How can I execute .jar file in folder from the Windows cmd without specifying its name.
I have tried below command (as there is only 1 jar in that folder I used *) but its not working(Error: Unable to access jarfile *.jar
).
java -jar *.jar
I am not sure it would be a good idea to just run everything in a directory, but you could:
FOR %A IN ("*.jar") DO (java -jar "%~A")
So what you appear to be asking is how to run the command
% java -jar somelongname.jar
as
% java -jar *.jar
to avoid some typing. That's not possible, because neither the Windows CMD shell or the java command is going to expand the *.jar wildcard pattern.
(On Linux / Unix / MacOS, the shell does wildcard expansion before passing arguments to a command. On Windows, it is the responsibility of the command to do this. In practice, it appears that the java command only expands wildcards in the arguments that are going to be passed to your application; see Stop expanding wildcard symbols in command line arguments to Java)
So if you want to avoid typing those pesky characters on Windows, you will need to do something like:
write a simple BAT file to run "java -jar somelongname.jar", or
write a clever BAT file to identify and run a JAR file that matches "*.jar", or
use Powershell.
For what it is worth, I think what you are trying to do is rather dangerous. This is a bit like typing "di*" to run the "dir". What if there is some other more dangerous command on the PATH that is going to match instead of "dir"?
I have the following text in a BAT files so I java program in windows. I was wondering how you can do this in linux.
File 1:
"C:\Program Files (x86)\Java\jdk1.6.0_23\bin\javac.exe" -sourcepath src -classpath bin;deps\jml-1.0b3-full.jar;deps\mail.jar -d bin src/*.java
File 2:
"C:\Program Files (x86)\Java\jdk1.6.0_23\bin\java.exe" -Xmx1536m -classpath bin;deps\jml-1.0b3-full.jar;deps\mail.jar HelloWorld
Id really appreciate it if someone would convert those to linux commands.
Thanks :D
get rid of the quotations - spaces in linux are not a usual practice, so the quotes are not needed
change ; to : as classpath separator
change slashes to /
make the file .sh
that should be it.
The options to java and javac are the same on all platforms. The only things that will change are the file path, for eg. deps\mail.jar might become deps/mail.jar; and the classpath separator, : (colon) instead of ; (semicolon).
check if java SDK (java-1.6.0-openjdk-devel or Sun) exists and in the PATH
which javac
Next put the same options
javac -sourcepath ..
I have a java program that I would like to be able to run from anywhere on my machine. I would like to run it from my Cygwin command prompt. I've made scripts to call the java program. I added the location of the java program to the classpath, and the scripts work when I run them from the java program's directory. However, when I try to run from any other directory, I get:
java.lang.NoClassDefFoundError: commandprogram/CommandProgram
This is my script:
#!/bin/sh
CWD=`dirname "$0"`
java -cp "$CWD/classes;$CWD/lib/AJarFile.jar" commandprogram/CommandProgram
Changing the java line to the following:
java -cp "$CWD/classes;$CWD/classes/commandprogram;$CWD/lib/AJarFile.jar" CommandProgram
produces the same results.
add your directory to classpath example:
java -classpath commandprogram CommandProgram
or
java -classpath directory_to_program Program
After trying just about everything I could think of, I echoed out the command and saw that there was mixing of Cygwin paths and Windows paths. The solution was to change the script to:
#!/bin/sh
CWD=`dirname "$0"`
CWD=`cygpath -w "$CWD"`
java -cp "$CWD/classes;$CWD/lib/AJarFile.jar" commandprogram/CommandProgram
Then CWD changed to "C:\Program Files\..." instead of "/cygdrive/c/Program\ Files/..."
I had previously encountered this problem and solved it with the cygpath -w solution, but then changed my script slightly and didn't notice that the path problem came back.
you have to use a dot to separate packages, not a slash.
java -cp "$CWD/classes;$CWD/lib/AJarFile.jar" commandprogram.CommandProgram
The usual way of running a java file is to save it in the Java/Bin folder and Run cmd
C:\Program Files\Java\jdk1.7.0_05\bin> javac filename.java && java classname
If you save the file in different directory such as D:, you can use the following on the cmd prompt:
D:\Project java> set path=%path%;C:Program Files\Java\jdk1.7.0_05\bin
I want to distribute a command-line application written in Java on Windows.
My application is distributed as a zip file, which has a lib directory entry which has the .jar files needed for invoking my main class. Currently, for Unix environments, I have a shell script which invokes the java command with a CLASSPATH created by appending all files in lib directory.
How do I write a .BAT file with similar functionality? What is the equivalent of find Unix command in Windows world?
You want to use the for loop in Batch script
#echo off
setLocal EnableDelayedExpansion
set CLASSPATH="
for /R ./lib %%a in (*.jar) do (
set CLASSPATH=!CLASSPATH!;%%a
)
set CLASSPATH=!CLASSPATH!"
echo !CLASSPATH!
This really helped me when I was looking for a batch script to iterate through all the files in a directory, it's about deleting files but it's very useful.
One-line batch script to delete empty directories
To be honest, use Jon's answer though, far better if all the files are in one directory, this might help you out at another time though.
Why would you use find? Presumably you know all the libraries your jar file needs ahead of time, so why not just list them?
Alternatively, you could always use -Djava.ext.dirs=lib and let it pick up everything that way.
Since 6.0, Java supports wildcard classpaths.
Java command-line classpath syntax
A variation if you want the CLASSPATH setting to hold outside of the script and you didn't start the shell with /V, you can do something like this:
#echo off
FOR /R ./lib %%a in (*.jar) DO CALL :AddToPath %%a
ECHO %CLASSPATH%
GOTO :EOF
:AddToPath
SET CLASSPATH=%1;%CLASSPATH%
GOTO :EOF
Useful for setting up your environment when you are just playing around on the command line.
Sounds like an executable JAR could work for you. If you're distributing a ZIP with all the JARs your main routine needs, and you really execute it in a command shell, perhaps you could create an executable JAR with the Main-Class and Classpath defined in the manifest. All your users have to do is double click on the JAR and Bob's your uncle.
SET CLASSPATH=""
FOR /R /lib %%a in (*.jar) DO CALL :AddToPath %%a
echo %CLASSPATH%
java -cp %CLASSPATH% com.a.b.MyClass
pause
:AddToPath
SET CLASSPATH=%1;%CLASSPATH%
GOTO :EOF
What I do when I release command-line JARs for Windows/Linux is to embed all the JAR libraries inside my JAR using the ANT command:
<fileset dir="${module.classes.dir}"/>
<zipfileset src="${endorsed.lib.dir}/myLibrary.jar"/>
In this way the libraries are melt together with your class files.
It is not the best way of binary reusability, but if the libraries are not so heavy, you have your application executed by simply calling java -jar myApp.jar in any OS.
You may use a bit different way if you use Maven to build your project.
Application Assembler Maven Plugin is intended for creating Java apps launchers. This plug-in generate bat file launcher for you and put dependencies in specified folder. Optionally you launcher is able to register you application as a service\demon.
This is a topic about using it
This plug-in helps keeping you CM parts of your project DRY.
Is there an easier way to do this?
Yes;
Since version 6, you can use class path wildcards.
javac -cp .;../lib/* yourJavaCodeDir/YourJavaFile.java
See also this related Q&A.
An option to make things portable is to set the classpath relative to the location of the batch file itself.
In Windows (assuming Java 6+, classes in ./bin, jars in ./lib):
#java -classpath %~dp0/bin;%~dp0/lib/* ClassName %1
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