How can I programatically determine the Linux distribution?

Question

There are various ways of finding out what Linux distribution is currently running, using commands such as:

uname -a
cat /etc/*-release
cat /etc/*-version
lsb_release -ri

Each of them provides me with human-readable information about my system, but they don't seem reliable nor can be easily parsed. Is there some script that can determine popular Linux distributions with accuracy and output something simple like Ubuntu / Slackware / CentOS / Mandriva / etc?

Answer 1

1

Not sure what all you want it to show.

$ lsb_release -is

will give you just the Distribution

link

answered 13 Jul '11, 14:05

DMon
11●1
accept rate: 0%

That only works on some distributions. For example, CentOS doesn't have the lsb_release executable.

(13 Jul '11, 14:58) fardelian

CentOS 6 does. Just tried it on there. But you are correct not every Distro does it the same however lsb seems to be more common in the newest batch of updates.

(13 Jul '11, 15:20) DMon

Furthermore not all distributions report something useful using lsb_release. My ArchLinux configuration prints "n/a" for example.

(13 Jul '11, 15:58) Jazz ♦

Answer 2

0	Here's one. You could enhance this script by first using the other ways of finding out which distribution is running before using uname. A combination of those commands should do the trick for any distribution. link answered 13 Jul '11, 16:04 Jazz ♦ 781●1●3●12 accept rate: 33%

Answer 3

Here's a skeletan of a Korn/Bash function to do it:

function get_distro { if [[ -r /etc/redhat-relase ]]; then set -- $(cat /etc/redhat-relase) case $1 in Red) echo Redhat;; Fedora) echo Fedora;; CentOS) echo CentOS;; *) echo ??? esac elif [[ -r /etc/debian_version ]]; then echo Debian else echo UNKNOWN fi }

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How can I programatically determine the Linux distribution?

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