Bash script: no output; no error!

Question

I am testing following code that intends to get input from user and after checking some conditions store it to a new variable. However, this code is not giving me any output or error upon running in the shell terminal. I have given execution permission to the file by sudo chmod +x test.sh.

#!/bin/bash counter="0" iso_val="100" while [ "$counter" == "3" ] do echo -n "Enter ISO [ 100-800 ]: "; read iso if (( "$iso" < "100" )) || (( "$iso" > "800" )); then echo "Error!" elif [ "$iso" == "" ]; then echo "Error!" else iso_val=$iso break fi counter=$[$counter + 1] done 

Answer 1

Your script as written runs without error and would not produce output.

A quick note: you should temporarily add set -x at the beginning of your bash scripts for getting a good start on debugging.

The issue is the line:

while [ "$counter" == "3" ] 

The do block following that line would only be run if $counter was equal to 3, but $counter will not increment unless that do block is ran.

You would need something outside the block you wish to execute to increment your counter. i.e.

#!/bin/bash counter="0" iso_val="100" while [ "$counter" -lt 3 ] do echo -n "Enter ISO [ 100-800 ]: "; read iso if [ -z "$iso" ] || [ "$iso" -lt 100 ] || [ "$iso" -gt 800 ]; then echo "Error!" else iso_val=$iso break fi counter=$(($counter + 1)) done 

I set a limit to how many times while is ran, otherwise the script does not exit on error. I also changed your string comparisons to integer comparisons where I could.

Edit: New script with clarified requirement from comments, "ask up to three times on error"