command line - 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?

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