Is there a way to exit a Bash script, but not quitting the terminal?

Yes; you can use return instead of exit. Its main purpose is to return from a shell function, but if you use it within a source-d script, it returns from that script.

As §4.1 "Bourne Shell Builtins" of the Bash Reference Manual puts it:

return [n]

Stop executing a shell function or sourced file and return the value n to its caller. If n is not supplied, the return value is the exit status of the last command executed. […]

[…]

The return status is non-zero if return is supplied a non-numeric argument or is used outside a function and not during the execution of a script by . or source.

You can add an extra exit command after the return statement/command so that it works for both, executing the script from the command line and sourcing from the terminal.

Example exit code in the script:

   if [ $# -lt 2 ]; then
     echo "Needs at least two arguments"
     return 1 2>/dev/null
     exit 1
   fi

The line with the exit command will not be called when you source the script after the return command.

When you execute the script, return command gives an error. So, we suppress the error message by forwarding it to /dev/null.

Instead of running the script using . run2.sh, you can run it using sh run2.sh or bash run2.sh.

A new sub-shell will be started. To run the script then, it will be closed at the end of the script, leaving the other shell opened.

For what it’s worth, I've been using the following for years in all my scripts; it works like a charm in all use cases I've encountered—I have extensive shell scripts that are sourced at startup, plus many Bash functions I regularly use. I remember trying return long back and having issues with it.

kill -INT $$
# Here's the definition...
# SIGINT    2    Interrupt from keyboard

Actually, I think you might be confused by how you should run a script.

If you use sh to run a script, say, sh ./run2.sh, even if the embedded script ends with exit, your terminal window will still remain.

However, if you use . or source, your terminal window will exit/close as well when subscript ends.

For more detail, please refer to What is the difference between using sh and source?.

I had the same problem and from the previous answers and from what I understood, this ultimately worked for me:

1. Have a shebang line that invokes the intended script, for example,

#!/bin/bash uses bash to execute the script

I have scripts with both kinds of shebang's. Because of this, using sh or . was not reliable, as it lead to a mis-execution (like when the script bails out having run incompletely)

The answer therefore, was

2. • Make sure the script has a shebang, so that there is no doubt about its intended handler.

   • chmod the .sh file so that it can be executed. (chmod +x file.sh)

   • Invoke it directly without any sh or .

     (./myscript.sh)

To write a script that is secure to be run as either a shell script or sourced as an rc file, the script can check and compare $0 and $BASH_SOURCE and determine if exit can be safely used.

Here is a short code snippet for that

[ "X$(basename $0)" = "X$(basename $BASH_SOURCE)" ] && \
    echo "***** executing $name_src as a shell script *****" || \
    echo "..... sourcing $name_src ....."

This is just like you put a run function inside your script run2.sh. You use exit code inside run while source your run2.sh file in the bash tty. If the give the run function its power to exit your script and give the run2.sh its power to exit the terminator. Then of cuz the run function has power to exit your teminator.

    #! /bin/sh
    # use . run2.sh

    run()
    {
        echo "this is run"
        #return 0
        exit 0
    }

    echo "this is begin"
    run
    echo "this is end"

Anyway, I approve with Kaz it's a design problem.

I improved the answer of Tzunghsing, with more clear results and error redirection, for silent usage:

#!/usr/bin/env bash

echo -e "Testing..."

if [ "X$(basename $0 2>/dev/null)" = "X$(basename $BASH_SOURCE)" ]; then
    echo "***** You are Executing $0 in a sub-shell."
    exit 0
else
    echo "..... You are Sourcing $BASH_SOURCE in this terminal shell."
    return 0
fi

echo "This should never be seen!"

Or if you want to put this into a silent function:

function sExit() {
    # Safe Exit from script, not closing shell.
    [ "X$(basename $0 2>/dev/null)" = "X$(basename $BASH_SOURCE)" ] && exit 0 || return 0
}

...

# ..it has to be called with an error check, like this:
sExit && return 0

echo "This should never be seen!"

Please note that:

• if you have enabled errexit in your script (set -e) and you return N with N != 0, your entire script will exit instantly. To see all your shell settings, use, set -o.
• when used in a function, the 1st return 0 is exiting the function, and the second return 0 is exiting the script.

It's correct that sourced vs. executed scripts use return vs. exit to keep the same session open, as others have noted.

Here's a related tip, if you ever want a script that should keep the session open, regardless of whether or not it's sourced.

The following example can be run directly like foo.sh or sourced like . foo.sh/source foo.sh. Either way it will keep the session open after "exiting". The $@ string is passed so that the function has access to the outer script's arguments.

#!/bin/sh
foo(){
    read -p "Would you like to XYZ? (Y/N): " response;
    [ $response != 'y' ] && return 1;
    echo "XYZ complete (args $@).";
    return 0;
    echo "This line will never execute.";
}
foo "$@";

Terminal result:

$ foo.sh
$ Would you like to XYZ? (Y/N): n
$ . foo.sh
$ Would you like to XYZ? (Y/N): n
$ |
(terminal window stays open and accepts additional input)

This can be useful for quickly testing script changes in a single terminal while keeping a bunch of scrap code underneath the main exit/return while you work. It could also make code more portable in a sense (if you have tons of scripts that may or may not be called in different ways), though it's much less clunky to just use return and exit where appropriate.

I think that this happens because you are running it on source mode with the dot:

. myscript.sh

You should run that in a subshell:

/full/path/to/script/myscript.sh

See 'source'.

Also make sure to return with the expected return value. Else if you use exit when you will encounter an exit, it will exit from your base shell since source does not create another process (instance).

The following trick allows the script to gracefully exit without terminating the shell, whether sourced or direct execution:

Instead of exit, use this:

[[ $PS1 ]] && return $? || exit $?

Alternatively, define this function and call it instead of exit:

    __EXIT__()
   {
       [[ $PS1 ]] && return $1 || exit $1
   }
    ....
    __EXIT__ $?

If a command succeeded successfully, the return value will be 0. We can check its return value afterwards.

Is there a “goto” statement in bash?

Here is some dirty workaround using trap which jumps only backwards.

#!/bin/bash
set -eu
trap 'echo "E: failed with exitcode $?" 1>&2' ERR

my_function () {
    if git rev-parse --is-inside-work-tree > /dev/null 2>&1; then
        echo "this is run"
        return 0
    else
        echo "fatal: not a git repository (or any of the parent directories): .git"
        goto trap 2> /dev/null
    fi
}

my_function
echo "Command succeeded"  # If my_function failed this line is not printed

Related:

• https://stackoverflow.com/a/19091823/2402577
• How to use $? and test to check function?

I couldn't find solution so for those who want to leave the nested script without leaving terminal window:

# this is just script which goes to directory if path satisfies regex
wpr(){
    leave=false
    pwd=$(pwd)
    if [[ "$pwd" =~ ddev.*web ]]; then
        # echo "your in wordpress instalation"
        wpDir=$(echo "$pwd" | grep -o  '.*\/web')
        cd $wpDir
        return
    fi
    echo 'please be in wordpress directory'
    # to leave from outside the scope
    leave=true
    return
}

wpt(){
    # nested function which returns $leave variable
    wpr
    
    # interupts the script if $leave is true
    if $leave; then
        return;
    fi
    echo 'here is the rest of the script, executes if leave is not defined'
}

If your terminal emulator doesn't have -hold, you can sanitize a sourced script and hold the terminal with:

#!/bin/sh
sed "s/exit/return/g" script >/tmp/script
. /tmp/script
read

Otherwise, you can use $TERM -hold -e script.

I don't have any idea whether this is useful for you or not, but in Z shell (executable zsh), you can exit a script, but only to the prompt if there is one, by using parameter expansion on a variable that does not exist, as follows.

${missing_variable_ejector:?}

Though this does create an error message in your script, you can prevent it with something like the following.

{ ${missing_variable_ejector:?} } 2>/dev/null