Bash If statement within while loop executes even when condition not met-CodePudding

I have been using a bash script within ChromeOS to record screen brightness every 5 minutes as well as change and record the brightness if it changes.

#!/bin/sh

#This script will collect brightness every 5 minutes
#If brightness is not at desired setting, script will correct and note time changed

i=13
time=$(date  %H%M)
bright=$(backlight_tool --get_brightness_percent)

echo "start time is "$time >> brightness_log.txt
while [ $i -ne 1 ]; do

    sleep 300 
    backlight_tool --get_brightness_percent >> brightness_log.txt
    i=$(( $i-1 ))

        if [ $bright != 60.5 ]; then
            backlight_tool --set_brightness_percent=60.5
            echo "Brightness changed at "$time >> brightness_log.txt
        else
            break
        fi 
done
exit

The script works, however even when brightness is 60.5, the if statement will execute and echo "Brightness changed at " with the same repeated time.

If the brightness IS equal to 60.5, why is the if statement executing and not skipping to the break?

CodePudding user response：

You never update the variable bright it's set outside of the loop. Move the assignment inside the while loop, and it will work.

while [ "$i" -gt 1 ]; do
  bright=$(backlight_tool --get_brightness_percent)
  ..
done

Please validate your script with shellcheck, as you have many pitfalls present, one being not quoting parameter expansion:

$ shellcheck myscript
 
Line 10:
echo "start time is "$time >> brightness_log.txt
                     ^-- SC2086 (info): Double quote to prevent globbing and word splitting.

CodePudding user response：

Within the while loop I inserted

echo "$bright" | grep "60.5" -q 
gcode=$?

I then changed the if statement to

if [[ $gcode -eq 0 ]]; then

The if statement now only executes as intended.