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I am trying to run .jar file for my java code from a .sh shell script file. the jar file name contains "." which is making the Cygwin terminal think it is a directory. Here is the command and the results:
java -jar ./lib/javax.json-1.0.jar
Result:
no main manifest attribute, in lib\javax.json-1.0.jar
Then:
error: package javax.json does not exist
import javax.json.Json;
With this mark ^ below the period (right after javax).
How can I solve it? I am working on Windows 10. Thanks!
EDIT:
I have written many forms of the .sh file to get it run, but it won't run. The current one is:
# !bin/bash
java -jar ./lib/javax.json-1.0.jar
java -jar ./lib/javax.json-api-1.0.jar
javac ./src/TimeTester.java
java TimeTester
Does this look good?
I am getting the following error:
.\src\TimeTester.java:22: error: package javax.json does not exist
import javax.json.Json; (With this ^ below the '.')
AND:
.\src\TimeTester.java:159: error: cannot find symbol
private static JsonObject getJsonFromString(String jsonStr){
And many similar lines in the error.. Any help?
EDIT 2:
This is my current file:
javac -cp ./lib/javax.json-1.0.jar:./lib/javax.json-api-1.0.jar ./src/TimeTester.java
java -cp ./lib/javax.json-1.0.jar:./lib/javax.json-api-1.0.jar:./src TimeTester
But I am getting:
.\src\TimeTester.java:22: error: package javax.json does not exist
import javax.json.Json;
^
With With this (^) under the last dot (.Json)
EDIT 3:
The current .sh file is:
#!/usr/bin/env bash
cd src
javac -cp '../lib/javax.json-1.0.jar;../lib/javax.json-api-1.0.jar' TimeTester.java
java -cp '../lib/javax.json-1.0.jar;../lib/javax.json-api-1.0.jar' TimeTester
The first command (javac) works and generates the .class file. BUT, the second command (java) does not work and it gives the following error:
Error: Could not find or load main class TimeTester
Your help is really appreciated!
Final EDIT:
Thanks for Jim, the shell script now works. Now I got a java execution error:
java.io.FileNotFoundException: .\in_input\in.txt (The system cannot find the path specified)
Thanks
TL;DR It is a pain to use Cygwin with programs written for Windows because of the conflicting command-line shell conventions between bash and cmd.exe. To compile and run Java programs it is much better to use an IDE such as Eclipse or Netbeans.
However, if you must...
None of this works because you are trying to pass Linux-style paths to the Windows JVM. However you seem to have a more basic misunderstanding:
# !bin/bash
java -jar ./lib/javax.json-1.0.jar
java -jar ./lib/javax.json-api-1.0.jar
javac ./src/TimeTester.java
java TimeTester
I am surmising that you think the first two statements make the libraries available to the compiler for the third javac line. This is not true, those two lines attempt to execute the jar file, which of course fails since the jar does not contain a main class
What you should be doing is providing those two library paths as arguments to the -cp option of the javac command.
This is where it gets quite tricky, as you are mixing a Linux-style shell emulator with a Windows JVM. Paths that are intended for the shell must remain in Linux style, while paths that are going to be consumed by the JVM must be converted to Windows format, and path strings for the JVM must be delimited with semicolon (Windows style) instead of colon (Linux style). That introduces a further complication since the semicolon in Cygwin (Linux) is the delimiter for multiple commands on one line, so the path string must be quoted to prevent the semicolon from breaking things.
Also problematic is the naming of the class to be compiled. You have not shown us the package declaration of the Java file, but I'm assuming it's in the default package (i.e. there is no package declaration and it's not package src;). In that case you should be in the src directory, not one directory above.
Finally, once you specify -cp, you must also add the current directory to the classpath on Windows if you want it to be included, otherwise it will not find your newly-compiled .class file.
So the compile and execute commands should be
javac -cp '../lib/javax.json-1.0.jar;../lib/javax.json-api-1.0.jar' TimeTester.java
java -cp '.;../lib/javax.json-1.0.jar;../lib/javax.json-api-1.0.jar' TimeTester
For simple relative paths the Windows JVM will accept forward slashes, but if you have absolute Linux paths (i.e. /cygdrive/c/..., or with the cygdrive path set to /, paths like /c/user/...) the JVM will not understand them and they will need to be translated using cygpath.
None of your 4 commands work:
java -jar ./lib/javax.json-1.0.jar does not work because javax.json-1.0.jar is not an executable jar file.
java -jar ./lib/javax.json-api-1.0.jar does not work because javax.json-api-1.0.jar is not an executable jar file.
javac ./src/TimeTester.java does not work because your class requires classes from the javax.json package to be on the classpath, and you haven't set the classpath. Classes from the javax.json package are found in the javax.json-1.0.jar file.
java TimeTester does not work because the compilation failed.
To fix all that, remove the first two lines, and specify the classpath on the other two lines, e.g.
javac -cp ./lib/javax.json-1.0.jar:./lib/javax.json-api-1.0.jar ./src/TimeTester.java
java -cp ./lib/javax.json-1.0.jar:./lib/javax.json-api-1.0.jar:./src TimeTester
Notice that you also had to list ./src on the classpath when executing your program.
I haven't been able to find a good answer to this yet, but when using javac, I find that for classpath I can't seem to get it to work using a relative path to the current working directory, although it seems to work for the target directory and source file location. Here is what we're currently using:
javac java/src/com/<our-company>/ecommerce/<vendor-tool>/executor/*.java java/src/com/<our-company>/ecommerce/<vendor-tool>/util/*.java -d ./target/java/bin -cp /c/Git/<our-api>/java/lib/some.jar:/c/Git/<our-api>/java/lib/another.jar -verbose
And we would like to use something like this:
javac java/src/com/<our-company>/ecommerce/<vendor-tool>/executor/*.java java/src/com/<our-company>/ecommerce/<vendor-tool>/util/*.java -d ./target/java/bin -cp java/lib/some.jar:java/lib/another.jar -verbose
or perhaps:
javac java/src/com/<our-company>/ecommerce/<vendor-tool>/executor/*.java java/src/com/<our-company>/ecommerce/<vendor-tool>/util/*.java -d ./target/java/bin -cp ./java/lib/some.jar:./java/lib/another.jar -verbose
The first example seems to work, but the other two get an error that the packages within those jars cannot be found. Since we are using Git, I would prefer the whole build path be relative to allow for wherever the individual's Git repo is located locally--as well as be flexible concerning which operating system they may be using (this is being run in a shell which should operate the same for each of the platforms).
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
I have to get some kinks out of a shell script for work, and one of the line looks like this:
-cp: this is the classpath
This is the set of classes that are used when running a specific class.
In your example; OrganT.Tune.Mix OrganT must be a class in the classpath (in this case, inside the OrganT.jar
Read the documentation, can be found here
Just a hint - under linux and mac you can use the
man <command goes here>
comman in the terminal/shell to display all parameters and usage information available for the specific command.
-cp stands for classpath. The CLASSPATH variable is one way to tell applications, including the JDK tools, where to look for user classes.
java -classpath .;YourJarFile.jar
I think you want to run a script for including the class path and execute the jar.
To do this in any text editor type java -jar YourJarFile.jar and save it, with extention (anyName.sh) assuming you have got linux flavour. Make it executable using the command chmod 775 anyName.sh
For windows type java -jar YourJarFile.jar, and save it with extention (anyName.bat)
I want to execute my program without using an IDE.
I've created a jar file and an exectuable jar file. When
I double click the exe jar file, nothing happens, and when I try to use the command in cmd it gives me this:
Error: Unable to access jarfile <path>
I use the command: java -jar Calculator.jar
How I created the jar:
Right click on project folder (Calculator)
Select
Click on Java Folder and select "Exectuable Jar File", then select next
Launch Configuration: Main - Calculator
Create Export Destination
Hit "Finish" and profit! Well, not really.
I had encountered this issue when I had run my Jar file as
java -jar TestJar
instead of
java -jar TestJar.jar
Missing the extension .jar also causes this issue.
Fixed
I just placed it in a different folder and it worked.
[Possibly Windows only]
Beware of spaces in the path, even when your jar is in the current working directory. For example, for me this was failing:
java -jar myjar.jar
I was able to fix this by givng the full, quoted path to the jar:
java -jar "%~dp0\myjar.jar"
Credit goes to this answer for setting me on the right path....
I had this issue under CygWin in Windows. I have read elsewhere that Java does not understand the CygWin paths (/cygdrive/c/some/dir instead of C:\some\dir) - so I used a relative path instead: ../../some/dir/sbt-launch.jar.
I had the same issue when trying to launch the jar file. The path contained a space, so I had to place quotes around. Instead of:
java -jar C:\Path to File\myJar.jar
i had to write
java -jar "C:\Path to File\myJar.jar"
Just came across the same problem trying to make a bad USB...
I tried to run this command in admin cmd
java -jar c:\fw\ducky\duckencode.jar -I c:\fw\ducky\HelloWorld.txt -o c:\fw\ducky\inject.bin
But got this error:
Error: unable to access jarfile c:\fw\ducky\duckencode.jar
Solution
1st step
Right click the jarfile in question. Click properties.
Click the unblock tab in bottom right corner.
The file was blocked, because it was downloaded and not created on my PC.
2nd step
In the cmd I changed the directory to where the jar file is located.
cd C:\fw\ducky\
Then I typed dir and saw the file was named duckencode.jar.jar
So in cmd I changed the original command to reference the file with .jar.jar
java -jar c:\fw\ducky\duckencode.jar.jar -I c:\fw\ducky\HelloWorld.txt -o c:\fw\ducky\inject.bin
That command executed without error messages and the inject.bin I was trying to create was now located in the directory.
Hope this helps.
None of the provided answers worked for me on macOS 11 Big Sur. The problem turned out to be that programs require special permission to access the Desktop, Documents, and Downloads folders, and Java breaks both the exception for directly opened files and the permission request popup.
Fixes:
Move the .jar into a folder that isn’t (and isn’t under) Documents, Desktop, or Downloads.
Manually grant the permission. Go to System Preferences → Security and Privacy → Privacy → Files and Folders → java, and check the appropriate folders.
I had a similar problem and I even tried running my CMD with administrator rights, but it did not solve the problem.
The basic thing is to make sure to change the Directory in cmd to the current directory where your jar file is.
Do the following steps:
Copy jar file to Desktop.
Run CMD
Type command cd desktop
Then type java -jar filename.jar
This should work.
Edit: From JDK-11 onwards ( JEP 330: Launch Single-File Source-Code Programs )
Since Java 11, java command line tool has been able to run a single-file source-code directly. e.g.
java filename.java
If you are using OSX, downloaded files are tagged with a security flag that prevents unsigned applications from running.
to check this you can view extended attributes on the file
$ ls -l#
-rw-r--r--# 1 dave staff 17663235 13 Oct 11:08 server-0.28.2-java8.jar
com.apple.metadata:kMDItemWhereFroms 619
com.apple.quarantine 68
You can then clear the attributes with
xattr -c file.jar
It can also happen if you don't properly supply your list of parameters. Here's what I was doing:
java -jar test#gmail.com testing_subject file.txt test_send_emails.jar
Instead of the correct version:
java -jar test_send_emails.jar test#gmail.com testing_subject file.txt
This worked for me.
cd /path/to/the/jar/
java -jar ./Calculator.jar
For me it happens if you use native Polish chars in foldername that is in the PATH.
So maybe using untypical chars was the reason of the problem.
sometime it happens when you try to (run or create) a .jar file under /libs folder by right click it in android studio. you can select the dropdown in top of android stuio and change it to app. This will work
My particular issue was caused because I was working with directories that involved symbolic links (shortcuts). Consequently, trying java -jar ../../myJar.jar didn't work because I wasn't where I thought I was.
Disregarding relative file paths fixed it right up.
In my case the suggested file name to be used was jarFile*.jar in the command line. The file in the folder was jarFile-1.2.3.jar . So I renamed the file to jarFile. Then I used jarFile.jar instead of jarFile*.jar and then the problem got resolved
It can happen on a windows machine when you have spaces in the names of the folder. The solution would be to enter the path between " ".
For example:
java -jar c:\my folder\x.jar -->
java -jar "c:\my folder\x.jar"
To avoid any permission issues, try to run it as administrator. This worked for me on Win10.
I know this thread is years ago and issue was fixed too. But I hope this would helps someone else in future since I've encountered some similar issues while I tried to install Oracle WebLogic 12c and Oracle OFR in which its installer is in .jar format. For mine case, it was either didn't wrap the JDK directory in quotes or simply typo.
Run Command Prompt as administrator and execute the command in this format. Double check the sentence if there is typo.
"C:\Program Files\Java\jdk1.xxxxx\bin\java" -jar C:\Users\xxx\Downloads\xxx.jar
If it shows something like JRE 1.xxx is not a valid JDK Java Home, make sure the System variables for JAVA_HOME in Environment Variables is pointing to the correct JDK directory. JDK 1.8 or above is recommended (2018).
A useful thread here, you may refer it: Why its showing your JDK c:program files\java\jre7 is not a valid JDK while instaling weblogic server?
For me it happen because i run it with default java version (7) and not with compiled java version (8) used to create this jar.
So i used:
%Java8_64%\bin\java -jar myjar.jar
Instead of java 7 version:
java -jar myjar.jar
I had a similar problem where TextMate or something replaced the double quotes with the unicode double quotes.
Changing my SELENIUM_SERVER_JAR from the unicode double quotes to regular double quotes and that solved my problem.
this is because you are looking for the file in the wrong path
1. look for the path of the folder where you placed the file
2. change the directory cd in cmd use the right path
I use NetBeans and had the same issue. After I ran build and clean project my program was executable. The Java documentation says that the build/clean command is for rebuilding the project from scratch basically and removing any past compiles. I hope this helps. Also, I'd read the documentation. Oracle has NetBeans and Java learning trails. Very helpful. Good luck!
Maybe you have specified the wrong version of your jar.
I finally pasted my jar file into the same folder as my JDK so I didn't have to include the paths. I also had to open the command prompt as an admin.
Right click Command Prompt and "Run as administrator"
Navigate to the directory where you saved your jdk to
In the command prompt type: java.exe -jar <jar file name>.jar
Keep the file in same directory where you are extracting it. That worked for me.
This is permission issue, see if the directory is under your User.
That's why is working in another folder!
Rename the jar file and try
Explanation :
yes, I know there are many answers still I want to add one point here which I faced.
I built the jar and I moved it into the server where I deploy (This is the normal process)
here the file name which I moved already existed in the server, here the file will override obviously right. In this case, I faced this issue.
maybe at the time of overriding there can be a permission copy issue.
Hope this will help someone.
Have you tried to run it under administrator privoleges?
meaning, running the command in "Run As" and then select administrator with proper admin credentials
worked for me
I was trying this:
After giving the file read, write, execute priviledges:
chmod 777 java-repl.jar
alias jr="java -jar $HOME/Dev/java-repl/java-repl.jar"
Unable to access bla bla..., this was on Mac OS though
So I tried this:
alias jr="cd $HOME/Dev/java-repl/ && java -jar java-repl.jar"
This did not work "Unable to access jarfile"
"C:\Program Files\java\jdk-13+33-jre\bin\javaw.exe" -jar "C:\Program Files\Maxim Integrated Products\1-Wire Drivers x64\ OneWireViewer.jar"
This does work
"C:\Program Files\java\jdk-13+33-jre\bin\javaw.exe" -jar "C:\Program Files\Maxim Integrated Products\1-Wire Drivers x64\OneWireViewer.jar"
The difference is the single space in front of OneWireViewer.jar not withstanding that it is surrounded with quotes and even has other spaces.