How to get Ubuntu distribution's full code name?

$ lsb_release -c
Codename:   trusty

$ cat /etc/issue
Ubuntu 14.04 LTS \n \l

$ cat /etc/lsb-release
DISTRIB_ID=Ubuntu
DISTRIB_RELEASE=14.04
DISTRIB_CODENAME=trusty
DISTRIB_DESCRIPTION="Ubuntu 14.04 LTS"

Output's of the above commands shows only the partial code name (ie, trusty). How do I get the full codename (trusty tahr) of my installed Ubuntu system?

Solution 1:

Using no external tools:

You can just source (the source command is a dot .) the /etc/os-release and you'll have access to all the variables defined there:

$ . /etc/os-release
$ echo "$VERSION"
14.04, Trusty Tahr

Edit. If you want to remove the 14.04, part (as asked by terdon), you could:

$ . /etc/os-release
$ read _ UBUNTU_VERSION_NAME <<< "$VERSION"
$ echo "$UBUNTU_VERSION_NAME"
Trusty Tahr

Note that this is a bit clunky, since on other distributions, the VERSION field can have different format. E.g., on my debian,

$ . /etc/os-release
$ read _ UBUNTU_VERSION_NAME <<< "$VERSION"
$ echo "$UBUNTU_VERSION_NAME"
(wheezy)

Then, you could imagine something like this (in a script):

#!/bin/bash

if [[ -r /etc/os-release ]]; then
    . /etc/os-release
    if [[ $ID = ubuntu ]]; then
        read _ UBUNTU_VERSION_NAME <<< "$VERSION"
        echo "Running Ubuntu $UBUNTU_VERSION_NAME"
    else
        echo "Not running an Ubuntu distribution. ID=$ID, VERSION=$VERSION"
    fi
else
    echo "Not running a distribution with /etc/os-release available"
fi

Solution 2:

My variant on what's already offered:

. /etc/os-release; echo ${VERSION/*, /}

The shortest, Bashiest answer to date.

If you don't care to load /etc/os-release's contents into your current environment, you can fake bash into thinking it's loading a script fairly easily:

bash <(cat /etc/os-release; echo 'echo ${VERSION/*, /}')

Solution 3:

Grep:

$ grep $(lsb_release -rs) /usr/share/python-apt/templates/Ubuntu.info | grep -m 1 "Description: Ubuntu " | cut -d "'" -f2
Trusty Tahr

Explanation:

• lsb_release -rs -> Prints your installed Ubuntu version.

• grep $(lsb_release -rs) /usr/share/python-apt/templates/Ubuntu.info -> Grab all the lines which contains your release version, in my case it's 14.04.

• grep -m 1 "Description: Ubuntu " -> Again grabs only the matched first line(because of -m flag) which contains the string Description: Ubuntu.

• cut -d "'" -f2 -> prints the field number 2 according to the delimiter single quote '

Awk:

$ awk -v var=$(lsb_release -rs) '$3~var {print $4" "$5;exit;}' /usr/share/python-apt/templates/Ubuntu.info | cut -d"'" -f2
Trusty Tahr

Explanation:

Declaring and assigning Variable in awk is done through -v parameter.So the value of lsb_release -rs command is assigned to the variable var which inturn helps to print field 4 ,field 5 from the lines contain the string 14.04 and exists if its found an one.Finally the cut command helps to remove the single quotes.

Solution 4:

The command you are looking for is:

grep -oP '(?<=VERSION\=\"(\d\d)\.(\d\d)\,\ )(.*?)(?="$)' /etc/os-release

This is very ugly and not optimized. I'm sure there should be an easier method and this has some issues.

Solution 5:

Answer relevant for at least ubuntu 16.04 and above:

lsb_release -cs

If for some reason lsb_release is not available

cat /etc/os-release | grep UBUNTU_CODENAME | cut -d = -f 2