Launch Java with a specific argument each time - java

I have a program which allows me to define the java executable (/usr/bin/java), but does not allow me to add specific arguments to the executable.
I want to be able to run Java with a specific argument each time in order to enable Security Manager.
So far, I have tried to add the argument after /usr/bin/java, so it looks like
java=/usr/bin/java -Djava.security.manager -Djava.security.policy=/home/java.policy
That did not work as the program probably checks to see if a file exists. Another way I tried was to make a bash script called java which contained:
/usr/bin/java -Djava.security.manager -Djava.security.policy=/home/java.policy $*
I then set the java path to /home/java (Location of my script). That however did not work either. Is there some sort of way I can do this?
Thanks.

Put your java call in a shell script java.sh:
#!/bin/bash
/usr/bin/java -Djava.security.manager -Djava.security.policy=/home/java.policy $#
Change permissions with chmod u+x java.sh, then call your program with java=./java.sh (adapt path for your script as needed).
Notes about executable bit and shebang line
Both the shebang line (#!/bin/bash) and execute permission are important here. Without them, system calls of the exec* family will fail because the kernel does not know what to do with the file or because execution is rejected due to the missing executable bit.
This is different when run directly from a shell (./java.sh), because most shells have some compatibility feature for that case so they will run a script in a shell if exec* fails. The execute permission must be set though.
The only case where neither is needed is if you give your script as argument to the shell: bash java.sh.

Your second approach would be ok, but your problem is likely to be that your Multicraft application is not finding your script. More so, it's the environment of your web server (Apache?) that may need to have PATH altered to be able to find your java wrapper script.

Related

Can I use -cp argument twice when executing Java?

I am trying to run a Java application which has many dependencies. In the past I have use the following command to launch the application
java -cp "program.jar:jar1.jar:jar2.jar:jar3.jar:[...]" program
However as the list of dependencies have grown, I have moved them into an arguments file, the contents of this file are:
-cp "\
program.jar:\
jar1.jar:\
jar2.jar:\
jar3.jar:\
[...]"
And I am running the application with
java #arguments-file program
Everything up to this point works fine.
Sometimes I end up with beta versions of program.jar, they share all of the same dependencies, but program.jar is renamed program-beta.jar.
So to run the jar the following command would be used
java -cp "program-beta.jar:jar1.jar:jar2.jar:jar3.jar:[...]" program
or more specifically, I would use an environment variable, so that the same script can be used, and the variable would be set to either program.jar, or program-beta.jar, depending on the circumstance
java -cp "$PROGRAM_JAR:jar1.jar:jar2.jar:jar3.jar:[...]" program
Now that I am using an arguments file I was hoping to be able to be able to do something like:
java -cp "$PROGRAM_JAR" #arguments-file program
However using -cp twice causes one of the two to be ignored, resulting in a java.lang.ClassNotFoundException exception.
Is there any way around this that allows me to specify one jar file by name, but abstract away all of the others so that my java command isn't thousands of characters?
This will be running entirely on Linux, so any command line "magic", such as using grep is fine, so long as the resulting code is easily readable
You could just write two bash scripts production.sh and beta.sh that contain a reference on program.jar and program-beta.jar, respectively.
Also, the classpath can contain wildcards (see man-page), so if you can ensure that on disk exists only one of the two versions, you can write it like this:
java -cp "program*:jar1.jar:jar2.jar:jar3.jar:[...]"
In the long term, you might think about building/running it with Maven or Gradle, depending on your requirements.

Where to put files for a command-line Java program?

I have written a Java program that takes in arguments and then executes. I pass in these arguments from the command line (I am on a Macbook Pro using Terminal, using the bash shell). Let's say the name of my program is prgm. I want to be able to say "prgm " from any directory in the shell and then have that program execute. So I figure I need to write a bash script to reference the Java files and take in arguments, and put that bash script somewhere in my PATH. Where do I put the bash file, and where do I put my Java files? Also, am I right to assume that I only need the .class (binary) Java files?
Step-by-step:
Assuming that the name of the Java executable if myjavaprog.
Assuming that the name of your bash script is myscript.
Make sure myscript is calling myjavaprog using absolute path and the desired arguments.
call echo $PATH and you will see a bunch of paths: /some/path1:/some/other/path2:...
Put your bash script in whatever path you want from the ones returned by echo $PATH.
Go to a random path.
Call you bash script bash myscript. See the execution of myjavaprog.
Tips:
If java program is for personal use only, put it in a path starting with /usr/ or even in your $HOME directory (and add that location to your PATH)
If java program must be shared with other users, put it in an accessible place, so that other users don't need to modify their PATH variable.

Java JAR file does not execute in startup script in Ubuntu 14.04

The following process normally works for my startup scripts. However, when I introduce a command to execute a JAR file, it does not work. This script works while I am logged in. However, it does not work as a startup script.
In /etc/init.d I create a bash script (test.sh) with the following contents:
#!/bin/bash
pw=$(curl http://169.254.169.254/latest/meta-data/instance-id)
pwh=$(/usr/bin/java -jar PWH.jar $pw &)
echo $pwh > test.txt
Make script executable
In /etc/rc.local, I add the following line:
sh /etc/init.d/test.sh
Notes:
I make a reference to the script in /etc/rc.local, because this script needs to run last after all services have started.
Please do not ask me to change the process (i.e., create script in /etc/init.d/ and reference it from /etc/rc.local), because it works for my other startup scripts.
I have tried adding nohup in front of java command, and it still did not work.
Thanks
As written, there is insufficient information to say what is going wrong. There are too many possibilities to enumerate.
Try running those commands one at a time in an interactive shell. The java command is probably writing something to standard error, and that will give you some clues.

Java runtime exec

I am trying to do something using system exec in Java
Runtime.getRuntime().exec(command);
Surprisingly everything that is related with paths, directories and files is not working well
I don't get why and just want to know is there any alternatives?
The alternative is to use the ProcessBuilder class, which has a somewhat cleaner interface, but your main problem is probably related to how the OS processes command lines, and there isn't much Java can do to help you with that.
As noted above, cd is a shell builtin. i.e. it's not an executable. You can determine this using:
$ which cd
cd: shell built-in command
As it's not a standalone executable, Runtime.exec() won't be able to do anything with it.
You may be better off writing a shell script to do the shell-specific stuff (e.g. change the working directory) and then simply execute that shell script using Runtime.exec(). You can set PATH variables etc. within your script and leave Java to simply execute your script.
One thing that catches people out is that you have to consume your script's stdout/stderr (even if you throw it away). If you don't do this properly your process will likely block. See this SO answer for more details.
The exec() method can take three arguments. The third is the directory your subprocess should use as its working directory. That solves your "cd" problem, anyway.

Do commands run from current directory in a shell script?

In a bash shell script I tried these two versions:
java -jar abc.jar&
and
CMD="java -jar abc.jar&"
$CMD
The first verison works, and the second version complains that abc.jar cannot be found. Why?
Commands do run from current directory in a shell script.
This is why the first command in your test script worked.
The second command may not work because either java isn't in your ${PATH} or abc.jar isn't in your ${CLASSPATH}. You could echo these environment variables or set +x to debug your bash script.
Bash (and others) won't let you do backgrounding (&) within the value of a variable (nor will they let you do redirection that way or pipelines). You should avoid putting commands into variables. See BashFAQ/050 for some additional information.
What is the actual error message you're getting? I bet it's something like "abc.jar& not found" (note the ampersand) because the ampersand is seen as a character in the filename.
Also, the current directory for the script is the directory that it is run from - not the directory in which it resides. You should be explicit about the directory that you want to have your file in.
java -jar /path/to/abc.jar&

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