I have to create two instances of jBoss 5.1.0 GA. In order to do that I have to execute the following in start-jboss.sh:
find . -exec /opt/novell/idm/jboss/bin/run.sh -Djboss.service.binding.set=ports-01 -c IDMProv -b 0.0.0.0 \; -exec /opt/novell/idm/jboss/bin/run.sh -Djboss.service.binding.set=ports-02 -c reporting_module -b 0.0.0.0 \;
The first instance runs but the second instance is not running.
Any help will be truly appreciated.
What is the exit code from running the 1st instance? (If it is non-zero, the 2nd -exec primary will not be run.)
How many files are in the current directory?
How many times do you want to run these two commands?
It is very strange to run two commands for every file found in a directory hierarchy without giving those commands the name of a file to work on as an operand???
I have tried using the following
find / -name "IDMProv" -type d -exec /opt/novell/idm/jboss/bin/run.sh -Djboss.service.binding.set=ports-01 -c IDMProv -b 0.0.0.0 \;
find / -name "reporting_module" -type d -exec /opt/novell/idm/jboss/bin/run.sh -Djboss.service.binding.set=ports-02 -c reporting_module -b 0.0.0.0 \;
But it is not working either. Only the first instance is starting. Is there any way to execute without using find.
Are you sure you are not missing brasses {} in your -exec syntax?
What you are doing makes absolutely no sense to me!
I repeat: How many hundreds of times do you want to execute these two commands? Please explain in English why you want to run so many identical copies of these commands.
Do these commands need to be given the pathname of a directory as an operand?
What exit code do these commands return when run from the command line without find?