case + how to implement equal or less or greater in case syntax

Question

My target is to verify a range of number with (only with case + esac), and print the range. So for example:

• If the number is between 0 and 80, print >=0<=80
• If the number is between 81 and 100 then print >=81<=100
• etc.

The problem with my script below print only >=0<=90 only if the number between 0 and 9. How to fix my script, so that it will print correct output according to the number range?

#!/bin/ksh
read number 
case $number in 
 [0-80])  echo ">=0<=80";; 
 [81-100]) echo ">=81<=100";; 
 [101-120]) echo ">=101<=120";;
 [121-300]) echo ">=121<=300";;
esac

Answer 1

case is only for pattern matching, it won't do arithmetic evaluation (except maybe if you consider zsh's <x-y> extended pattern matching operator).

You could match numbers with patterns like:

case $(($number)) in
  ([0-9]|[1-7][0-9]|80) echo ">=0<=80";;
  (8[1-9]|9[0-9]|100) echo ">=81<=100";;
  ... and so on
esac

But you can easily see that it's not the right tool.

The [...] matches one character against the list of specified characters, so [121-300] matches for any character that is either 1, 2, 1 to 3, 0 or 0, so it's the same as [0-3] or [0123].

Use:

if [ "$number" -ge 0 ] && [ "$number" -le 80 ]; then
  echo ">=0<=80"
elif [ "$number" -ge 81 ] &&  [ "$number" -le 100 ]; then
  echo ">=81<=100"
elif ... and so on
  ...
fi

Answer 2

This is not very nice but you can use this :

 #!/bin/ksh

read number  

case $number in
[0-9]|[1-7][0-9]|80) echo  echo ">=0<=80";;
8[1-9]|9[0-9]|100) echo ">=81<=100";;
10[1-9]|11[0-9]|120) echo ">=101<=120";;
12[1-9]|130) echo ">=121<=300";;
esac