Comparing PHP version numbers using Bash?

Question

I have this script that should make sure that the users current PHP version is between a certain range, though it SHOULD work, there is a bug somewhere that makes it think t

Answer 1

Here's another solution that:

• does not run any external command apart from tr
• has no restriction on number of parts in version string
• can compare version strings with different number of parts

Note that it's Bash code using array variables.

compare_versions()
{
    local v1=( $(echo "$1" | tr '.' ' ') )
    local v2=( $(echo "$2" | tr '.' ' ') )
    local len="$(max "${#v1[*]}" "${#v2[*]}")"
    for ((i=0; i

The function returns:

• 0 if versions are equal (btw: 1.2 == 1.2.0)
• 1 if the 1st version is bigger / newer
• 2 if the 2nd version is bigger / newer

However #1 -- it requires one additional function (but function min is quite usable to have anyway):

min()
{
    local m="$1"
    for n in "$@"
    do
        [ "$n" -lt "$m" ] && m="$n"
    done
    echo "$m"
}

However #2 -- it cannot compare version strings with alpha-numeric parts (though that would not be difficult to add, actually).