Compare multiple string in case shell script

Question

I am comparing string in case statement as below : input variable can be D P KL ...

case $input in
      D|P|Q|L1|L2)
         val="Hello";;
      DD|DE|DF|CA)
          val="Hi" ;;
      MM|KL||TK|SZ)
         val="Bye" ;;
         echo $input

input variable not printing anything..

What is the value in $input? Also, the case statement is partial (not closed by esac), and you're not using $val anywhere. — Kusalananda, Mar 29 '18 at 07:39
Should always be case "$input". It seems to me that the esac is missing. Or shall $input be printed only in the last case? And why $input and not $val? Strange code... — Hauke Laging, Mar 29 '18 at 07:42
@HaukeLaging Quoting $input in the case statement is good practice, but actually not necessary. https://unix.stackexchange.com/questions/68694/when-is-double-quoting-necessary (note, I always quote the expansion in case statements) — Kusalananda, Mar 29 '18 at 07:51

score 8 · Accepted Answer · edited Mar 29 '18 at 08:40

There are two main issues in you script:

The case statement is not closed by esac.
The third pattern contains || which is a syntax error in most Bourne-like shells (use '' or "" or an expansion that resolves to an empty value to match on the empty string portably)

It's unclear what your script is actually doing, so I'm speculating a bit and wrote this:

#!/bin/sh

input="$1"

case "$input" in
      D|P|Q|L1|L2)
          val='Hello' ;;
      DD|DE|DF|CA)
          val='Hi' ;;
      MM|KL|""|TK|SZ)
          val='Bye' ;;
      *)
          echo 'error' >&2
          exit 1
esac

printf 'input was "%s", val is "%s"\n' "$input" "$val"

Testing it:

$ ./script.sh D
input was "D", val is "Hello"

$ ./script.sh MM
input was "MM", val is "Bye"

$ ./script.sh BOO
error

Compare multiple string in case shell script

1 Answers1