修改shell脚本以删除文件夹和文件

发布于 2024-09-17 17:50:15 字数 575 浏览 3 评论 0原文

我的 shell 脚本：

#!/bin/bash
if [ $# -lt 2 ]
then
    echo "$0 : Not enough argument supplied. 2 Arguments needed."
    echo "Argument 1: -d for debug (lists files it will remove) or -e for execution."
    echo "Followed by some path to remove files from. (path of where to look) "
    exit 1
fi

if test $1 == '-d'
then
    find $2 -mmin +60 -type f -exec ls -l {} \;
elif test $1 == '-e'
then
    find $2 -mmin +60 -type f -exec rm -rf {} \;
fi

基本上，这将在作为第二个参数提供的给定目录中查找文件，并列出（-d 表示参数 1）或删除（-e 表示参数 1）60 分钟前修改的文件。

我怎样才能修改它以删除文件夹？

原文

My shell script:

#!/bin/bash
if [ $# -lt 2 ]
then
    echo "$0 : Not enough argument supplied. 2 Arguments needed."
    echo "Argument 1: -d for debug (lists files it will remove) or -e for execution."
    echo "Followed by some path to remove files from. (path of where to look) "
    exit 1
fi

if test $1 == '-d'
then
    find $2 -mmin +60 -type f -exec ls -l {} \;
elif test $1 == '-e'
then
    find $2 -mmin +60 -type f -exec rm -rf {} \;
fi

Basically this will find files in a given directory provided as second argument and either list (-d for argument 1) or remove (-e for argument 1) files modified >60 minutes ago.

How can I rework this to also remove folders ?

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

有木有妳兜一样 2024-09-24 17:50:15

删除 -type f
将 ls -l 更改为 ls -ld

更改 1 将列出所有内容，而不仅仅是文件。这也包括链接。如果您不喜欢列出/删除文件和目录以外的任何内容，那么您需要单独列出/删除文件和目录，如下所示：

if test $1 == '-d'
then
    find $2 -mmin +60 -type f -exec ls -ld {} \;
    find $2 -mmin +60 -type d -exec ls -ld {} \;
elif test $1 == '-e'
then
    find $2 -mmin +60 -type f -exec rm -rf {} \;
    find $2 -mmin +60 -type d -exec rm -rf {} \;
fi

需要更改 2，因为目录上的 ls -l 会列出文件在目录中。

Remove -type f
changing ls -l to ls -ld

Change 1 will list everything and not just files. This includes links as well. If you are not fine with listing/deleting anything other than files and directories then you need to separately list/delete files and directories as:

if test $1 == '-d'
then
    find $2 -mmin +60 -type f -exec ls -ld {} \;
    find $2 -mmin +60 -type d -exec ls -ld {} \;
elif test $1 == '-e'
then
    find $2 -mmin +60 -type f -exec rm -rf {} \;
    find $2 -mmin +60 -type d -exec rm -rf {} \;
fi

Change 2 is needed as ls -l on a directory will list the files in the directories.

回复收藏 0 原文

撧情箌佬 2024-09-24 17:50:15

#!/bin/bash
if [ $# -lt 2 ]
then
    echo "$0 : Not enough argument supplied. 2 Arguments needed."
    echo "Argument 1: -d for debug (lists files it will remove) or -e for execution."
    echo "Followed by some path to remove files from. (path of where to look) "
    exit 1
fi

if test $1 == '-d'
then
    find $2 -mmin +60 -type d -exec ls -l {} \;
    find $2 -mmin +60 -type f -exec ls -l {} \;
elif test $1 == '-e'
then
    find $2 -mmin +60 -type d -exec rm -rf {} \;
    find $2 -mmin +60 -type f -exec rm -rf {} \;
fi

那应该对你有用。

#!/bin/bash
if [ $# -lt 2 ]
then
    echo "$0 : Not enough argument supplied. 2 Arguments needed."
    echo "Argument 1: -d for debug (lists files it will remove) or -e for execution."
    echo "Followed by some path to remove files from. (path of where to look) "
    exit 1
fi

if test $1 == '-d'
then
    find $2 -mmin +60 -type d -exec ls -l {} \;
    find $2 -mmin +60 -type f -exec ls -l {} \;
elif test $1 == '-e'
then
    find $2 -mmin +60 -type d -exec rm -rf {} \;
    find $2 -mmin +60 -type f -exec rm -rf {} \;
fi

That should work for you.

回复收藏 0 原文

~没有更多了~