shell: Validate for multiple alphabets in an if statement

Question

I know this was asked multiple times with respect to validating multiple variables. This however is a if condition to validate for single letters(alphabets).

I have a statement that goes like this. The arguments are passed from another script

action=$1
name=$2
gname=$3

echo "action is $action name is $name and gname is $gname" # making sure i got the args passed right

#if [ $action != B ] || [ $action != T ] || [ $action != R ] ; // This WORKS!!
if [ $action != [ B | R | T ]; then // Returns an error
echo "NADA"
exit
fi

if [ $name != SNAP1 ] || [ $name != SNAP2 ] || [ $name != SNAP3 ] ; 
echo "NADA"
exit
fi

I'd like to know how to compare more than just those 3 letters and 3 variables. What if i want to validate more than say 8 letters, is there a better way to write the if statement, how can i shorten the comparison?

Are you limited to POSIX sh statements? You could use `case` for pattern comparison. Can you use Bash statement? You could use `[[ ]]` pattern matching. — Benjamin W., Jul 10 '18 at 18:17
In ksh you can use : if [[ $action = !(B|R|T) ]]; then........ — Andre Gelinas, Jul 10 '18 at 19:55

score 0 · Answer 1 · answered Jul 10 '18 at 18:17

0

You can use case:

case "$action" in
   [BRT]) ;;
    *) echo "NADA" ;;
esac

answered Jul 10 '18 at 18:17

anubhava

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shell: Validate for multiple alphabets in an if statement

1 Answers1