Shell Programming and Scripting

Folks,
Below is a basic synopsis of the problem. I have a script that I need to check for some env vars and fail (exit the script) if they are not there. At the same time I need to set some default env vars. To do this I must run the script from the parent shell or source the script. Doing this means my call to exit kills my shell. Any ideas?

--
#!/bin/sh

Defined()
{
[ "${!1-one}" == "${!1-two}" ]
}

Fail()
{
echo $(date +%H:%M:%S) ERROR: $1
exit 1
}

if ! Defined FOO; then
Fail "FOO is not defined"
fi

export BAR=/foo/bar

# end script

~$ . ./test.sh # Need to run it this way for it to export my env vars

This exits my running shell


Is there any way to workaround this, such as goto in perl? Or other way to export my env vars without having to run in the parent shell? Thanks in advance.

You can also use structure to avoid calling either exit or return.

Code:

...
if ! Defined FOO; then
    echo "$(date +%H:%M:%S) ERROR:  FOO is not defined"
else
    export BAR=/foo/bar
fi

A word from member : https://www.unix.com/members/302085081.html

1.
Regarding "Quitting a bash script": The thread is closed so I couldn't post there, but I did want to share this.

Code:

      exit

exits the shell, which is usually not what you want if the script was
Code:

source


d.

An alternative is to use

Code:

      kill -SIGINT $$

which sends an interrupt to the shell process (whose pid is stored in the $$ variable).

You can also use this test to determine if a bash script is being run sourced or was run as its own command:

Code:

x"${BASH_SOURCE[0]}" == x"$0"

If this is true, then it isn't sourced, if it's false (the scriptname is not $0) it was sourced.