Bash script help in getting the first digit from a number

Question

How do I get the first digit from a number? For example, 25 - to get only the "2" from the 25.

This is what I tried:

echo -n "Enter age: "    
read age    
echo $(s:0:1)

Answer 1

You're using the wrong variable s instead of age and parameter expansion works with curly braces ${...}:

read -p "Enter age: " age
echo "${age:0:1}"

Answer 2

The most basic and POSIX compatible way (assuming no sign) is:

echo "${age%"${age#?}"}"

In some shells (bash,ksh,zsh):

echo "${age:0:1}"

If the value could contain a sign, remove it first:

$ age=+23
$ age=${age#[+-]}
$ echo "${age%"${age#?}"}"
2

Answer 3

As you mention bash in the title:

#!/bin/bash

read -p 'Enter a number: '
printf '%.1s\n' "$REPLY"

This prints only the first character of the string $REPLY (read by read).

To store this in another variable, othervar, use

printf -v othervar '%.1s' "$REPLY"

You could also just use a simple variable substitution, as mentioned by others,

printf '%s\n' "${REPLY:0:1}"

where 0 is the zero-based position in the string where we want to start pulling out data, and 1 is the length of the substring that we want.

You can assign ${REPLY:0:1} to another variable directly:

othervar=${REPLY:0:1}

Answer 4

POSIX compatible solution:

#!/usr/bin/env sh

printf "Enter age: "
read -r age
echo "$age"  
echo first digit: "$(echo "$age" | cut -c1)"

Example:

$ ./script.sh
Enter age: 25
25
first digit: 2