Compare multi-digit version numbers in bash

Question

Trying to write a script that searches for the version of the Application then returns the value. My problem is the value is three to four intergers long (example 4.3.2).

I have searched for a while and can't find any syntax that would allow you to use a != or -ge for anything higher than a number with periods in it. Just wondering if anyone has a better way or I will just keep adding for every version release.

What I want

else if [ $version1 -ge "9.0.8" ]; then

How it is written now

vercheck=`mdls -name kMDItemVersion /Applications/iMovie.app`
version=`echo ${vercheck:17}`
version1=`echo ${version:1:5}`

[...]

else if [ $version1 = "9.0.8" ]; [ $version1 = "9.1.1" ]; then
    echo "You already have this version or a higher version installed"
    exit 0

Answer 1

Actually, comparing version numbers is pretty straightforward (at least as long as they are strictly numeric) as they are hierarchically structured left to right. A sequential comparison in that same order will yield a clear result.

The following bash function will return 0 (true) if two version numbers are not equal, 1 (false) if they are, as long as the variables $version_1 and $version_2 both contain only an arbitrary number of digit groups separated by periods:

function versions_not_equal {
    while [[ $version_1 != "0" || $version_2 != "0" ]]; do
        (( ${version_1%%.*} != ${version_2%%.*} )) && return 0
        [[ ${version_1} =~ "." ]] && version_1="${version_1#*.}" || version_1=0
        [[ ${version_2} =~ "." ]] && version_2="${version_2#*.}" || version_2=0
    done
    false
}

Implementing other comparisons, like greater or equal, is as simple as changing the comparison operator of the arithmetic evaluation (i.e. (( ${version_1%%.*} >= "${version_2%%.*}" ))).