Bash - how to set variable within an if [] condition and use outside [duplicate]

Question

I'm trying to get my head around this. I have a basic Bash script:

#!/bin/bash

do_exit=0

while [ "$do_exit" == 0 ] ; do

        echo $do_exit

        do_exit=1

done

This works. However, when I want to set do_exit=1 inside an if condition:

#!/bin/bash

do_exit=0

while [ "$do_exit" == 0 ] ; do

    free=`free -m | grep Mem | awk '{print $4}'`

    if [ "$free" -gt 0 ]
    then

        echo $do_exit

        do_exit=1

    fi & sleep 5;

done

$do_exit is ALWAYS 0 ... so never exists. What am I doing wrong? I'm have the same issue with "breaking" and "exit 0" when I want to kill te script - it just carries on!)

I must be misunderstanding how variables work in Bash? In Perl (my normal programing language), you can set variables anywhere (with the exception of explicitly setting them in a function)

Answer 1

There are a couple of errors.

1. you echo do_exit before you set it to 1
2. you have fi & sleep 5; which makes the if statement a child process, thus you can't use the value in the parent process
3. -gt is for numeric comparisment, not strings.

Below should work for you.

#!/bin/bash

do_exit=0

while [ "$do_exit" == 0 ] ; do

    free=`free -m | grep Mem | awk '{print $4}'`

    if [[ $free -gt 0 ]]; then
        do_exit=1
        echo ${$do_exit}
    fi
    sleep 5;

done

You can break while loops as well. Check this:

#!/bin/bash

while true; do

    free=$(free -m | grep Mem | awk '{print $4}')

    if [[ $free -gt 0 ]]; then
        break;
    fi
    sleep 5;
done