[SOLVED] Regex digit matching in bash

Issue

I'm writing a way to check if input is a 4 digit pin code. I cant figure out how digit matching \d works in bash. ^\d{4}$ is more elegant solution than ^[0-9][0-9][0-9][0-9]$ but it never matches. Is there some way that I can use ^\d{4}$?

Example:

#!/bin/bash -e

REGEX='^[0-9][0-9][0-9][0-9]$'
ALTREGEX='^\d{4}$'
CHECK="1234"
if [[ ! $CHECK =~ $REGEX ]]
then
    echo "$REGEX didnt match"
else 
    echo "$REGEX matched"
fi
if [[ ! $CHECK =~ $ALTREGEX ]]
then
    echo "$ALTREGEX didnt match"
else 
    echo "$ALTREGEX matched"
fi

I've assumed there's some issue with the variables so I've changed the variable literacy around with no success.

Solution

Bash uses basic regular expressions. \d is part of a Perl-compatible regular expression (PCRE).

Quantifiers such as {n} work the same in BRE.

In this case you could use ^[0-9]{4}$

Answered By - mashuptwice

Answer Checked By - Terry (WPSolving Volunteer)