How to hide output of first command when using the || operator and exit?

Question

This hides output from the first command, and prints Oops to stderr if it fails. So far, so good.

#!/usr/bin/env bash
invalid_command > /dev/null 2>&1 || >&2 echo "Oops"
echo hi

That outputs this:

Oops
hi

---

But I need to exit as well as printing a message if the first command failed. So I tried using parenthesis.

#!/usr/bin/env bash
invalid_command > /dev/null 2>&1 || ( >&2 echo "Oops" ; exit )
echo hi

Here's the output of that:

Oops
hi

But now the exit doesn't work because it's doing it in a subshell, causing it to print hi, even though I wanted the script to exit.

So, how do I get Bash to echo some text and exit if a specific command failed using the || operator? I'm aware that I can use an if one-liner to do that, but I'd prefer to not have to use a full if statement if I can avoid it.

#!/usr/bin/env bash
if [ "$(invalid_command > /dev/null 2>&1 ; printf $?)" != "0" ]; then >&2 echo 'Oops' ; exit 1; fi

Answer 1

Use command grouping (notice the ; on the end of exit).

#!/usr/bin/env bash
invalid_command > /dev/null 2>&1 || { >&2 echo "Oops" ; exit; }
echo hi

Return this on my terminal.

$ ./test.sh 
Oops

If you have a lot command, you can trap all errors.

#!/usr/bin/env bash
on_error() {
    >&2 echo "Oops"
    exit 1
}
trap 'on_error' ERR
invalid_command > /dev/null 2>&1
echo hi

This will print Oops and exit immediately, if any command on the script is failing.