Propagate exist status from a child shell script to a parent.
Hi All,
I have a parent shell script A and a child shell script B.
1). If a command i.e a mysqdump fails in shell script B fails then I trap the error with this code
Code:
if [[ $?>0 ]]
then func_exit "Failed to export the cleaned DB1.${MYDBNAME} database to the ${MYARCHIVEDIR} directory"
else
func_logwriter "Successfully exported the cleaned DB1.${MYDBNAME} database to the ${MYARCHIVEDIR} directory"
fi
the func_exit writes various messages to log and calls exit 1
2). The parent script A called script B and also monitors the return status in the same way as above. However, the func_exit possess the following code:
For some reason the func_exit in script A is not being called on a failure in script B. Hence I don't get an email.
Please could anyone advise if the exit status propagation between the parent and child script is correct. I am using Debian.
Many thanks for your help.
Where is the wait statement in the parent process? I do not see one.
The parent creates the child, then (immediately or a little later) waits for the child. That is how the parent process knows about the child's exit status.
Code:
# parent example
child. sh &
# using your func_exit()
# this line waits for the child pid ( $! variable) then calls func_exit if error
wait $! || func_exit
I have requirement where I need to fail parent if any one of the child process fails. Here is the code snippet
for i in 1 2 3 4 5 6 7 8 9 10
do
child_script $i &
done
wait
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