[SOLVED] CentOS8 if condition being ignored

witchkinkofangmar · 08-14-2020, 02:14 PM

Trying to differentiate between CentOS7 and CentOS8 and the CentOS8 condition is being skipped and I'm not sure why.

Code:

release=$(cat /etc/redhat-release | awk '{print $1,$2}')
release8="cut -c22 /etc/redhat-release"

if [[ -d "$sepDir" ]]; then
  echo "$(tput setaf 1)$(tput setab 7)Symantec Endpoint Protection is already installed.$(tput sgr 0)"
elif [[ ! -d "$sepDir" && $release == 'Red Hat' ]]; then
  echo "$(tput setaf 1)$(tput setab 7)RHEL Install.$(tput sgr 0)"
  sepInstallRHEL
elif [[ ! -d "$sepDir" && "$release8" == '8' ]]; then
  echo "CentOS 8 install"
  sepInstallCentOS8
elif [[ ! -d "$sepDir" && $release == 'CentOS Linux' || $release == 'CentOS release' ]]; then
  echo "$(tput setaf 1)$(tput setab 7)CentOS Install.$(tput sgr 0)"
  sepInstallCentOS
else
  echo "An error has occurred. Please contact your Systems Administrator."
fi

boughtonp · 08-14-2020, 02:50 PM

You're setting release8 to the string "cut -c22 /etc/redhat-release" instead of evaluating it like you do for the variable one line previous.

Using Bash's -x mode should make this obvious when you run the script.

Also, grabbing the 22nd character of every line doesn't seem like a great way to identify version.

Do those OSes not use /etc/os-release, where you can simply grab the value of the VERSION_ID line, e.g.

Code:

grep -oP '(?<=VERSION_ID=")\d+' /etc/os-release