How to avoid a bash script from failing when -e option is set?

Question

I have a bash script with -e option set, which fails the whole script on the very first error.

In the script, I am trying to do an ls on a directory. But that path may or may not exist. If the path does not exist, the ls command fails, since the -e flag is set.

Is there a way by which I can prevent the script from failing?

As a side note, I have tried the trick to do an set +e and set -e before and after that command and it works. But I am looking for some better solution.

Answer 1

You can "catch" the error using || and a command guaranteed to exit with 0 status:

ls $PATH || echo "$PATH does not exist" 

Since the compound command succeeds whether or not $PATH exists, set -e is not triggered and your script will not exit.

To suppress the error silently, you can use the true command:

ls $PATH || true 

To execute multiple commands, you can use one of the compound commands:

ls $PATH || { command1; command2; } 

or

ls $PATH || ( command1; command2 ) 

Just be sure nothing fails inside either compound command, either. One benefit of the second example is that you can turn off immediate-exit mode inside the subshell without affecting its status in the current shell:

ls $PATH || ( set +e; do-something-that-might-fail )