how to bash script menu in one row only

Question

I am creating a menu in a script (#!/bin/bash), and it works as I needed:

PS3='[ menu ] choose partition tool:' options=("fdisk" "cfdisk" "continue") select opt in "${options[@]}" do case $opt in "fdisk") fdisk /dev/sda ;; "cfdisk") cfdisk ;; "continue") break ;; *) echo "invalid option $REPLY" ;; esac done 

this is the result on screen:

1) fdisk 2) cfdisk 3) continue [ menu ] choose partition tool: 

how can I transform the layout like this (?):

[ menu ] choose partition tool (1) fdisk (2) cfdisk (3) continue: 

in a nutshell, I know that it is enough to write the options in PS3= like this: PS3='[ menu ] choose partition tool (1) fdisk (2) cfdisk (3) continue:' but I want to know how not to bring up the options list (up on description).

Answer 1

You have to write the code for your custom version of select yourself. Something like this might work:

#!/bin/bash # PS3='[ menu ] choose partition tool:' options=('fdisk' 'cfdisk' 'continue') while : do printf '%s' "$PS3" for ((i = 0; i < ${#options[@]}; i++)) do printf ' (%d) %s' $((i+1)) "${options[i]}" done read -p ': ' opt case "$opt" in 1|"fdisk") fdisk /dev/sda ;; 2|"cfdisk") cfdisk ;; 3|"continue") break ;; *) echo "invalid option $opt" ;; esac done