How can I check that two input values are correct in a bash script?

Question

I am writing a bash script and want to add some simple validation tests. The input should be either 1 or 2 and the script should ask for input again until the correct input is given.

I was thinking of doing this using an until loop but it didn't work. I tied until [ $group -eq 1 ] (where $group is the value given by the user) but obviously I also need a 2nd input. How can I do this correctly?

Answer 1

until
  printf 'Please enter 1 or 2: '
  IFS= read -r line || exit # exit on EOF
  [ "$line" = 1 ] || [ "$line" = 2 ]
do
  printf >&2 '"%s" is neither 1 nor 2, try again.\n' "$line"
done

With IFS= and -r, the input line is stored as-is in $line. You may want to omit the IFS= so that blanks (assuming you've not modified $IFS) are automatically stripped from the beginning or end of the input.

If the input has to be the 1 or 2 strings only, then you want to use =, not -eq.

If you want to allow other expressions of the 1 or 2 numbers, like 01, 1+1, 1.0, 100e-2, exp(0), RANDOM (sometimes), then you could use -eq, but note that not all shells accept all types of expressions above (ksh93 will accept them all, but bash would only accept things like 1 (leading and/or trailing blanks), or 0001 (leading zeros)), and that means you'll get error messages for inputs that don't form valid arithmetic expressions.

With shells that interpret arithmetic expressions, that's also dangerous as that can change variable values (like for an input like PATH=123) or even run arbitrary commands (like for inputs like a[0$(cmd>&2)]).

Depending on the shell, you could also get false positives on 18446744073709551617 or 4294967297 (or any other multiple of 2³² or 2⁶⁴ + 1 or 2) as most shells use 64bit or 32bit integer numbers.

Whatever you do, do not use the -o and -a binary test/[ operators. Those should be banished as they make the parsing of the [ command potentially ambiguous and in practice unreliable (and are now marked obsolete in the POSIX spec).

For instance:

$ line='!' sh -c '[ "$line" = 1 -o "$line" = 2 ]'
sh: line 0: [: too many arguments

And remember to quote your variables ("$line", not $line which would undergo split+glob).

Answer 2

This should work, combine them with or (-o):

#init group variable
group=0
#read until value is either 1 or 2
until [ $group -eq 1 -o $group -eq 2 ]; 
    do 
    #your script...

    #read user input and put it in group
    read -p "Which group?" group 
done

You can also use || as or operator:

until [ $group -eq 1 ] || [ $group -eq 2 ]

Although in some cases, for example when you use the error code ($?) the first options is better, see this answer.

Answer 3

while IFS= read -r; do
    case "$REPLY" in
        1|2) break ;;
        *)   echo 'Error' >&2 ;;
    esac
done

This code snipped reads a value from standard input and then tests it against 1 and 2. If matching either string, the loop exits, otherwise a terse diagnostic message is produced and the loop continues.

Answer 4

Just add the second condition to the expression using the or operator -o.