Shell脚本对文件进行计数，然后删除最旧的文件 [英] Shell script to count files, then remove oldest files

问题描述

我是Shell脚本的新手，因此在这里需要一些帮助.我有一个充满备份的目录.如果我有10个以上的备份文件，我想删除最旧的文件，以便仅剩下10个最新的备份文件.

I am new to shell scripting, so I need some help here. I have a directory that fills up with backups. If I have more than 10 backup files, I would like to remove the oldest files, so that the 10 newest backup files are the only ones that are left.

到目前为止，我知道如何对文件进行计数，这似乎很容易，但是如果计数超过10，该如何删除最旧的文件?

So far, I know how to count the files, which seems easy enough, but how do I then remove the oldest files, if the count is over 10?

if [ls /backups | wc -l > 10]
    then
        echo "More than 10"
fi

推荐答案

尝试一下:

ls -t | sed -e '1,10d' | xargs -d '\n' rm

这应该处理文件名中的所有字符(换行符除外).

This should handle all characters (except newlines) in a file name.

这是怎么回事?

• ls -t以修改时间的降序列出当前目录中的所有文件.即，最近修改的文件是第一个，每行一个文件名.
• sed -e '1,10d'删除前10行，即10个最新文件.我用它代替tail，因为我永远不记得需要tail -n +10还是tail -n +11.
• xargs -d '\n' rm收集每行输入(不包含换行符)，并将每行作为参数传递给rm.

• ls -t lists all files in the current directory in decreasing order of modification time. Ie, the most recently modified files are first, one file name per line.
• sed -e '1,10d' deletes the first 10 lines, ie, the 10 newest files. I use this instead of tail because I can never remember whether I need tail -n +10 or tail -n +11.
• xargs -d '\n' rm collects each input line (without the terminating newline) and passes each line as an argument to rm.

与任何此类物品一样，请在安全的地方进行实验.

As with anything of this sort, please experiment in a safe place.

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