UNIX for Dummies Questions & Answers

We have a batch Unix process that runs during the day and it is getting an exit code 11 from Unix. It finishes a sqlplus step and gets the exit code 11 before it starts the next step. This used to happen once a year and now is happening more often (but not every time the process runs). We have investigated other processes that are running at that time but have not found anything. Any ideas?

Well, a few clues would be useful here!Most importantly, what have you tried so far?


You have effectively asked me "Why has my car broken down" with no symptoms other that "It's stopped"

You will need to give us more to get something useful back.



Robin

I sense a slight misunderstanding here and before we go further i want to clear it up: "exit codes" are the result of a system call which ends a program and sets a certain exit code. It is common that an exit code of "0" indicates success and any non-zero exit code describes some or the other error condition. What this error condition might be is not standardized, so one program might give back "2" for "for not found" and another for something different.

The bottom line here is: not "UNIX gives back an exit code" but the application sets it itself upon exiting. UNIX just reports it. You could easily write a shell script yourself which sets any arbitrary exit code by using the "exit" keyword with some (signed short short) integer:

Code:

#!/bin/sh

i=125

# ... arbitrary code ....

exit $i

Run this and query the exit code:

Code:

# /path/to/script ; echo $?
125

You could enter any value -127-+127 and get the respective result. Therefore you need to consult the documentation of your application and find out what exit code "11" means. This depends entirely on the application you use and could be anything.

I hope this helps.

bakunin