unix - How to combine AND and OR condition in bash script for if condition?

Question

I was trying to combine logical AND & OR in a bash script within if condition. Somehow I am not getting the desired output and it is hard to troubleshoot. I am trying to validate the input parameters passed to a shell script for no parameter and the first parameter passed is valid or not.

if [ "$#" -ne 1 ] && ([ "$1" == "ABC" ] || [ "$1" == "DEF" ] || [ "$1" == "GHI" ] || [ "$1" == "JKL" ]) 
then
echo "Usage: ./myscript.sh [ABC | DEF | GHI | JKL]"
exit 1
fi

Can anyone point out what is going wrong here?

Answer 1

The immediate problem with your statement is one of logic: you probably meant to write:

if [ "$#" -ne 1 ] || ! ([ "$1" = "ABC" ] || [ "$1" = "DEF" ] || [ "$1" = "GHI" ] || [ "$1" = "JKL" ]) 
then
  echo "Usage: ./myscript.sh [ABC | DEF | GHI | JKL]" >&2
  exit 1
fi

That is: abort, if either more than 1 argument is given OR if the single argument given does NOT equal one of the acceptable values.

Note the ! to negate the expression in parentheses and the use of the POSIX-compliant form of the string equality operator, = (rather than ==).

However, given that you're using Bash, you can make do with a single [[ ... ]] conditional and Bash's regular-expression matching operator, =~:

if [[ $# -ne 1 || ! $1 =~ ^(ABC|DEF|GHI|JKL)$ ]] 
then
  echo "Usage: ./myscript.sh [ABC | DEF | GHI | JKL]" >&2
  exit 1
fi

If POSIX compliance is not required, [[ ... ]] is preferable to [ ... ] for a variety of reasons. In the case at hand, $# and $1 didn't need quoting, and || could be used inside the conditional.

Note that =~ as used above works in Bash 3.2+, whereas the implicit extglob syntax used in anubhava's helpful answer requires Bash 4.1+;
in earlier versions you can, however, explicitly enable (and restore to its original value after) the extglob shell option: shopt -s extglob.