Declaring a variable name (not value) in bash with string substitution

Question

I have a set of variables whose values I need to modify in a loop, if a match is found in an array which I am reading later. The array elements are substrings of the variable names, so I thought of writing a generic for loop, which would modify the variable values on the fly, attempting string substitution in the variable names, but that doesn't work. Here is the code:

CUT_APPLE=false
CUT_GUAVA=false
CUT_MANGO=false

AVAILABLE_FRUITS=("APPLE" "GUAVA")

for i in "${AVAILABLE_FRUITS[@]}"; do
CUT_$i=true
done
**CUT_APPLE=true: command not found**                               
**CUT_GUAVA=true: command not found**

Is there any way using which I can get interpolation in variable name declaration to work?

Don't use all-upper-case variable names like that, see [correct-bash-and-shell-script-variable-capitalization](https://stackoverflow.com/questions/673055/correct-bash-and-shell-script-variable-capitalization). — Ed Morton, Jul 03 '21 at 16:20

score 4 · Accepted Answer · edited Jul 04 '21 at 10:39

If your Bash version is greater than 4, it supports associative arrays which make it easy to check whether a boolean key exists:

#!/usr/bin/env bash

# Associative array to store keys of available fruits
declare -A cut_fruit=()

declare -a available_fruits=('APPLE' 'GUAVA' 'RED BANANA')

for fruit in "${available_fruits[@]}"; do
  # Create key for fruit in Associative array if available
  # shellcheck disable=SC2034 # cut_fruit is indeed used
  cut_fruit[$fruit]=
done

# Check if cut fruit is available by checking key existence
for fruit in 'APPLE' 'BANANA' 'RED BANANA' 'MANGO'; do
  if [[ -v cut_fruit[$fruit] ]]; then
    printf 'Cut %s is available.\n' "$fruit"
  else
    printf 'There is no cut %s.\n' "$fruit"
  fi
done

Sample output:

Cut APPLE is available.
There is no cut BANANA.
Cut RED BANANA is available.
There is no cut MANGO.

Here is a little game to play with associative arrays keys and boolean checks:

#!/usr/bin/env bash

print_basket() {
  # Print remaining keys (content of basket)
  for fruit in "${!basket[@]}"; do
    printf '%d %s\n' "${basket[$fruit]}" "${fruit,,}"
  done
}

cat <<'EOF'
--:{# Guess the fruits in my basket #}:--
EOF

declare -a all_fruits=(
  'apple' 'banana' 'cherry' 'lemon' 'mango'
  'orange' 'peach' 'plum' 'red banana' 'sweat chestnut')
# Number of fruits in the basked
selection_size=3

declare -a fruits_selection
mapfile -t fruits_selection < <(
  # Shuffle a selection of fruits
  printf '%s\n' "${all_fruits[@]}" | shuf -n "$selection_size"
)

declare -A basket
# Set selected fruits into basked
for fruit in "${fruits_selection[@]}"; do
  # Uppercase the key for case insensitive match later
  basket[${fruit^^}]=$((RANDOM % 9 + 1))
done

# Cheating
declare -p basket

max_tries=3
tries_left=$max_tries
guessed=0
while [ $guessed -lt $selection_size ] && [ $tries_left -gt 0 ]; do
  printf 'Enter the name of a fruit: '
  read -r fruit
  # Uppercase fruit for case insensitive match
  fruit="${fruit^^}"
  if [[ -v basket[$fruit] ]]; then
    printf 'Yes, there is %d %s in my basket!\n' \
      "${basket[$fruit]}" "${fruit,,}"
    guessed=$((guessed + 1))
    # Remove fruit from basket
    unset "basket[$fruit]"
  else
    tries_left=$((tries_left - 1))
    printf 'No, there is no %s in my basket!\nTries left: %d\n' \
      "${fruit,,}" "$tries_left"
  fi
done

case $guessed in
  "$selection_size")
    printf '* * * Congratulations, you guessed all fruilts! * * *\n'
    ;;
  0)
    printf 'Sorry you could not guess any of the fruits!\nBasket contained:\n'
    print_basket
    ;;
  *)
    printf 'Nice! You guessed %d fruits out of %s!\nUnguessed fruits:\n' \
      "$guessed" "$selection_size"
    print_basket
    ;;
esac

Declaring a variable name (not value) in bash with string substitution

1 Answers1