What does -cp mean in the terminal when running a .jar file? - java

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)

Related

Javac command in cmd

I have problems compiling Java programs from command line interfaces (both command prompt and NetBeans terminal).
I added the jdk directory to the PATH system variable but not all commands are recognizable, some commands (in the jdk1.6.0_37\bin folder) are recognized (like: java, javaw, packger)and others I'd have to type the full directory name for it to work (like: jar, javac, javah, javap) (this applies for both cmd and NB).
I don't know why this should be, all of these files are .exe , all are Java Platform SE binary, the only difference I can see is the icon of the files, the ones that work have the Java logo (Coffee Mug) and the rest have the standard .exe logo. Any help would be appreciated. Thanks
You have to add the jdk1.6.0_37\bin directory to the PATH variable. That is where all the commands that you want to run are in. After you add the directory to the PATH variable, open a new command prompt and run those commands again.
I'm searched many answers that suggest me to type in cmd:
set path = "%path%;c:program files\java\jdk1.7.0\bin"
but this is WRONG!
the right solution is that you leave "set" and just type
path = %path%;c:program files\java\jdk1.7.0\bin
P/s: of course you have to replace "jdk1.7.0" folder by your current java version folder

Can you run a Java program in Linux using the name only, without the "java" command?

If I am writing HelloWorld, is there a way I can run the program from any directory by just typing HelloWorld? Sort of the same way once you set up Ant, you can just run Ant from any directory?
Just for some details, we are creating a CLI based toolkit for a customer, and just curious if we can compile it, install it, and just have them run it using the toolkit name.
You can always create a shell script, call it HelloWorld and make it run java with your JAR.
You'll then need to chmod the script to make it executable, and place it somewhere in your $PATH.
The script would like something like:
#!/bin/bash
cd /path/to/helloworld
java -jar HelloWorld.jar "$#"
or
#!/bin/bash
java -jar /path/to/helloworld/HelloWorld.jar "$#"
depending on your exact requirements.
Common solution for your problem is to create a separate launcher application, which is non-java application that runs your Java program. Launcher can be written in some compilable language such as C/C++ and compiled into native executable. Also it can be written in some interpreted language such as Unix shell, perl, python etc and made executable by adding #!/path/to/interpreter line at the beginning of launcher file and setting executable flag on it. Also there are several utilities that can generate launcher for your program such as launch4j or jsmooth.
On Linux (specifically), you could use the /proc filesystem (see proc(5) man page) and its binfmt_misc (actually the /proc/sys/fs/binfmt_misc/register pseudo-file and other pseudofiles under /proc/sys/fs/binfmt_misc/) to register java as the handler for .class or .jar files. Read the Documentation/binfmt_misc.txt file in the kernel source for gory details.
Then any executable .jar file would be interpreted by java (or jexec)
I'm not sure it is worth the effort. I find that wrapping your Java program in some shell script is much more easy (and more portable, because few Linux systems actually use binfmt_misc, and your customer may need some sysadmin skills to enable it).

What causes "Unable to access jarfile" error?

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.

Deploying a Java project on Linux system

I developed a project using Java and now I've to deliver it to client who is using Linux. Which executable file format will be delivered and how to make that?
Executable file format?
If you're delivering a Java app, give them a jar file (and associated libs).
Provide a shell script to set up its environment and execute it.
For example, assuming I define ROOT_DIR as my app's install directory, and so on:
CLASSPATH="${ADD_JARS}:${CLASSPATH}:${ROOT_DIR}/lib/myApp.jar:\
${ROOT_DIR}/lib/jibx/jibx-run.jar:\
${ROOT_DIR}/lib/jibx/xpp3.jar:\
${ROOT_DIR}/lib/bindings.jar:\
${ROOT_DIR}/lib/commons-lang-2.0.jar:\
${ROOT_DIR}/lib/forms-1.0.5.jar"
"${JAVACMD}" -Xmx256M -DanyDefsNeeded=foobar -Dbase.dir="${ROOT_DIR}" -cp "${CLASSPATH}" myApp.main.Launcher "$#"
What goes into the shell script depends totally on what your app actually needs to start up.
A jar. If it is not executable, then a script (.sh) to launch the jar.
Well basically what you wanna put in a .sh file is the commands you'd normally type at the console to run your jar file. They should be separated by a new line (i.e. each on a separate line in the .sh file).
The most basic you can go is add something like this to your sh file:
java -Xms=64m -Xmx=256m -jar myJar.jar -classpath [dependencies dir]/dep1.jar,[dependencies dir]/dep2.jar
beyond this you can do more exotic stuff, like parametrise some environment variables, get command line argumens from when the .sh is launched and pass them to the jar executatble etc. Look up "bash scripting" for advanced stuff:
http://tldp.org/HOWTO/Bash-Prog-Intro-HOWTO-2.html#ss2.1
You might have better luck using Launch4J, IzPack or other installer that has cross-platform capabilities. This might be a better first option than trying to understand the intricacies and idiosyncrasies of the different Linux distributions and shells.
If your app. has a GUI, the best user experience for installation/deployment can be had by way of Java Web Start. Note that JWS can deploy apps. to Windows, *nix and Mac. and avoids all the maintenance woes of generating 3 separate (platform specific) executables.

Can an executable .jar file be called without having to use its full path?

I have a .jar file that I would like to be able to call without having to use a full file path to its location.
For example, if the .jar file is located at: /some/path/to/thearchive.jar
I'd like to be able to run it with:
java -jar thearchive.jar
instead of:
java -jar /some/path/to/thearchive.jar
when I'm elsewhere in the directory tree. In my specific case, I'm running a Mac with OS X 10.5.7 installed. Java version "1.5.0_16". I tried adding "/some/path/to" to PATH, JAVA_HOME and CLASSPATH, but that didn't work.
So, how do I setup to run a .jar from the command line without having to use its full path?
UPDATE: Another item to deal with would be arguments. For example:
java -jar /some/path/to/thearchive.jar arg1 arg2
This can have an effect on the way the question is dealt with as mentioned in the answers below.
You can add a variable to hold the directory:
export JARDIR=/some/path/to
java -jar $JARDIR/thearchive.jar
I'm not sure you can do it from environment variables implicitly.
No you can't.
Running a jar with -jar does not involve any kind of classpath mechanism since the jar file is the classpath.
Alternatively use a shell alias to launch the jar or a small script file.
According to Sun:
java -jar app.jar
To run the application from jar file that is in other directory, we need to specify the path of that directory as below: java -jar path/app.jar
where path is the directory path at which this app.jar resides.
So either out the path in a "standard" environment variable or define a wrapper which would be in your PATH
I don't believe so. If you have the jar specified in your CLASSPATH you could just call java with the main class specified. (i.e java com.test.Main) Alternatively you could create an alias in you shell to execute the command
alias execJar="java -jar /some/path/to/thearchive.jar"
Or another alternative is to create a wrapper script to execute it.
The Java system itself does not give you a way to specify something like JAR_PATH (a list of places to look for jar files). The other answers given use the MAC/Unix shell capabilities:
Setting an environment variable
Setting an alias
Possibly using a symbolic link (to the file or to the directory).
What might be helpful is to find out why specifying the entire path is a problem. That may guide us as to which answer is best or possibly find a completely different solution to your problem.
To run a .jar file without typing the full path you can put it in your classpath and run it by typing:
java fullclassname arg1 arg2
Mac OSX Developer Library recommends 'additional jar files that need to be placed on the system classpath be placed in the /Library/Java/Extensions folder. You can also put them in your own Library/Java/Extensions folder, but you will probably have to create the Java and Extensions folders.
If you do not know the full name of the main class in your .jar file, you can expand it and look in the MANIFEST.MF file in the META-INF folder. The Main-Class: line will tell you.
So, for example, to run the saxon9he.jar put it in /Library/Java/Extensions and you can type (from whichever folder you want)
java net.sf.saxon.Transform arg1 arg2...
Almost as short as typing java -jar jarfile.jar arg1 arg2, and you don't need to change any environment variables.
In short, if the jar is in your classpath, use the classname and you don't need the pathname.
Since there is no extra command line option for the location of jars or an environment variable is taken into account I am also not aware of an easy solution but would be highly interested in it as well.
A different approach could be to use a zsh wrapper script to get such a behaviour:
~/.scripts/java # .scripts at a prior position in your $PATH variable than java itself
#!/usr/bin/env zsh
# get -jar option and remove from $# (-D option)
zparseopts -D jar:=jarname
if [ -e $JAR_PATH/$jarname[2] ];
then
java -jar $JAR_PATH/$jarname[2] $#
elif [ -e $jarname[2] ];
then
java -jar $jarname[2] $#
else
java $#
fi
An advantage of zparseopts is that it can strip off the -jar option but all other options are retained within $#.
A further improvement would be to extend bash-completion or zsh-completion for the java command option -jar. For instance bash-completion of java -jar restricts file listings to *.jar files. For convenient usage someone could extend this by not only looking into current path but into $JAR_PATH. As a starting point see following unix.sx question.
But this solution doesn't look too good either.

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