork; -d c:\home\classes c:\home\files*.java

for javac, there are options and arguments
arg: it takes argument as path of source file
options: we require for basic compilation
-sourcepath: the path of dependent source files
-d: directory path of output classes
javac -sourcepath './src' -d './bin' -verbose './src/App.java'