6

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

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0

5 Answers 5

Reset to default
16

Use seq with the -s option:

seq -s, $counter -1 1
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6 Comments

Looks like this has the same problem of a trailing comma
@Kevin: counter=5; seq -s, $counter -1 1 outputs 5,4,3,2,1 for me.
5,4,3,2,1, (without a newline) for me, on OSX
The version of seq that ships with OS X does seem buggy, although the following hack seems to remove the trailing comma: seq -s, -t$'\b' 10 1. The backspace will overwrite the trailing comma.
This was a HUGE mega help. Made life so much easier.
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6

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
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1 Comment

Not safe if you don't validate the input value. Consider counter having the value 3}; echo arbitrary code; : {3.
2

One way

read -p "Enter a Number great than 1: " counter
echo -n "$counter"
((counter--))
until ((counter < 1)); do
echo -n ",$counter"
((counter--))
done
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2 Comments

+1. It's also possible to do the reverse, using a condition of ((counter <= 1)) and printing 1 after the end of the loop.
Oooh oooh! Okay okay so don't actually try to get it to be one statement but instead a ton of smaller statements
1

A slightly awkward construction using an array, the seq command, and a subshell to localize a change to the IFS parameter will work.

read -p "Enter a Number great than 1: " counter
range=( $(seq $counter -1 1) )
( IFS=,; echo "${range[*]}" )
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1 Comment

Hm, kind of pointless, since I forgot about the -s option to seq and didn't read choroba's answer closely before posting.
1

Here is another way … kinda influenced by chepner's solution but not using seq:

Content of script.sh:

#!/bin/bash

read -p "Enter a Number great than 1: " counter
range=( $(eval echo {$counter..1}) )
( IFS=,; echo "${range[*]}" )

Test:

$ bash script.sh
Enter a Number great than 1: 5
5,4,3,2,1
$ bash script.sh
Enter a Number great than 1: 30
30,29,28,27,26,25,24,23,22,21,20,19,18,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1
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