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

I wrote this inelegant function a while back for a similar problem. vers_limit will return 0 if arg1 is less than or equal to arg2, non-0 otherwise:

vers_limit()
{
VERNEW=$1
VERLMT=$2

CHKNEW=$VERNEW
CHKLMT=$VERLMT

PASSED=

while :
do
        PARTNEW=${CHKNEW%%.*}
        PARTLMT=${CHKLMT%%.*}
        if [[ $PARTNEW -lt $PARTLMT ]]
        then
                PASSED=GOOD
                break
        elif [[ $PARTNEW -gt $PARTLMT ]]
        then
                PASSED=
                break
        else
                NXTNEW=${CHKNEW#*.}
                if [[ $NXTNEW == $CHKNEW ]]
                then
                        if [[ $NXTNEW == $CHKLMT ]]
                        then
                                PASSED=GOOD
                                break
                        else
                                NXTNEW=0
                        fi
                fi
                NXTLMT=${CHKLMT#*.}
                if [[ $NXTLMT == $CHKLMT ]]
                then
                        NXTLMT=0
                fi
        fi
        CHKNEW=$NXTNEW
        CHKLMT=$NXTLMT
done
test "$PASSED"
}

This is not as compact as some of the other solutions here, nor does it provide 3-way status (i.e., less, equal, greater), though I believe one can order the arguments, interpret the pass/fail status, and/or call twice to accomplish any desired result. That said, vers_limit does have certain virtues:

1. No calls to external utilities such as sort, awk, gawk, tr, etc.

2. Handles numeric versions of arbitrary size (up to the shell's limit for integer calculations)

3. Handles an arbitrary number of "dotted" parts.