Check system name and kernel version in script

Question

How do I make something like:

!#/bin/bash
if (system name == Jessie)
#script
if (system name == Wheezy)
#script
else
echo "It won't work on your Raspberry Pi version, please update"

if (kernel version == >= 4.4)
#script
if (kernel version == <= 4.4)
#script
else
echo "Bad kernel version"

Answer 1

You can try something like this. You may have to enhance this but the below script will work for you syntactically and also will tell which files to lookup and verify.

#!/usr/bin/bash

debian_version=`egrep "jessie|wheezy" /etc/os-release`
if [ `echo $debian_version|grep -c "jessie"` == 1 ]
then
   echo "jesse"
elif [ `echo $debian_version|grep -c "wheezy"` == 1 ] 
then
   echo "wheezy"
else
   echo "Some other version"
fi 


kernel_version=`uname -a|awk '{print $3}'`
if [ $kernel_version == "4.4" ]
then
    echo "version is 4.4"
fi

Answer 2

Here's something which takes a different twist on /etc/os-release, which has a standardized format based around setting shell variables, meaning you can source it and use them. First have a look at the actual file to get the idea, then consider:

#!/bin/bash

source /etc/os-release

case $VERSION_ID in
"7")
    echo "Raspbian 7 is wheezy."
    ;;
"8")
    echo "Raspbian 8 is jessie."
    ;;
"9")
    echo "Raspbian 9 is stretch."
    ;;
*)
    echo "This is $NAME, not Raspbian."
    ;;
esac

It doesn't include anything about the kernel for which parsing uname -r (or /proc/version) is more appropriate. Also, it could be made more robust by actually checking $NAME to make sure this isn't some other Debian/Raspbian derived distro using the same version numbering (although what you are doing will likely work with them anyway as they are parallel, which is why there's a Debian 6 but no Raspbian 6).

BTW stay away from Raspbian 9, it is still considered "testing".