Hi I am using below code in build step in jenkin
#!/bin/bash
pwd
cd ./eclipse-workspace/SoapUiTest
pwd
javac -classpath "lib/* -d ./bin ./src/defaultPackage/*.java"
java -cp "bin;lib/* org.testng.TestNG testng.xml"
while running job i am getting below error
javac: no source files
Usage: javac
use -help for a list of possible options
Usage: java [-options] class [args...]
(to execute a class)
or java [-options] -jar jarfile [args...]
(to execute a jar file)
where options include:
-d32 use a 32-bit data model if available
-d64 use a 64-bit data model if available
-server to select the "server" VM
-zero to select the "zero" VM
-dcevm to select the "dcevm" VM
The default VM is server,
because you are running on a server-class machine.
-cp <class search path of directories and zip/jar files>
-classpath <class search path of directories and zip/jar files>
A : separated list of directories, JAR archives,
and ZIP archives to search for class files.
An obvious error is the mixure of flags in flag values in javac command:
#!/bin/bash
pwd
cd ./eclipse-workspace/SoapUiTest
pwd
javac -classpath "lib/*" -d "./bin" "./src/defaultPackage/*.java"
java -cp "bin;lib/* org.testng.TestNG testng.xml"
javac command explanation: you specify where to find additional class binaries that the compilation depends on, where to output the compiled files and finally where to read the sources from.
Related
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
I have a folder Called TutorialFolder. Inside this, i have tutorial_class folder and WordCount.java file.
When I run wordcount program, it is giving below error.
hduser#ubuntu:~/Desktop/TutorialFolder$ javac -classpath ${HADOOP_CLASSPATH}-d '/home/hduser/Desktop/TutorialFolder/tutorial_class' '/home/hduser/Desktop/TutorialFolder/WordCount.java'
javac: invalid flag: /home/hduser/Desktop/TutorialFolder/tutorial_class
Usage: javac <options> <source files>
use -help for a list of possible options
The problem here is with the variable ${HADOOP_CLASSPATH} which is not set or empty. Thus the command is interpreted as,
javac -classpath -d /home/hduser/Desktop/TutorialFolder/tutorial_class /home/hduser/Desktop/TutorialFolder/WordCount.java
Fix the $HADOOP_CLASSPATH variable, the command should work.
Or, Try with hadoop classpath command
javac -classpath `hadoop classpath` -d /home/hduser/Desktop/TutorialFolder/tutorial_class /home/hduser/Desktop/TutorialFolder/WordCount.java
I'm following this tutorial on how to build an Android Plugin for Unity
I'm currently at the part where the author tells me to do the following in command line:
1.> javac CompassActivity.java -classpath C:\Program Files (x86)\Unity\Editor\Data\PlaybackEngines\androidplayer\bin\classes.jar
-bootclasspath C:\android-sdk-windows\platforms\android-8\android.jar -d .
2.> javap -s com.yourcompany.yourgamename.CompassActivity
3.> jar cvfM ../Compass.jar com/
However when I type the following line:
javac CompassActivity.java -classpath C:\Program Files (x86)\Unity\Editor\Data\PlaybackEngines\androidplayer\bin\classes.jar
I get the following message:
javac: invalid flags: (x86)\Unity\Editor\Data\PlaybackEngines\androidplayer\bin\classes.jar
usage: javac <options> <source files>
use -help for a list of possible options
So I've tried retyping the line putting my path of the file in angled brackets, placing a dot in between classpath and the start of my file location, but I keep getting the same issue.
Am I using classpath wrong?
If so, what is the correct way I should be doing it?
I should add that the console does point to the correct folder location. That was the first thing I've checked.
There are spaces in the path to classes.jar, you must enclose it using ", or shell will consider it as three distinct parameters (C:\Program, Files and (x86)\Unity\Editor\Data\PlaybackEngines\androidplayer\bin\classes.jar"):
javac CompassActivity.java -classpath "C:\Program Files (x86)\Unity\Editor\Data\PlaybackEngines\androidplayer\bin\classes.jar"
You must try the command like:
usage: javac <options> <source files>
javac -classpath "C:\Program Files (x86)\Unity\Editor\Data\PlaybackEngines\androidplayer\bin\classes.jar" CompassActivity.java
First Check your System is 32-bit or 64-bit
check it out full steps for Config and run:http://introcs.cs.princeton.edu/java/15inout/windows-cmd.html
USe
javac -cp filepath
or you also try to set the classpath first by command
set classpath="filepath"
Then u can try with a command
java filepath
I believe this is how I can compile and run a file that uses external library. I'm using Windows.
top level directory
|
|-log4-1.2.17.jar
|-MyApp.java
|-com
|-foo
|-Bar.java
Compiling
javac -cp log4j-1.2.17.jar;. com\foo\Bar.java
javac -cp log4j-1.2.17.jar;"com\foo";. MyApp.java
Executing
java -cp log4j-1.2.17.jar;"com\foo";. MyApp
Compiling itself failed.
simple batch script, for compiling all your project
set COMPILED_CLASSES=.\
set TEMP_FILE=temp
dir .\*.java /s /B > %TEMP_FILE%
javac -classpath log4j-1.2.17.jar;%COMPILED_CLASSES% -d %COMPILED_CLASSES% #%TEMP_FILE%
rm %TEMP_FILE%
add it to top level dir and run
EDIT
step by step
javac ./com/foo/Bar.java -classpath log4j-1.2.17.jar
next
javac ./MyApp.java -classpath log4j-1.2.17.jar;./
run
java -classpath log4j-1.2.17.jar;./ MyApp
Include current directory in java classpath
java -cp log4j-1.2.17.jar;. MyApp
Why do you have to include current directory:
The default class path is the current directory. Setting the CLASSPATH variable or using the -classpath command-line option overrides that default, so if you want to include the current directory in the search path, you must include "." in the new settings.
Y need to include the local directory. If you want to do it in the current directory it would be something like:
javac -cp .;log4j-1.2.17.jar Bar
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'