How to get Ubuntu distribution's full code name?

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

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/*, /}')

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.

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.

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

Here are some more choices. They all parse the /etc/os-release file which, on my 13.10, looks like this:

NAME="Ubuntu"
VERSION="13.10, Saucy Salamander"
ID=ubuntu
ID_LIKE=debian
PRETTY_NAME="Ubuntu 13.10"
VERSION_ID="13.10"
HOME_URL="http://www.ubuntu.com/"
SUPPORT_URL="http://help.ubuntu.com/"
BUG_REPORT_URL="http://bugs.launchpad.net/ubuntu/"

All of the solutions below will parse the second line to produce Saucy Salamander.

1. grep

   grep -oP 'VERSION=.* \K\w* \w*' /etc/os-release
   

   The -o means "print only the matching part of the line" and the -P enables Perl Compatible Regular Expressions. This lets us use \K which discards whatever was matched up to that point, which combined with -o means "print only what matches after the \K. So, the actual regular expression used will match the last two words of a line that contains VERSION=.

2. awk

   awk -F'[" ]' '/VERSION=/{print $3,$4}'  /etc/os-release
   

   Setting the fields separator to " and space means that the 3d and 4rth fields of the line containing VERSION= are the string we're after.

3. sed

   sed -nr '/VERSION=/{s/.* (\w* \w*)"/\1/;p}' /etc/os-release
   

   The -n switch suppresses normal output, no lines will be printed. The -r allows extended regular expressions. Then, on lines that match VERSION=, we will delete everything except the last two words. The p at the end means print.

4. perl

   perl -lne '/VERSION=.*\b(\w+ \w+)/ && print $1' /etc/os-release
   

   The -n means 'process every input line with the script given by -e". The -l adds a newline character to every print call (and some other stuff which is not relevevant here). The regular expression matches the last two words (\b is a word boundary) and prints them if the line contains VERSION=.

5. coreutils

   grep VERSION= /etc/os-release | cut -d ' ' -f 2-  | tr -d '"' 
   

   Here, we just grep the relevant line and use cut setting the field separator to space (-d ' ') and printing everything from the 2nd field to the end of the line. The tr -d command will delete the " from the end.

6. Pure shell (shamelessly stealing @gniourf_gniourf's clever source idea):

   . /etc/os-release && echo ${VERSION//[0-9,. ]/ }
   

   The . sources the file which makes the variables available to the shell and I use bash's string manipulation capabilities to remove the version numbers.

. /etc/os-release 
echo $VERSION

Using regex you could do this:

grep 'VERSION' /etc/os-release | grep -oP "[a-zA-Z]+ [a-zA-Z]+"

Explanation:

This searches the line with VERSION in the file /etc/os-release. Then it finds 2 successive words seperated by a space (Perl regex). The -o flag keeps only what matches the search.

You have to use [a-zA-Z] instead of `w` to avoid including digits.