Command to yield just the linux distribution name only

Question

Suppose I have a set of pre-requisites I need to install for any of a number of linux distributions. In this case, I'd need to switch my install expression against the distribution.

I am aware of uname -v, and I am aware that I can inspect the output of this command with some friable string expression and logic. However what I really need is:

V=$(something-like-uname)
echo $V

"Debian"
# or
"Gentoo"
# or
"Redhat"
# or 
"Arch"
# etc.

What is this command? (Does it exist?)

AFAIK there is no posix standard for this so it's going to be different from distribution to distribution. There are plenty of examples out there of code that determines distribution though. — jesse_b, Dec 07 '19 at 20:25
What about packaging using the various common package systems, like apt and yum, instead of trying to write an installer that does the right thing? Or write a Linuxbrew formula? — Kusalananda, Dec 07 '19 at 20:26

score 2 · Accepted Answer · answered Dec 07 '19 at 20:35

I'd always check lsb_release -i first and then check /etc/os-release as the fallback. It should be available on many distributions but you can never count on it being everywhere of course. It's Freedesktop standard and is present on systemd systems. On Slackware it shows:

$ cat /etc/os-release
NAME=Slackware
VERSION="14.2"
ID=slackware
VERSION_ID=14.2
PRETTY_NAME="Slackware 14.2"
ANSI_COLOR="0;34"
CPE_NAME="cpe:/o:slackware:slackware_linux:14.2"
HOME_URL="http://slackware.com/"
SUPPORT_URL="http://www.linuxquestions.org/questions/slackware-14/"
BUG_REPORT_URL="http://www.linuxquestions.org/questions/slackware-14/"

And on Raspbian:

$ cat /etc/os-release
PRETTY_NAME="Raspbian GNU/Linux 10 (buster)"
NAME="Raspbian GNU/Linux"
VERSION_ID="10"
VERSION="10 (buster)"
VERSION_CODENAME=buster
ID=raspbian
ID_LIKE=debian
HOME_URL="http://www.raspbian.org/"
SUPPORT_URL="http://www.raspbian.org/RaspbianForums"
BUG_REPORT_URL="http://www.raspbian.org/RaspbianBugs"

The field you're interested here is probably ID.

xenoid · Answer 2 · 2019-12-07T20:33:30.490

1

I would use:

grep -Po "(?<=^ID=).+" /etc/os-release | sed 's/"//g'

edited Dec 07 '19 at 20:33

answered Dec 07 '19 at 20:25

xenoid

8,888

This file doesn't exist on my centos machine, or on my debian machine. – jesse_b Dec 07 '19 at 20:26
-bash: lsb_release: command not found – jesse_b Dec 07 '19 at 20:28
Edited the answer to use os-release that seems to exist on Debian, Ubuntu, and Centos – xenoid Dec 07 '19 at 20:34
/etc/os-release doesn't seem to be on my debian machine – jesse_b Dec 07 '19 at 20:35
It is in my debian and centos docker images. – xenoid Dec 07 '19 at 20:41
1

Yeah my debian is pretty old, debian 6. I think OPs issue is much more complicated than a single command though. – jesse_b Dec 07 '19 at 20:43

score 1 · Answer 3 · answered Dec 07 '19 at 20:26

1

lsb_release -i might do what you want if it is installed. Otherwise the trick it to check for files like /etc/redhat-release, /etc/debian_version, etc.

$ lsb_release -i 
Distributor ID: Fedora

answered Dec 07 '19 at 20:26

Mark McKinstry

15,393

Command to yield just the linux distribution name only

3 Answers3