Bash + print default character after user prompt

Question

-I made a bash function like this:

find_stuff () {
list=`find $SEARCHPATH -type f -user $USERID -mmin -$SOMETIME -not -path '*/\.*' -exec ls -1rt {} \+`
     if [ -n "$list" ]; then
        for i in $list; do
            echo "Found item: "$i""; printf 'Is this ok? y/n [y]:'
            read arg
            case "arg" in
                y|\Y)
                    do_stuff
                    ;;
                n|\N)
                    continue
                    ;;
                *)
                    echo "Please type 'y'=yes or 'n'=no, -exiting..."
                    exit 0
                    ;;
            esac
        done
     else
         echo "No items found, exiting..."
         exit 0
     fi
}

Now I want to expand the function for the enduser like this:

./finditems.sh

Trying to find stuff and then do some stuff after confirmation.

Found item: /dir/mod.item

Is this ok? y/n [y]:y

---

How can I print the "user removable" y character on the user's shell prompt?

I tried some variations with read -p, but it didn't give me the result that I want.

Answer 1

This implementation seems to complex. I have found another solution: By adding double quotes to y|\Y|"") I have eliminated the need for a 'y' character and questionable or unnecessary code. By pressing 'return' after the "Is this ok? y/n [y]:" statement the for-loop continues as expected.