I'm working on a simple bash script that should be able to run on Ubuntu and CentOS distributions (support for Debian and Fedora/RHEL would be a plus) and I need to know the name and version of the distribution the script is running (in order to trigger specific actions, for instance the creation of repositories). So far what I've got is this:
OS=$(awk '/DISTRIB_ID=/' /etc/*-release | sed 's/DISTRIB_ID=//' | tr '[:upper:]' '[:lower:]')
ARCH=$(uname -m | sed 's/x86_//;s/i[3-6]86/32/')
VERSION=$(awk '/DISTRIB_RELEASE=/' /etc/*-release | sed 's/DISTRIB_RELEASE=//' | sed 's/[.]0/./')
if [ -z "$OS" ]; then
OS=$(awk '{print $1}' /etc/*-release | tr '[:upper:]' '[:lower:]')
fi
if [ -z "$VERSION" ]; then
VERSION=$(awk '{print $3}' /etc/*-release)
fi
echo $OS
echo $ARCH
echo $VERSION
This seems to work, returning ubuntu or centos (I haven't tried others) as the release name however. However, I have a feeling that there must be an easier, more reliable way of finding this out... No -- is that true?
It doesn't work for RedHat. /etc/redhat-release contains : Redhat Linux Entreprise release 5.5
So, the version is not the third word, you'd better use :
OS_MAJOR_VERSION=`sed -rn 's/.*([0-9])\.[0-9].*/\1/p' /etc/redhat-release`
OS_MINOR_VERSION=`sed -rn 's/.*[0-9].([0-9]).*/\1/p' /etc/redhat-release`
echo "RedHat/CentOS $OS_MAJOR_VERSION.$OS_MINOR_VERSION"
