delete last 3 files

I have a little problem. Every time I deploy a release, I move the old one into an old-folder. The problem is: is there a way to delete all files in this old-folder except the 3 last files? kind of an opposite of ls -tr | tail -n 3?

I took half an hour and hacked a little script the easy and st00pid way:


#!/bin/bash
#
# Name: del-except-3
# Author: pes
# Date: 08.04.2010
# Description
# Deletes all files in ${searchpath} except the last 3
#

debug=1
searchpath=$1

if [ ! -d ${searchpath} ]; then
    echo "${searchpath} does not exist"
    exit 1
fi

cd ${searchpath}

last3=`ls -tr | tail -3`

set -- ${last3}

last3_1=$1
last3_2=$2
last3_3=$3

if [ ${debug} -ne 0 ]; then
    echo "last 3 files in ${searchpath}"
    echo " 1: $last3_1"
    echo " 2: $last3_2"
    echo " 3: $last3_3"
fi

for i in `ls -1`; do
    if [ "${i}" != "${last3_1}" ]; then
        if [ "${i}" != "${last3_2}" ]; then
            if [ "${i}" != "${last3_3}" ]; then
                echo "rm ${searchpath}/${i}"
                rm ${i}
            fi
        fi
    fi
done

is there a smart combination of shell commands? someting that does not look that stupid?