Bash: default boolean value in getopts

Question

I want to include a default option in my script where if the user uses this option, set the flag to true or it should be false by default. It seems that the script is not accepting false or true as boolean value. How can I make it boolean?

flag=

instructions() {
  echo " -a File name" >&2
  echo " -f optional boolean" flag=${flag:-false}
}

while getopts ":a:fi" option; do
  case "$option" in
    a ) file=$OPTARG;;
    f ) flag=true;;
    u )  
       instructions
       ;;
    \?)
      echo "Not valid -$OPTARG" >&2
      instructions
      ;;
    : ) echo "args required";;
  esac
done

if [[ "$flag" != true || "$flag" != false ]]; then
  echo "Not a boolean value"
fi

Answer 1

Check this, I made some fixes to your script (commented in the code) along with a proper formatting.

#!/bin/bash

# Set the default value of the flag variable
flag=false

instructions() {
  echo "Usage: $0 [ -a FILE ] [ -f ]" >&2
  echo " -a File name" >&2
  echo " -f optional boolean flag=${flag:-false}" >&2
}


# If the script must be executed with options, this checks if the number of arguments
# provided to the script is greater than 0
if [ $# -eq 0 ]; then
    instructions
    exit 1
fi

while getopts ":a:fi" option; do
  case "${option}" in
    a ) 
       file="$OPTARG"
       ;;
    f ) 
       flag=true
       ;;
    i ) # "u" is not a valid option
       instructions
       exit 0
       ;;
    \?)
       echo "Option '-$OPTARG' is not a valid option." >&2
       instructions
       exit 1
       ;;
    : )
       echo "Option '-$OPTARG' needs an argument." >&2
       instructions
       exit 1
       ;;
  esac
done

# Since a variable can't have 2 values assigned at the same time, 
# you should use && (and) instead of || (or)
if [[ "$flag" != true ]] && [[ "$flag" != false ]]; then
  echo "Not a boolean value"
fi

exit 0