verify value of a variable being part of a list

Question

I want to make sure the value of a variable (STATE) is part of a list (ALL_STATES).

#!/usr/bin/bash
STATE='somevalue'
ALL_STATES=( lirum larum loeffelstiel )
echo state id is: "${STATE}"
echo all states are: "${ALL_STATES[@]}"
while ${STATE} not in ${ALL_STATES} ; do
      echo ${STATE} needs a valid value
      read -p "please provide the state ID: " STATE
done
echo state id is: "${STATE}"

I am getting ./test_while_loop.sh: line 8: somevalue: command not found though. So there seems to be something wrong with my condition, just what?

Always paste your script into https://shellcheck.net, a syntax checker, or install shellcheck locally. Make using shellcheck part of your development process. — waltinator, Oct 24 '23 at 15:35
Please read correct-bash-and-shell-script-variable-capitalization. — Ed Morton, Oct 25 '23 at 14:04
Most of the shells' special variables are uppercase, but since zsh has one called path, one has to check their shell's manual anyway if they want to avoid clashing with names their shell holds special. The rest of the stuff about capitalization styles (e.g. snake_case vs. CamelCase) is just opinion-based... — ilkkachu, Oct 25 '23 at 14:12

score 0 · Answer 1 · answered Oct 24 '23 at 15:47

The error in your script is that in bash, the while keyword must be followed by a command, it does not interpret expressions. Also, <value> in <array> is not a valid expression either.

Instead you could turn the list into an associative array. Consider following the example given in this answer:

https://unix.stackexchange.com/a/177589/573867

verify value of a variable being part of a list

1 Answers1