I never used to have this problem and when I went to run Java programs (in a terminal) I would simply say,
java className
and the program would run. But lately when I try this I get the error,
could not find or load main class className
To fix this I've been adding -cp ./ my curring working directory to the classpath and then the program runs. I'm not sure what to add to my environment variables so that the current working directory will always be on the class path thus avoiding this longer script,
java -cp ./ className
According to this StackOverflow post the standard implementation from Oracle looks in the current working directory for the class name first.
(I am using Java 8 and Windows 8.1)
The classpath, if not found on the command line, and not found in the CLASSPATH environment variable, defaults to the current directory.
-cp classpath
Specifies a list of directories, JAR files, and ZIP archives to search for class files. Separate class path entries with semicolons (;)
Specifying -classpath or -cp overrides any setting of the CLASSPATH environment variable.
If -classpath and -cp are not used and CLASSPATH is not set, then the user class path consists of the current directory (.).
If you specify -cp, then you must explicitly include . (the current directory) in the classpath, or it will not be included.
If not using -cp doesn't work, but explicitly stating -cp . does work, then you must have the CLASSPATH environment variable defined, and it must not explicitly include .. That would explain why overriding CLASSPATH by including . with the -cp option works.
I'm trying to run Java program using cmd.
When I compile that file using
javac helloworld.java
It works perfectly but when I try to run that file using
java helloworld
I get an error:
couldn't find or load main class.
even though my I put my javac path in class path in system variables and javac working correctly.
After searching on how I can fix it I found that I can use
java -cp . helloworld
because it let you to find .class file.
and it works but I know that I can run java program without this -cp so what this for and how I can run my program without it?
-cp specifies the Java classpath for the JRE you are attempting to start. Look for an environment variable CLASSPATH and add '.'.
-cp is used to set the classpath for the jar file, this flag is same as importing a jar file to eclipse and then use it.
If you want to run without this flag, use should set the classpath first beforing running.
export CLASSPATH=path/to/your/jarfile.jar
If you already have some classpath set
export CLASSPATH=$CLASSPATH:path/to/your/jarfile.jar
If you want to include current directory
export CLASSPATH=$CLASSPATH:path/to/your/jarfile.jar:.
As other have mentioned, you can set the CLASSPATH. However, a much better approach is to bundle the .class files in a JAR ans user java -jar nameofthejar.jar.
I created a Java project to call a Web service.
It has one Main java file and another class file.
I have used some jar files for HTTP client.
In Eclipse it runs fine.
I need to run the Java program in command prompt by passing some arguments.
In command prompt I went to src folder containing main java and sub class java file and gave the following command
javac mainjava.java
I'm getting following error
mainjava.java:14: cannot find symbol
symbol : class SubClass
here SubClass is my another java class file used to call the web service.
How to run the program by passing arguments?
javac is the Java compiler. java is the JVM and what you use to execute a Java program. You do not execute .java files, they are just source files.
Presumably there is .jar somewhere (or a directory containing .class files) that is the product of building it in Eclipse:
java/src/com/mypackage/Main.java
java/classes/com/mypackage/Main.class
java/lib/mypackage.jar
From directory java execute:
java -cp lib/mypackage.jar Main arg1 arg2
A very general command prompt how to for java is
javac mainjava.java
java mainjava
You'll very often see people doing
javac *.java
java mainjava
As for the subclass problem that's probably occurring because a path is missing from your class path, the -c flag I believe is used to set that.
You can use javac *.java command to compile all you java sources. Also you should learn a little about classpath because it seems that you should set appropriate classpath for succesful compilation (because your IDE use some libraries for building WebService clients). Also I can recommend you to check wich command your IDE use to build your project.
All you need to do is:
Build the mainjava class using the class path if any (optional)
javac *.java [ -cp "wb.jar;"]
Create Manifest.txt file with content is:
Main-Class: mainjava
Package the jar file for mainjava class
jar cfm mainjava.jar Manifest.txt *.class
Then you can run this .jar file from cmd with class path (optional) and put arguments for it.
java [-cp "wb.jar;"] mainjava arg0 arg1
HTH.
javac only compiles the code. You need to use java command to run the code. The error is because your classpath doesn't contain the class Subclass iwhen you tried to compile it. you need to add them with the -cp variable in javac command
java -cp classpath-entries mainjava arg1 arg2 should run your code with 2 arguments
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