bash：删除所有超过 1 个月的文件，但保留周一的文件
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/20034415/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
bash: Delete all files older than 1 month, but leave files from Mondays
提问by freento
I have a lot of daily backup archives. To manage disk usage, I need a bash script that will delete all files older than 1 month, but keep all files created on Mondays, even if they are older than 1 month.
我有很多日常备份档案。为了管理磁盘使用情况，我需要一个 bash 脚本来删除所有超过 1 个月的文件，但保留在星期一创建的所有文件，即使它们超过 1 个月。
For example, this will delete all files last modified more than 30 days ago:
例如，这将删除 30 多天前最后修改的所有文件：
find /path/to/files* -type f -mtime +30 -delete
But I don't really know how to keep files created on Mondays.
但我真的不知道如何保留周一创建的文件。
回答by janos
Slightly simpler and more cautious version of @JoSo's answer:
@JoSo 回答的稍微简单和谨慎的版本：
find /path/to/files -type f -mtime +30 \
-exec sh -c 'test $(date +%a -r "") = Mon || echo rm ""' -- {} \;
The differences:
区别：
- Using
date -rto get the last modification date of a file directly - Using
%ato work with more comprehensible weekday names - Just echo the
rm "$1"first to review what will be deleted. If looks good, then either stick| shat the end to really execute, or remove theecho
- 使用
date -r直接获取文件的最后修改日期 - 使用
%a工作更加理解工作日名称 - 只需回显
rm "$1"第一个即可查看将要删除的内容。如果看起来不错，那么要么坚持| sh最后真正执行，要么删除echo
However, @JoSo is right to point out that date +%ais locale dependent, so these versions would be indeed safer:
然而，@JoSo 正确地指出这date +%a是依赖于语言环境的，所以这些版本确实更安全：
find /path/to/files -type f -mtime +30 \
-exec sh -c 'test $(date +%u -r "") = 1 || echo rm ""' -- {} \;
find /path/to/files -type f -mtime +30 \
-exec sh -c 'test $(LC_TIME=C date +%a -r "") = Mon || echo rm ""' -- {} \;
回答by Jo So
As findto my knowledge has no weekday check, you need to call an external program.
据find我所知没有工作日检查，你需要调用一个外部程序。
find /path/to/files -type f -mtime +30 \
-exec sh -c \
'[ "$(date +%u -d @"$(stat -c %Y "")")" != 1 ] && rm ""' -- {} \;
Update: Using the -rswitch to date(Kudos to Janos) and only testing, not deleting inside the shell command probably yields the cleanest possible version:
更新：使用-r切换到date(Kudos to Janos) 和只测试，而不是在 shell 命令中删除可能会产生最干净的版本：
find /path/to/files -type f -mtime +30 \
-exec sh -c 'test "$(date +%u -r "")" != 1' -- {} \; \
-print # or -delete instead
回答by user2196728
you can use
您可以使用
stat -c %y yourfile
to get the date the file was created.
获取文件的创建日期。
Then extract the date with cut to have a var like myvar=yyyymmdd
然后使用 cut 提取日期以具有类似 myvar=yyyymmdd 的 var
and finally use
最后使用
date +%u --d $myvar
date +%u --d 20130822
it will return the day of week, if it returns 1 it was monday. In my case it return 4 because 22/08/2013 was a Thursday
它将返回星期几，如果返回 1 则是星期一。就我而言，它返回 4，因为 22/08/2013 是星期四
Edit : if you can get all these working as a simple command line as Jo So suggest, it's better !
编辑：如果你能像 Jo So 建议的那样将所有这些作为一个简单的命令行工作，那就更好了！
回答by brablc
What about avoiding subprocesses in loop:
如何避免循环中的子进程：
find /path/to/files -type f -mtime +30 -printf '%Ta %p\n' \
| grep -v ^Mon | cut -c5- | tr "\n" "##代码##" | xargs -0 rm -v
