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ksh case statement issue

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# 1  
Old 02-02-2017
ksh case statement issue
Hi. I wrote the following case statement to replace a series of 'ELIF' statements as it looks better and is easier to maintain. However, for some reason the commands don't fully work in this format. Take option 1. It should call a script that runs in the background but it doesn't work. Can anyone see what I'm doing wrong?
Many thanks.

Code:
case ${_OPTION} in
                  1) print "Running option 1: $file\n"; $(nohup ${BASE}/run_test ${file} > /dev/null 2>&1 &) ; _RC=$? ; [[ $_RC -ne 0 ]] && exit 1
                        ;;
                  2) print "Running option 2: $file\n"; ${BASE}/run_test ${file}; _RC=$?; [[ $_RC -ne 0 ]] && exit 1
                        ;;
                  3) print "Running option3"
                        ;;
                  *)
                        print "[ERROR] Invalid number specified. Exiting ...\n "; exit 1
                esac

# 2  
Old 02-02-2017
Code:
1) print "Running option 1: $file\n"; { nohup ${BASE}/run_test ${file} > /dev/null 2>&1 ; } & ; _RC=$? ; [[ $_RC -ne 0 ]] && exit 1";";";

This User Gave Thanks to rdrtx1 For This Post:
user052009
# 3  
Old 02-02-2017
... or perhaps this:-
Code:
                  1) print "Running option 1: $file\n"; (nohup ${BASE}/run_test ${file} > /dev/null 2>&1) & _RC=$? ; [[ $_RC -ne 0 ]] && exit 1

I've dropped the leading $ from $(nohup and changed the trailing &) ; to be ) &


I assume that you just want to know that the task launched in the background successfully, not that the task finished successfully.



Robin
These 2 Users Gave Thanks to rbatte1 For This Post:
drl user052009
# 4  
Old 02-02-2017
Thanks guys. Both methods worked!
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