Question

Good evening,
I want to write a script that will say if its argument is a number or not, however I get this:

a: 4: Syntax error: "(" unexpected (expecting ")")

At first I tried like this:

    #!/bin/bash
    case "$1" in
        0x* ) echo hex ;;
        ''|*[!0-9]*) echo dec ;;
        * ) echo wtf ;;
    esac

Later I have found this bit:

    if ! [[ "$yournumber" =~ ^[0-9]+$ ]] ; then
       exec >&2; echo "error: Not a number"; exit 1
    fi

Well, none of them works, I call for help!

Answer 1

I’m not sure what you mean by *[!0-9]*, but the following seems to work for me.

#!/bin/bash

shopt -s extglob

case "$1" in
    '')
        echo missing ;;
    0x+([0-9a-fA-F]))
        echo hex ;;
    +([0-9]))
        echo dec ;;
    *)
        echo wtf ;;
esac

Note that extglob lets me use the more complex pattern match to find “hex” and “dec” numbers. You can find info on this in the bash man page.