In a Bash script, how can I exit the entire script if a certain condition occurs?

I'm writing a script in Bash to test some code. However, it seems silly to run the tests if compiling the code fails in the first place, in which case I'll just abort the tests.

Is there a way I can do this without wrapping the entire script inside of a while loop and using breaks? Something like a dun dun dun goto?

Try this statement:

exit 1

Replace 1 with appropriate error codes. See also Exit Codes With Special Meanings.

Use set -e

#!/bin/bash

set -e

/bin/command-that-fails
/bin/command-that-fails2

The script will terminate after the first line that fails (returns nonzero exit code). In this case, command-that-fails2 will not run.

If you were to check the return status of every single command, your script would look like this:

#!/bin/bash

# I'm assuming you're using make

cd /project-dir
make
if [[ $? -ne 0 ]] ; then
    exit 1
fi

cd /project-dir2
make
if [[ $? -ne 0 ]] ; then
    exit 1
fi

With set -e it would look like:

#!/bin/bash

set -e

cd /project-dir
make

cd /project-dir2
make

Any command that fails will cause the entire script to fail and return an exit status you can check with $?. If your script is very long or you're building a lot of stuff it's going to get pretty ugly if you add return status checks everywhere.

A SysOps guy once taught me the Three-Fingered Claw technique:

yell() { echo "$0: $*" >&2; }
die() { yell "$*"; exit 111; }
try() { "$@" || die "cannot $*"; }

These functions are *NIX OS and shell flavor-robust. Put them at the beginning of your script (bash or otherwise), try() your statement and code on.

Explanation

(based on flying sheep comment).

• yell: print the script name and all arguments to stderr:

  • $0 is the path to the script ;
  • $* are all arguments.
  • >&2 means > redirect stdout to & pipe 2. pipe 1 would be stdout itself.

• die does the same as yell, but exits with a non-0 exit status, which means “fail”.

• try uses the || (boolean OR), which only evaluates the right side if the left one failed.

  • $@ is all arguments again, but different.

If you will invoke the script with source, you can use return <x> where <x> will be the script exit status (use a non-zero value for error or false). But if you invoke an executable script (i.e., directly with its filename), the return statement will result in a complain (error message "return: can only `return' from a function or sourced script").

If exit <x> is used instead, when the script is invoked with source, it will result in exiting the shell that started the script, but an executable script will just terminate, as expected.

To handle either case in the same script, you can use

return <x> 2> /dev/null || exit <x>

This will handle whichever invocation may be suitable. That is assuming you will use this statement at the script's top level. I would advise against directly exiting the script from within a function.

Note: <x> is supposed to be just a number.

I often include a function called run() to handle errors. Every call I want to make is passed to this function so the entire script exits when a failure is hit. The advantage of this over the set -e solution is that the script doesn't exit silently when a line fails, and can tell you what the problem is. In the following example, the 3rd line is not executed because the script exits at the call to false.

function run() {
  cmd_output=$(eval $1)
  return_value=$?
  if [ $return_value != 0 ]; then
    echo "Command $1 failed"
    exit -1
  else
    echo "output: $cmd_output"
    echo "Command succeeded."
  fi
  return $return_value
}
run "date"
run "false"
run "date"

Instead of if construct, you can leverage the short-circuit evaluation:

#!/usr/bin/env bash

echo $[1+1]
echo $[2/0]              # division by 0 but execution of script proceeds
echo $[3+1]
(echo $[4/0]) || exit $? # script halted with code 1 returned from `echo`
echo $[5+1]

Note the pair of parentheses which is necessary because of priority of alternation operator. $? is a special variable set to exit code of most recently called command.

I have the same question but cannot ask it because it would be a duplicate.

The accepted answer, using exit, does not work when the script is a bit more complicated. If you use a background process to check for the condition, exit only exits that process, as it runs in a sub-shell. To kill the script, you have to explicitly kill it (at least that is the only way I know).

Here is a little script on how to do it:

#!/bin/bash

boom() {
    while true; do sleep 1.2; echo boom; done
}

f() {
    echo Hello
    N=0
    while
        ((N++ <10))
    do
        sleep 1
        echo $N
        #        ((N > 5)) && exit 4 # does not work
        ((N > 5)) && { kill -9 $$; exit 5; } # works 
    done
}

boom &
f &

while true; do sleep 0.5; echo beep; done

This is a better answer but still incomplete a I really don't know how to get rid of the boom part.

You can close your program by program name on follow way:

for soft exit do

pkill -9 -x programname # Replace "programmname" by your programme

for hard exit do

pkill -15 -x programname # Replace "programmname" by your programme

If you like to know how to evaluate condition for closing a program, you need to customize your question.