Get Ubuntu version number only with bash

Question

I execute command cat /etc/os-release in order to get info like

NAME="Ubuntu"
VERSION="18.04 LTS (Bionic Beaver)"
ID=ubuntu
ID_LIKE=debian
PRETTY_NAME="Ubuntu 18.04 LTS"
VERSION_ID="18.04"
HOME_URL="https://www.ubuntu.com/"
SUPPORT_URL="https://help.ubuntu.com/"
BUG_REPORT_URL="https://bugs.launchpad.net/ubuntu/"
PRIVACY_POLICY_URL="https://www.ubuntu.com/legal/terms-and-policies/privacy-policy"
VERSION_CODENAME=bionic
UBUNTU_CODENAME=bionic

I want to get only 18.04 from VERSION_ID

I tried like

cat /etc/os-release | grep VERSION_ID=

which returns

VERSION_ID="18.04"

I don't know how to take that version number. Can advice me how to do that ?

How about lsb_release -sr in Bash/Terminal?

stephanmg
– stephanmg

2020-04-21 14:29:07 +00:00
Commented Apr 21, 2020 at 14:29 — stephanmg
– stephanmg, Commented Apr 21, 2020 at 14:29

John Kugelman · Accepted Answer · 2020-04-21 14:22:29Z

7

Source it and you can access the variable directly:

. /etc/os-release
echo "$VERSION_ID"

answered Apr 21, 2020 at 14:22

John Kugelman

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JRFerguson · Accepted Answer · 2020-04-21 14:33:25Z

4

Using lsb_release is an ideal way to access distribution information. To get only the version, use:

lsb_release -sr

If lsb_release isn't installed on your system (if I recall correctly) do:

sudo apt-get install lsb-core

answered Apr 21, 2020 at 14:33

JRFerguson

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1 Comment

stephanmg Over a year ago

Should be provided with the installation of the OS version already. Didn't think my comment was worth an answer. :)

anubhava · Accepted Answer · 2020-04-21 14:23:41Z

1

You may use awk:

awk -F= '$1 == "VERSION_ID" {gsub(/"/, "", $2); print $2}' /etc/os-release

18.04

answered Apr 21, 2020 at 14:23

anubhava

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