Regex matches on regxr but not in bash

Asked Aug 27 '20 at 10:09

Active Aug 27 '20 at 10:18

Viewed 41 times

I have a regex:

^(\d+\.)?(\d+\.)?(\*|\d+)-SNAPSHOT$

This matches 1.30.0-SNAPSHOT using regexr:

I now tried to check this in bash:

#!/bin/bash
REGEX="^(\d+\.)?(\d+\.)?(\*|\d+)-SNAPSHOT$"
VERSION="1.30.O-SNAPSHOT"
if [[ $VERSION =~ $REGEX ]]
then
  echo "yes"
else
  echo "not"

This prints not

asked Aug 27 '20 at 10:09

mealesbia

1

Bash doesn't have `\d`. Use `[0-9]` instead. – M. Nejat Aydin Aug 27 '20 at 10:10
1

Beside using `[0-9]` instead of `\d`, you have the capital letter `O` instead of `0` after `30.` Is it a typo? – Wiktor Stribiżew Aug 27 '20 at 10:12
Use: `'^[0-9]+(\.[0-9]+)*-SNAPSHOT$'` as regex – anubhava Aug 27 '20 at 10:15
See https://ideone.com/8vlHBz – Wiktor Stribiżew Aug 27 '20 at 10:18

0 Answers0