Aborting a shell script if any command returns a non-zero value?

Question

I have a Bash shell script that invokes a number of commands. I would like to have the shell script automatically exit with a return value of 1 if any of the commands return a non-zero value. Is this possible without explicitly checking the result of each command? e.g. dosomething1 if [[ $? -ne 0 ]]; then exit 1 fi dosomething2 if [[ $? -ne 0 ]]; then exit 1 fi

Answer 1

Not that I know of ...

I could recommend something that works, but is a bit clumsy - see below.

You would be defining a function that would in turn be used to launch every command in your script:

function launch {

typeset cmd="$1"

shift

$cmd $@ || exit 1

}

launch ls -l A*

launch grep "pattern" A*

launch date

Answer 2

This notation will accomplish what you want without explicitly interrogating the $? variable after each command invocation:

command || {

echo "Some Error Diagnostic ..."

exit 1

}

The || operator allow the statements on the right to be executed only if the statement(s) on the left exit with a non-0 status (that is, if what is executed on the left fails). The || operator is the inverse of the && operator.

I suppose if you wanted to, you can encapsulate this in a function, and then leverage that function for everything launched command in your script.

See below.

function myExec {

command="$1"

shift

args="$@"

$command $args || {

echo "$command $args: FAILED" >&2

exit 1

}

return $?

}

And then the significant commands in your script can be executed like this ...

myExec command arg1 arg2 ....