Bash question about bash variable and testing for said variable

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1 
#!/bin/bash
var = $(< /sys/class/leds/input3::numlock/brightness)
if $var = 0 
then
	notify-send 'Keyboard alert' 'Numlock Off.' --icon=dialog-information
else
	notify-send 'Keyboard alert' 'Numlock On.' --icon=dialog-information
fi
exit 0

Can anyone tell me what is wrong with that as I have searching and searching as well as changing code over and over to no avail.
Now it complains that
var is not a command
well duh, it’s a variable that I want to test against.
The reason for this script is because my computer does not, sadly, have an LED for Numlock.
So any suggestions would be appreciated

2

Well… ok that’s not stackoverflow here, but that should be more correct:

#!/usr/bin/env bash

var=$(cat /sys/class/leds/input3::numlock/brightness)
if   [[ $var -eq 0 ]]; then
	notify-send 'Keyboard alert' 'Numlock Off.' --icon=dialog-information
elif [[ $var -eq 1 ]]; then
	notify-send 'Keyboard alert' 'Numlock On.' --icon=dialog-information
fi

exit 0

You see there were a number of errors… variables for example must not have spaces between and conditions are like this: [[ expr != expr ]] in bash. Maybe have look at Bash scripting cheatsheet for a quick view. :wink:

What also should be mentioned is shellcheck to check your script for bad syntax.

3 Likes
3

Thanks for the help and I installed shellcheck and that is helping, at least somewhat.

PS: I joined stackoverflow. I need all the help I can get.

4

May I also suggest these two resources:

5

Solved. Kinda diff from original

#!/bin/bash
sleep 0.5
if cat /sys/class/leds/input3::numlock/brightness |grep "1"
then 
	notify-send 'Keyboard alert' 'Numlock On.' --icon=dialog-information
else
	notify-send 'Keyboard alert' 'Numlock Off.' --icon=dialog-information
fi
exit 0
6

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