Extract version using Regular expressions in bash

Question

All I need to do is extract the versioning information from the following file:

 my_archive_1.1.1.201_x86_64.tgz

I am trying to extract both the version number which is 1.1.1 and the release number which is 201. Normally I use python for these purposes, but I have been asked not to. How do I do it by just using bash? The filename will always be of the form

([A-Za-z_]+)_([0-9]+\.[0-9]+\.[0-9]+)\.([0-9]+)_x86_64\.tgz

The groups are in parenthesis. I need the second and third groups if you start counting from 1.

Answer 1

Use pure BASH:

s='my_archive_1.1.1.201_x86_64.tgz'
[[ $s =~ ^[^_]+_[^_]+_(([^.]+\.){2}[^.]+)\.([^_]+) ]] && \
        echo "${BASH_REMATCH[1]}, ${BASH_REMATCH[3]}"

OUTPUT:

1.1.1, 201

Using your own regex:

[[ $s =~ ([A-Za-z_]+)_([0-9]+\.[0-9]+\.[0-9]+).([0-9]+)_x86_64\.tgz ]] && \
        echo "${BASH_REMATCH[2]}, ${BASH_REMATCH[3]}"

Answer 2

You can use simple string substitutions to extract substrings. You don't really need regular expressions. As a bonus, this is portable to other POSIX shells. Whether this is simpler or not is a matter of taste, and also depends on the problem.

s='my_archive_1.1.1.201_x86_64.tgz'
# ${s%%_[0-9]*} is 'my-archive'
s=${s#${s%%_[0-9]*}_}
# s='1.1.1.201_x86_64.tgz'
s=${s%%_*}
# s='1.1.1.201'
release=${s##*.}
version=${s%."$release"}

You might also want to experiment with set:

s='my_archive_1.1.1.201_x86_64.tgz'
oldIFS=$IFS
IFS=_
set $s
# $1 = my, $2=archive, $3=1.1.1.201, $4=x86, $5=64.tgz
# Shift until $1 contains only numbers and periods
while $1; do
    case $1 in *[!.0-9]* ) shift ;; *) break ;; esac
done
IFS=.
set $1
version=$1.$2.$3
release=$4
IFS=$oldIFS