如何使用shell脚本从备份中删除多余的文件？

Question

备份 Shell 脚本

#!/bin/bash

backdest=/home/backup
date=$(date "+%F")

backupall="$backdest/arch-full-$date.tgz"
backuphome="$backdest/jary_p-$date.tgz"

tar -czpvf $backupall / --exclude=/home/* --exclude=/mnt/* --exclude=/media/* \
                        --exclude=/proc/* --exclude=/sys/* --exclude=/dev/*   \
                        --exclude=/tmp/* --exclude=/lost+found/*

tar -czpvf $backuphome /home/jary_p

数（5）次后
/home/backup 中有 Serveral(10) 个文件

$ls /home/backup
backup.sh
arch-full-2011-05-13.tgz
arch-full-2011-05-25.tgz
arch-full-2011-06-01.tgz
arch-full-2011-06-09.tgz
arch-full-2011-06-11.tgz
jary_p-2011-05-13.tgz
jary_p-2011-05-25.tgz
jary_p-2011-06-01.tgz
jary_p-2011-06-09.tgz
jary_p-2011-06-11.tgz

我怎样才能保留最新的 3 个文件(6) 并删除多余的文件？
谢谢
并且，抱歉我糟糕的英语。

Answer 1

ls -t $backdest/jary_p-*.tgz | tac | tail -n +3 | xargs rm

并使用 $backupall 的 glob 重复

Answer 2

我已经找到如何解决它

ls -htr *.tgz|head -n -6 | xargs rm