Exclude filename containing certain string, regexp

Question

Already saw some negative lookback answers here, for some reason i was not able to use ls -l with these.

Let's assume i have three files:

sample_one.zip
sample_two.zip
sample_three.zip

What i want is to simply ls -l (regex).zip and exclude filenames containing one and two. tried

   ls -l '*(?!one|two).zip'

but no luck!

It seems you can use shell globbing with `ls`, but not a regex. As for the lookarounds, you might try `grep` with `-P`: `ls | grep -P '(?<!one|two)\.zip$'` (if you have a GNU `grep`, it will work). — Wiktor Stribiżew, Jun 29 '17 at 09:22
i actually need it for "cp", and i don't have a grep option :( — aprin, Jun 29 '17 at 09:26
@WiktorStribiżew ls doesn't recognise shell globbing, it just expands before ls see's it. It takes literal strings only. — 123, Jun 29 '17 at 10:04
@123 Perhaps, I misinterpreted [this answer](https://stackoverflow.com/a/15346038/3832970). — Wiktor Stribiżew, Jun 29 '17 at 10:07

Raman Sailopal · Answer 1 · 2017-06-29T10:33:03.190

1

If you can utilize the find command, you could:

find -name "*.zip" . ! -regex ".*\(\one\|\two\).zip"

You could then use -exec to action the copy.

edited Jun 29 '17 at 10:33

answered Jun 29 '17 at 10:11

Raman Sailopal

1

You might want to add `-name '*.zip'` to make it more like what's being asked for. – Toby Speight Jun 29 '17 at 10:14

score 0 · Answer 2 · answered Jun 29 '17 at 10:06

0

Try rsync if that's a option

rsync -av --exclude '*one.zip' --exclude '*two.zip' /from /to

answered Jun 29 '17 at 10:06

papkass

2 Answers2