Question 26820 in CentOS Question posted in
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How to match one of two strings as a condition in bash – CentOS

firstly I had very simple script like this

#!/bin/sh
if cat /etc/redhat-release | grep -q 'AlmaLinux'; then
    echo "your system is supported"
    MY SCRIPT HERE
else
    echo "Unsupported OS"
    exit0;
fi

and it works but I want to add another correct value that will also return "your system is supported" and let me pass the script to go further

so for example if file /etc/redhat-release will contain AlmaLinux or Rockylinux 8 it will work for both AlmaLinux and Rockylinux but if it will contain Centos 6 it will not go further

I tried something along this:

#!/bin/sh
if cat '/etc/redhat-release' | grep -q 'AlmaLinux'|| | grep -q 'RockyLinux 8'; then
    echo "your system is supported"
else
    echo "Unsupported OS"
fi

but it gives me an error and I am even not sure if this is a right syntax.

Can anyone help me?

Tags:

3

Answers


  1. 0

    Maybe by using a regex pattern?

    #!/bin/sh
if cat '/etc/redhat-release' | grep -q -E 'AlmaLinux|RockyLinux 8'; then
    echo "your system is supported"
else
    echo "Unsupported OS"
fi
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  2. 0

    Try this:

    #!/bin/sh
grep -qE 'AlmaLinux|RockyLinux 8' /etc/redhat-release
if [ $? ]; then
  echo "your system is supported"
else
  echo "Unsupported OS"
fi
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  3. 0

    Other perfectly valid detection:

    #!/bin/sh

NAME=unknown
VERSION='??'
if . /etc/os-release && case $NAME in
  *Rocky*)
    case $VERSION in
      8*) ;;
      *) false ;;
    esac
    ;;
  *AlmaLinux*) ;;
  *) false ;;
esac
then
  printf 'Your system %s version %s is supportedn' "$NAME" "$VERSION"
else
  printf '%s %s is not supported!n' "$NAME" "$VERSION" >&2
  exit 1
fi
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