如何删除超过 (n) 天的文件但保留 (n) 个文件而不考虑年龄? [英] How can I delete files over (n) days old but leave (n) files regardless of age?

问题描述

我在 PHP 中编写了以下内容，但我想知道是否有一种优雅的方法可以在 Linux shell 脚本中执行此操作?基本上删除 (n) 天前的文件，但保留 (n) 最新的文件，无论年龄大小.

I wrote the following in PHP but I was wondering if there is an elegant way to do this in a Linux shell script? Basically delete files over (n) days old, but leave the (n) newest files regardless of age.

    PHP

    foreach (glob("backup/*.db") as $file) {
        $a[$file]=date("Y-m-d",filemtime($file));
    }
    $i=0;
    arsort($a);
    foreach($a as $file=>$date) {
        if ($i++>=10) {
            if ($date<=date("Y-m-d",strtotime("-10 days"))) {
                unlink($file);
                xmessage("PURGED: $file");
            }   
        }   
    }   

我的想法是用find -mtime +(n) exec rm"删除但只在不在head -n +(n)"中的文件中使用管道?但是head -n"似乎并没有像我想的那样做.谢谢.

My idea was to delete with "find -mtime +(n) exec rm" but only pipe in the files that are NOT in "head -n +(n)"? But "head -n" does not seem to do what I thought it would. Thanks.

SHELL SCRIPT

find -mtime +10 | ls -t *.DB.tar.gz | head -n -10

推荐答案

使用所有 GNU find、sort、awk 和 xargs 试试这个:

Try this using all GNU find, sort, awk, and xargs:

find . -type f -printf '%Ts %p' |
sort -k1,1nr -sz |
awk -v days=10 -v cnt=10 '
    BEGIN { RS=ORS=""; secs=systime()-(days*24*60*60) }
    (NR>cnt) && ($1>secs) { print gensub(/S+s+/,"",1) }
' |
xargs -0 ls --

完成测试后将 ls 更改为 rm 并确保它为您提供预期的输出.

Change ls to rm when you're done testing and sure it's giving you the expected output.

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