bash allow only positive integer or positive integer with decimal point

Question

price=\"1.11\"
case $price in
            \'\'|*[!0-9]*) echo \"It is not an integer.\";
            ;;
        esac

Output: It is not an integer

Answer 1

Doing this in a POSIX-compatible way is tricky; your current pattern matches non-integers. Two bash extensions make this fairly easy:

1. Use extended patterns

   shopt -s extglob
   case $price in
       # +([0-9]) - match one or more integer digits
       # ?(.+([0-9])) - match an optional dot followed by zero or more digits
       +([0-9]))?(.*([0-9]))) echo "It is an integer" ;;
       *) echo "Not an integer"
   esac
   

2. Use a regular expression

   if [[ $price =~ ^[0-9]+(.?[0-9]*)$ ]]; then
       echo "It is an integer"
   else
       echo "Not an integer"
   fi
   

(In theory, you should be able to use the POSIX command expr to do regular expression matching as well; I'm having trouble getting it to work, and you haven't specified POSIX compatibility as a requirement, so I'm not going to worry about it. POSIX pattern matching isn't powerful enough to match arbitrarily long strings of digits.)

---

If you only want to match a "decimalized" integer, rather than arbitrary floating point values, it is of course simpler:

1. The extended pattern +([0-9])?(.)
2. The regular expression [0-9]+\.?

Answer 2

Try this regular expression ^(0*[1-9][0-9]*(\.[0-9]+)?|0+\.[0-9]*[1-9][0-9]*)$