I want to remove files that have not the string '999' (without the '') in their name.

I have tried:

grep -vlr 999 . | xargs -0 rm -f --
find . -print0 | grep --null-data -v 999 | xargs -0 rm --

But none of them works. I am using macOS Sierra, with bash: 3.2.57.

  • 2
    grep -l causes it to list the files where a match was found (or not found with -v) in the content, not the filename. grep always matches against the content of the files you specify, never their names. – JoL May 16 '18 at 0:52

Using an extended globbing pattern in bash:

rm ./!(*999*)

This requires shopt -s extglob to be enabled (and for safety, also shopt -s failglob, so that no file with the unusual name !(*999*) gets removed by mistake if all filenames contain 999). The pattern !(*999*) will match any name in the current directory, except those names matching *999*. To also remove hidden files (files whose name starts with a dot), also enable the dotglob option.

To only care about regular files or symbolic links to regular files (not directories etc.):

for name in ./!(*999*); do [ -f "$name" ] && rm "$name"; done

The zsh shell equivalent to the above loop would be

rm ./(^(*999*))(.)

Your first command will not work since grep will look inside files. It would remove all files that have lines without 999 in them (had you added the --null option so it works with xargs -0).

Your second command will not work since grep on macOS does not support --null-data (it does however have a --null option, but only for when generating filename output). Also note that it would look for 999 anywhere in the file's path (including directory components), not only the file's name.

  • 1
    You'd also want to enable the failglob option or it could end up removing the file called !(*999*) if there's no other file whose name doesn't contains 999. – Stéphane Chazelas Dec 9 '19 at 13:04

Just invert the name condition in find :

find . -type f \! -name "*999*" 

Add -delete or -exec rm {} + to actually remove the matched files.

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