Bash Script to Get Operating System + Version as string - bash

Question

I want to make a bash script to take the system os and the version as a simple string.
Possible ways to get these info is from
/etc/issue
cat /etc/*-release
lsb_release -a
and probably some others which i dont know. The problem is that i want the bash script to work on Ubuntu 12,13,14 and CentOS. Some of the above does not work in these systems. For example the lsb_release does not work on CentOS and sometimes the /etc/issue is empty so i'm little confused about it.
As for the string i want to get it in this way (and save it to var). I will give examples.
If OS is Ubuntu 12.x i want to take it as ubuntu12
If OS is Ubuntu 13.x i want to take it as ubuntu13
If OS is CentOS 7.x i want to take it as centos7
Is that easy?
THANK YOU

Answer 1

Here is a bash solution. I tested on Ubuntu, but not on CentOS (I only have RHEL available now). But you can test the CentOS part and modify as needed.
#!/usr/bin/env bash
RELEASE=unknown
version=$( lsb_release -r | grep -oP "[0-9]+" | head -1 )
if lsb_release -d | grep -q "CentOS"; then
RELEASE=centos$version
elif lsb_release -d | grep -q "Ubuntu"; then
RELEASE=ubuntu$version
fi
echo $RELEASE
Or, without lsb_release on CentOS:
#!/usr/bin/env bash
RELEASE=unknown
if [ -f /etc/redhat-release ]; then
version=$( cat /etc/redhat-release | grep -oP "[0-9]+" | head -1 )
RELEASE=centos$version
elif [ -n $(which lsb_release 2> /dev/null) ] && lsb_release -d | grep -q "Ubuntu"; then
version=$( lsb_release -d | grep -oP "[0-9]+" | head -1 )
RELEASE=ubuntu$version
fi
echo $RELEASE
In any case, there's more than one way to skin this cat.