Generating a comma-separated sequence of numbers based on input in Bash

Question

I've found more than few things on here to help me as I'm learning to code in Bash and they all come close but not quite.

I need to take an input of a positive integer and print out on a single line down to one, all separated by commas, without a comma at the end of the last variable.

This is what I have so far:

#!/bin/bash
#countdown

read -p "Enter a Number great than 1: " counter

until ((counter < 1)); do
echo -n ",$counter"
((counter--))
done

It almost works out but I can't figure out how to prevent the comma in front and not have it behind the last variable.

EDIT: You guys are AMAZING. Poured over this book and learned more in ten minutes here than I did with an hour there.

So is there some sort of command I could use to ensure it was only one number entered and ensure it had to be positive?

Some way to put an if statement on the read to ensure its <= 1 and only one character?

I only have a background in some basic C coding, so I have the basics but translating them to BASH is harder than expected

Answer 1

Use seq with the -s option:

seq -s, $counter -1 1 

Answer 2

Probably simper way using brace expansion:

#!/bin/bash
#countdown

read -p "Enter a Number great than 1: " counter

eval printf "%s" {${counter}..2}, 1

Test:

Enter a Number great than 1: 10
10,9,8,7,6,5,4,3,2,1

To validate the input, you can use regular expressions:

#!/bin/bash
#countdown

read -p "Enter a Number great than 1: " counter

if [[ ${counter} =~ ^[1-9][0-9]*$ ]]
then
  eval printf "%s" {${counter}..2}, 1
fi