shell script - how to use variables for a shell script without defining the variable in the script

Question

Raspbian (Jessy) - root@Raspberry Pi - Putty

In the Terminal i type in

finalanswer=0

now i got a script with this code

#!/bin/bash
source /lib/lsb/init-functions

echo $finalanswer              #just as a test
if [ ! "$finalanswer" = "0" ]
then
        rm -r mnt/objects/all
        log_warning_msg "All Files has been deleted" || true
        touch its_over.txt
else
        let finalanswer=1
        log_action_msg "Var finalanswer was 0. setting back to 1" || true
fi

there is a cronjob that starts this script every hour

sooo. somewhere there must be an error. because he is reading the Variable $finalanswer as nothing.

that means variables that has been defined outside of this script will not work?

how do i fix this?

Answer 1

Shell variables are not inherited by child processes. If you want a variable to be inherited, it has to be an environment variable. You create environment variables using the export command.

export finalanswer=0

or

finalanswer=0
export finalanswer

You can also export a variable just for the duration of a command by putting the assignment at the beginning of the command:

finalanswer=0 /path/to/script

Note that variables you assign in your shell will not be accessible to cron jobs. Variables can only be exported to processes that are descended from the shell, and processes run by cron are not related to your shell process. If you want to set a variable for use in a cron job, you can put the assignment into the crontab file itself.

Answer 2

You can define a variable for a single command by placing its definition before the command you wish to run:

$ VARIABLE=hunter perl -E 'say $ENV{VARIABLE}'
hunter

you can do the same thing for a cron entry:

*/10 * * * * VARIABLE=hunter <command>