使用 shell 脚本删除前 10 个最大的常规文件

Question

I'm trying to delete the first largest regular files from the given directory, but it doesn't work for files which contain whitespace caracters.

My code (it works if the files doesn't contain whitespace caracters):

find mydir -type f -exec du -ahb {} + | sort -n -r | cut -f2 | head -n 10 | xargs rm -i

I also tried this , but it gives an error message:

find mydir -type f -exec du -ahb {} + -print 0 | sort -n -r | cut -f2 | head -n 10 | xargs -0 rm -i

Answer 1

The following should work at least with GNU coreutils 8.25 and newer:

find mydir -type f -exec du -0b {} + | sort -znr | cut -zf2 | head -zn 10 |  xargs -0pn 1 rm

I made sure every command handled and outputted NUL bytes ( \0 ) separated records rather than linefeed separated records:

• du outputs NUL-separated records with -0
• sort , cut and head handle and output NUL-separated records with -z
• xargs handles NUL-separated records with -0

Additionally, I removed the interactive mode of rm and asked xargs to handle that instead ( -p ), because xargs didn't provide a prompt to rm when invoking it. I had to limit the number of parameters given at once to rm to 1 for this to work ( xargs ' -n 1 parameter). There might be a way to preserve the -i and provide rm with an interface to your prompt, but I don't know how.

Last point: I removed du 's -h uman-readable mode because it would have made the sort often fail and it didn't serve any purpose since the filesizes were never displayed to an human.