Shell script: list all entries that don't contain a number

Question

I tried

#!/bin/bash
ls * [!0-9] * .*

but that doesn't work - I still get some files that contain a number.

score 3 · Accepted Answer · answered Feb 01 '14 at 22:02

3

If

shopt extglob

says

extglob         on

then you can try

ls !(*[0-9]*)

Use the following to enable this option

shopt -s extglob

answered Feb 01 '14 at 22:02

Nik O'Lai

+1. If you wanted to robustly process all matching files inside the script, you can use `for f in !(*[0-9]*); do ...; done` – mklement0 Feb 01 '14 at 23:48

Red Cricket · Answer 2 · 2014-02-02T06:11:37.903

2

how about good ol grep like so ... ls | grep -v "[0-9]"

Note: removed the * as suggested by BMW.

edited Feb 02 '14 at 06:11

answered Feb 01 '14 at 21:57

Red Cricket

Nice alternative to having to deal with the `extglob` shell option. – mklement0 Feb 01 '14 at 23:21
1

what's the reason using ls with `*`? – BMW Feb 02 '14 at 05:06
@BMW you are correct. Just `ls | grep -v "[0-9]"` would have sufficed. – Red Cricket Feb 02 '14 at 06:10

2 Answers2