bash脚本按文件夹名称删除日期文件夹 [英] bash script to remove date folders by folder name
问题描述
我有一个由当前日期创建一个备份文件夹文件夹中的脚本。该脚本通过cron运行,每天一次,每一天。
有没有办法通过文件名来删除文件夹年长超过3天?像
日期-3?
脚本的作品:谢谢乔所以。此脚本创建日期的文件夹。 COM presses文件进行备份,坚持他们在您的备份目录,并清除了超过3天以上的备份: - )
#!/斌/庆典 CD /家庭/备份 的mkdir $(日期+%Y-%M-%D) CD /选择/ 焦油-pczf /家庭/备份/ $(日期+%Y-%M-%D)/opt.tar.gz code CD的/ var / 焦油-pczf /家庭/备份/ $(日期+%Y-%M-%D)/var.tar.gz工作CD /家庭/备份/
threedaysago =`日期-D4天前+%Y%M%D`对于在备份[0-9] [0-9] [0-9] [0-9] - [0-9] [0-9] - [0-9] [0-9]
做
backupdate =`回声$备份| TR -d -`#删除破折号 如果测试$ backupdate-lt$ threedaysago
然后
室射频$备份
科幻
DONE
threedaysago =`日期-D4天前+%Y%M%D`对于在备份[0-9] [0-9] [0-9] [0-9] - [0-9] [0-9] - [0-9] [0-9]
做
backupdate =`回声$备份| TR -d -`#删除破折号 如果测试$ backupdate-lt$ threedaysago
然后
室射频$备份
科幻
DONE
独立的mtime的工作,我可以告诉你,它不会在特别奇怪的角落情况下打破; - )
I have a script which creates folders in a backup folder by the current date. This script is run once a day, every day via cron.
Is there a way to remove the folders older than 3 days via folder name? something like
date -3 ?
Script that works: Thank you to Jo So. This script creates a folder by date. Compresses the files for backup, sticks them in your backup directory and clears out backups older than 3 days :-)
#!/bin/bash
cd /home/backups
mkdir $(date +%Y-%m-%d)
cd /opt/
tar -pczf /home/backups/$(date +%Y-%m-%d)/opt.tar.gz code
cd /var/
tar -pczf /home/backups/$(date +%Y-%m-%d)/var.tar.gz work
cd /home/backups/
threedaysago=`date -d "3 days ago" +%Y%m%d`
for backup in [0-9][0-9][0-9][0-9]-[0-9][0-9]-[0-9][0-9]
do
backupdate=`echo "$backup" | tr -d -` # remove dashes
if test "$backupdate" -lt "$threedaysago"
then
rm -rf "$backup"
fi
done
threedaysago=`date -d "3 days ago" +%Y%m%d`
for backup in [0-9][0-9][0-9][0-9]-[0-9][0-9]-[0-9][0-9]
do
backupdate=`echo "$backup" | tr -d -` # remove dashes
if test "$backupdate" -lt "$threedaysago"
then
rm -rf "$backup"
fi
done
Work independently of mtime, and I can tell you that it will not break under particularly strange corner cases ;-)
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