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How to make `trap` know if the EXIT is after successful program finish or because of premature as an error or something

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2

1

PROBLEM:

I have a shell program that I have been writing but I can't find out how to make sure that trap is trapping for cleanup at the end or because of a error in some command, it cleans up either way.

Here is the code:

################################### Successful exit then this cleanup ###########################################################3

successfulExit()

###############################################################################################################################33
####### Catch the program on successful exit and cleanup
trap successfulExit EXIT

QUESTION:

How can I make trap only trap EXIT on program finish?

Here is the full script:

debianConfigAwsome.5.3.sh

shell-script exit-status trap

share|improve this question

edited 2 hours ago

muru

37.6k589165

asked 7 hours ago

somethingSomethingsomethingSomething

1,850103460

add a comment | 

2

1

PROBLEM:

I have a shell program that I have been writing but I can't find out how to make sure that trap is trapping for cleanup at the end or because of a error in some command, it cleans up either way.

Here is the code:

################################### Successful exit then this cleanup ###########################################################3

successfulExit()

###############################################################################################################################33
####### Catch the program on successful exit and cleanup
trap successfulExit EXIT

QUESTION:

How can I make trap only trap EXIT on program finish?

Here is the full script:

debianConfigAwsome.5.3.sh

shell-script exit-status trap

share|improve this question

edited 2 hours ago

muru

37.6k589165

asked 7 hours ago

somethingSomethingsomethingSomething

1,850103460

add a comment | 

PROBLEM:

I have a shell program that I have been writing but I can't find out how to make sure that trap is trapping for cleanup at the end or because of a error in some command, it cleans up either way.

Here is the code:

################################### Successful exit then this cleanup ###########################################################3

successfulExit()

###############################################################################################################################33
####### Catch the program on successful exit and cleanup
trap successfulExit EXIT

QUESTION:

How can I make trap only trap EXIT on program finish?

Here is the full script:

debianConfigAwsome.5.3.sh

shell-script exit-status trap

share|improve this question

edited 2 hours ago

muru

37.6k589165

asked 7 hours ago

somethingSomethingsomethingSomething

1,850103460

################################### Successful exit then this cleanup ###########################################################3

successfulExit()

###############################################################################################################################33
####### Catch the program on successful exit and cleanup
trap successfulExit EXIT

share|improve this question

edited 2 hours ago

muru

37.6k589165

asked 7 hours ago

somethingSomethingsomethingSomething

1,850103460

share|improve this question

edited 2 hours ago

muru

37.6k589165

asked 7 hours ago

somethingSomethingsomethingSomething

1,850103460

edited 2 hours ago

muru

37.6k589165

edited 2 hours ago

muru

37.6k589165

asked 7 hours ago

somethingSomethingsomethingSomething

1,850103460

asked 7 hours ago

somethingSomethingsomethingSomething

1,850103460

1 Answer
1

active

oldest

votes

4

On entry to the EXIT trap, $? contains the exit status. That's the same value you'd find as $? after calling this script in another shell: either the argument passed to exit (truncated to the range 0–255) or the return status of the preceding command. In the case of an exit due to set -e, it's the return status of the command that triggered the implicit exit.

Usually you should save $? and exit again with the same status.

cleanup () 
 if [ -n "$1" ]; then
 echo "Aborted by $1"
 elif [ $status -ne 0 ]; then
 echo "Failure (status $status)"
 else
 echo "Success"
 fi

trap 'status=$?; cleanup; exit $status' EXIT
trap 'trap - HUP; cleanup SIGHUP; kill -HUP $$' HUP
trap 'trap - INT; cleanup SIGINT; kill -INT $$' INT
trap 'trap - TERM; cleanup SIGTERM; kill -TERM $$' TERM

share|improve this answer

answered 6 hours ago

GillesGilles

547k13011131631

add a comment | 

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4

On entry to the EXIT trap, $? contains the exit status. That's the same value you'd find as $? after calling this script in another shell: either the argument passed to exit (truncated to the range 0–255) or the return status of the preceding command. In the case of an exit due to set -e, it's the return status of the command that triggered the implicit exit.

Usually you should save $? and exit again with the same status.

cleanup () 
 if [ -n "$1" ]; then
 echo "Aborted by $1"
 elif [ $status -ne 0 ]; then
 echo "Failure (status $status)"
 else
 echo "Success"
 fi

trap 'status=$?; cleanup; exit $status' EXIT
trap 'trap - HUP; cleanup SIGHUP; kill -HUP $$' HUP
trap 'trap - INT; cleanup SIGINT; kill -INT $$' INT
trap 'trap - TERM; cleanup SIGTERM; kill -TERM $$' TERM

share|improve this answer

answered 6 hours ago

GillesGilles

547k13011131631

add a comment | 

4

On entry to the EXIT trap, $? contains the exit status. That's the same value you'd find as $? after calling this script in another shell: either the argument passed to exit (truncated to the range 0–255) or the return status of the preceding command. In the case of an exit due to set -e, it's the return status of the command that triggered the implicit exit.

Usually you should save $? and exit again with the same status.

cleanup () 
 if [ -n "$1" ]; then
 echo "Aborted by $1"
 elif [ $status -ne 0 ]; then
 echo "Failure (status $status)"
 else
 echo "Success"
 fi

trap 'status=$?; cleanup; exit $status' EXIT
trap 'trap - HUP; cleanup SIGHUP; kill -HUP $$' HUP
trap 'trap - INT; cleanup SIGINT; kill -INT $$' INT
trap 'trap - TERM; cleanup SIGTERM; kill -TERM $$' TERM

share|improve this answer

answered 6 hours ago

GillesGilles

547k13011131631

add a comment | 

On entry to the EXIT trap, $? contains the exit status. That's the same value you'd find as $? after calling this script in another shell: either the argument passed to exit (truncated to the range 0–255) or the return status of the preceding command. In the case of an exit due to set -e, it's the return status of the command that triggered the implicit exit.

Usually you should save $? and exit again with the same status.

cleanup () 
 if [ -n "$1" ]; then
 echo "Aborted by $1"
 elif [ $status -ne 0 ]; then
 echo "Failure (status $status)"
 else
 echo "Success"
 fi

trap 'status=$?; cleanup; exit $status' EXIT
trap 'trap - HUP; cleanup SIGHUP; kill -HUP $$' HUP
trap 'trap - INT; cleanup SIGINT; kill -INT $$' INT
trap 'trap - TERM; cleanup SIGTERM; kill -TERM $$' TERM

share|improve this answer

answered 6 hours ago

GillesGilles

547k13011131631

cleanup () 
 if [ -n "$1" ]; then
 echo "Aborted by $1"
 elif [ $status -ne 0 ]; then
 echo "Failure (status $status)"
 else
 echo "Success"
 fi

trap 'status=$?; cleanup; exit $status' EXIT
trap 'trap - HUP; cleanup SIGHUP; kill -HUP $$' HUP
trap 'trap - INT; cleanup SIGINT; kill -INT $$' INT
trap 'trap - TERM; cleanup SIGTERM; kill -TERM $$' TERM

share|improve this answer

answered 6 hours ago

GillesGilles

547k13011131631

answered 6 hours ago

GillesGilles

547k13011131631

answered 6 hours ago

GillesGilles

547k13011131631