按字母顺序排序时删除除 N 个文件之外的所有文件的 Bash 脚本

Question

It's hard to explain in the title.

I have a bash script that runs daily to backup one folder into a zip file. The zip files are named world YYYYMMDD .zip with YYYYMMDD being the date of backup. What I want to do is delete all but the 5 most recent backups. Sorting the files alphabetically will list the oldest ones first, so I basically need to delete all but the last 5 files when sorted in alphabetical order.

Answer 1

The following line should do the trick.

ls -F world*.zip | head -n -5 | xargs -r rm

• ls -F : List the files alphabetically
• head -n -5 : Filter out all lines except the last 5
• xargs -r rm : remove each given file. -r : don't run rm if the input is empty

Answer 2

How about this:

find /your/directory -name 'world*.zip' -mtime +5 | xargs rm

Test it before. This should remove all world*.zip files older than 5 days. So a different logic than you have.

Answer 3

我现在无法测试它，因为我没有 Linux 机器，但我认为它应该是：

rm `ls -A | head -5`

Answer 4

ls | grep ".*[\.]zip" | sort | tail -n-5 | while read file; do rm $file; done

• sort sorts the files
• tail -n-5 returns all but the 5 most recent
• the while loop does the deleting

Answer 5

ls world*.zip | sort -r | tail n+5 | xargs rm

sort -r will sort in reversed order, so the newest will be at the top

tail n+5 will output lines, starting with the 5th

xargs rm will remove the files. Xargs is used to pass stdin as parameters to rm.