As hinted at in this question, the answer is setting up a trap on the ERR signal - basically set -e corresponds to trap 'exit' ERR. So to e.g. log errors but continue execution, use

trap 'logger -t myscriptname "Command $BASH_COMMAND exited with code $?"' ERR

In bash, the variable $BASH_COMMAND contains the offending command. $? contains the exit code of the last command executed - note that if you want to perform multiple actions, e.g. logger ...; echo $? the $? will contain logger's exit code, not $BASH_COMMAND's one, so you may have to store it in a variable first, e.g.

trap 'EXITCODE=$?; logger -t "Command $BASH_COMMAND exited with code $EXITCODE"; echo "$BASH_COMMAND at line $LINENO exited with code $EXITCODE"' ERR

Of course you might be better off declaring a function and passing it the $? and $BASH_COMMAND instead of such a long line!

As hinted at in this question, the answer is setting up a trap on the ERR signal - basically set -e corresponds to trap 'exit' ERR. So to e.g. log errors but continue execution, use

trap 'logger -t myscriptname "Command $BASH_COMMAND exited with code $?"' ERR

In bash, the variable $BASH_COMMAND contains the offending command. $? contains the exit code of the last command executed - note that if you want to perform multiple actions, e.g. logger ...; echo $? the $? will contain logger's exit code, not $BASH_COMMAND's one, so you may have to store it in a variable first, e.g.

trap 'EXITCODE=$?; logger -t "Command $BASH_COMMAND exited with code $EXITCODE"; echo "$BASH_COMMAND at line $LINENO exited with code $EXITCODE"' ERR

Of course you might be better off declaring a function and passing it the $? and $BASH_COMMAND instead of such a long line!

As hinted at in this question, the answer is setting up a trap on the ERR signal - basically set -e corresponds to trap 'exit' ERR. So to e.g. log errors but continue execution, use

trap 'logger -t myscriptname "Command $BASH_COMMAND exited with code $?"' ERR

The variable $BASH_COMMAND contains the offending command while $? contains the exit code of the last command executed - note that if you want to perform multiple actions, e.g. logger ...; echo $? the $? will contain logger's exit code, not $BASH_COMMAND's one, so you may have to store it in a variable first, e.g.

trap 'EXITCODE=$?; logger -t "Command $BASH_COMMAND exited with code $EXITCODE"; echo "$BASH_COMMAND at line $LINENO exited with code $EXITCODE"' ERR

Of course you might be better off declaring a function and passing it the $? and $BASH_COMMAND instead of such a long line!

As hinted at in this question, the answer is setting up a trap on the ERR signal - basically set -e corresponds to trap 'exit' ERR. So to e.g. log errors but continue execution, use

trap 'logger -t myscriptname "Command $BASH_COMMAND exited with code $?"' ERR

In bash, the variable $BASH_COMMAND contains the offending command. $? contains the exit code of the last command executed - note that if you want to perform multiple actions, e.g. logger ...; echo $? the $? will contain logger's exit code, not $BASH_COMMAND's one, so you may have to store it in a variable first, e.g.

trap 'EXITCODE=$?; logger -t "Command $BASH_COMMAND exited with code $EXITCODE"; echo "$BASH_COMMAND at line $LINENO exited with code $EXITCODE"' ERR

Of course you might be better off declaring a function and passing it the $? and $BASH_COMMAND instead of such a long line!