(Crossposting note: I have asked this question one week ago at the JRuby mailing list, but didn't get any answer yet).
I have a jar file provided by someone else, no access to the source code. The jar file is in lib/other/appl.jar, the class is named Appl, and the package is com.abc.xyz
I would like to instantiate an Appl object from the JRuby irb, jirb_swing_ex.
(Of course my problem applies not only to jirb, but to running JRuby programs in general, but I explain it in the way I'm using it right now, just in case there are some peculiarities in Jirb which need special treatment).
THIS is the way it DOES work:
(1) Invoke jirb by:
java -jar jr/jruby-complete-1.7.27 jb/jirb_swing_ex
(2) Put the directory with the jar file into the load path:
$: << 'lib/other'
(3) Load the jar file
require 'appl.jar'
(4) Import the class
java_import com.abc.xyz.Appl
(5) Create the object
x = Appl.new
As I said, this works, and I can live with it if necessary, but I would prefer a simpler approach:
NOW TO MY QUESTION: Instead of fiddling around with load path and doing a require for the Jar file, I thought I could let Java already include the jar file. This is what I have tried:
java -cp lib/other/appl.jar -jar jr/jruby-complete-1.7.27 jb/jirb_swing_ex
The problem is: How can I get at my object? If I just use the class name com.abc.xyz.Appl, JRuby complains that the class not found (NameError: missing class name).
BTW, I have also tried forward slashes (since I'm on Windows), i.e.
java -cp lib\other\appl.jar -jar jr\jruby-complete-1.7.27 jb\jirb_swing_ex
but the same effect. I had expected that, having appl.jar in my class path, would make the classes available somehow, but I seem to miss something.
Running jirb or jirb_swing with custom class path
jirb and jirb_swing will use the value of JRUBY_CP environment variable (if present) to extend class path given to Java command line.
Example with commons-lang3 library taken from my local maven repository, using bash on Linux or macOS:
$ export JRUBY_CP=${HOME}/.m2/repository/org/apache/commons/commons-lang3/3.4/commons-lang3-3.4.jar
$ jirb
irb(main):001:0> org.apache.commons.lang3.mutable.MutableBoolean.new
=> #<Java::OrgApacheCommonsLang3Mutable::MutableBoolean:0x7c24b813>
Running JRuby programs with custom class path
To run a JRuby program that uses a third-party java library, this won't work:
java -cp lib/other/appl.jar -jar jr/jruby-complete-1.7.27 ...
You must use either -jar or -cp, you can't combine the two.
From java man page:
When you use this option [-jar], the JAR file is the source of all user classes, and other user class path settings are ignored.
In addition, you need to pass the main Java class, which is org.jruby.Main, and that class needs arguments: either a path to a Ruby script, or other command line arguments such as -e 'puts 2+2'.
So the command line structure is the following:
# Run script file:
java -cp path/to/jruby.jar:other/custom.jar org.jruby.Main path/to/script
# Run simple one-line Ruby program
java -cp path/to/jruby.jar:other/custom.jar org.jruby.Main -e 'some ruby here'
(on Windows please use ; instead of : as separator)
Actual example with same commons-lang3 library & OS:
$ alias myjruby="java -cp ${JRUBY_HOME}/lib/jruby.jar:${HOME}/.m2/repository/org/apache/commons/commons-lang3/3.4/commons-lang3-3.4.jar org.jruby.Main"
# Verifying base jruby code works with that:
$ myjruby -e 'puts 2+2'
4
# Now verifying it can use my 3rd-party lib:
$ myjruby -e 'puts org.apache.commons.lang3.mutable.MutableBoolean.new'
false
I'm trying to run a java library, so that I can build upon it and do my customization. The library is called jayu parses ASN files so that you can decode them. It can be downloaded here
There are a few test data to check the library in the "test" folder and mentioned in the Readme.txt file. There is a asn2csv batch file for windows but I'm using OSX mavericks. According to the Readme file, I need to run it by invoking the command:
ls $ASN_DATA_DIR/*.dat | xargs java -cp "./*.jar:." Path/To/Stream1.txt test.testdata.Stream1 $OUTPUT_DIR
OR
java -cp "*.jar;$PATH_TO_TEST_DIR" Stream1.txt test.testdata.Stream1 . Path/To/Stream1.dat
But whatever I try I always get Error: Could not find or load main class Stream1.txt
I'm not a Java programmer. What am I doing wrong here? The development of this seems to be inactive lately but it's still useful. So, I'm posting it here, hoping someone will help me run the example.
Edit: I've added the tree structure of the directory and the files
+ jayu
|--Readme.txt
|--commons-compiler.jar
|--janino.jar
|--jayu.jar (ASN parser)
|--AsnToCsv.bat (Command line Tool)
+--test (Contains test data for examples)
|
+ testdata
|
Stream1.txt (Grammar File)
Stream1.dat (ASN Data File)
Stream1.java {mapFile}
...
It is interpreting Stream1.txt as the Java class you are trying to execute, because it is treating at as your first argument to java. Your first argument should be the name of the class containing the main(), or, the executable JAR must be indicated with a -jar option.
This would seem to indicate that "*.jar;$PATH_TO_TEST_DIR" is evaluating to blank. Can you see if there are any .jar files in your current working directory? Also what is the value of $PATH_TO_TEST_DIR?
Another thing is that if you use the second form on OS X, you should have a : instead of a ; because it is a Unix-based OS, not Windows.
Update1
I had a slight error with my description of how to run an executable JAR. You use the -jar option, not -cp (I corrected it above). Since they put -cp in their invocation, I'm guessing they are not intending to target the executable JAR, but rather to name the main class. That to me says that test.testdata.Stream1 is that main class, which means the ordering they gave you is wrong. Try this:
java -cp "*.jar:$PATH_TO_TEST_DIR" test.testdata.Stream1 Stream1.txt . Path/To/Stream1.dat
or some other ordering that starts with:
java -cp "*.jar:$PATH_TO_TEST_DIR" test.testdata.Stream1 ..........
That is, that makes test.testdata.Stream1 the very first command line argument to java.
My environment is Windows 7 and JDK 1.7.
I have not set the CLASSPATH environment variable; echo %CLASSPATH% outputs nothing.
Java compiler: C:\Program Files\Java\jdk1.7.0_10\bin
Java source: is in D:\tmpmulu\Tj.java
I run the command like below:
C:\>"C:\Program Files\Java\jdk1.7.0_10\bin\javac.exe" -cp d:\tmpmulu\ d:\tmpmulu\Tj.java
It works. The command set the classpath and compiled the file.
But when I change the command to use . instead of d:\tmpmulu\ as my classpath:
C:\>"C:\Program Files\Java\jdk1.7.0_10\bin\javac.exe" -cp . d:\tmpmulu\Tj.java
It's also OK.
That confused me. The . means the current path, it should be c:\. How did it run successfully?
Another question is command like below:
C:\>"C:\Program Files\Java\jdk1.7.0_10\bin\javac.exe" -cp d:\tmpmulu\ Tj.java
As my thinking, the classpath is set to d:\tmpmulu\, it should find the Tj.java file. But the result is 'file not found Tj.java'.
Can anyone tell me details?
Well, presumably your code doesn't rely on having anything in the classpath, basically. If it only uses classes from the JDK, that's absolutely fine.
Note that the classpath is only used to find class files - not source code. That explains both the lack of failure when your source path isn't on the classpath, and then failure when you try to use the classpath to locate Tj.java.
First, if you're compiling with javac -cp . myClass.java, you can omit it the -cp . completely, since it's the default classpath.
Second, -cp should be used when you need to specify references to additional JARs file not included in the standard JDK library, such as a jdbc connector. If your class doesn't require any additional library, then it doesn't really matter what directory you tell javac to look into.
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 Mojave…
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