bash wildcard n digits

Question

So I've got the following files in the tmp directory:

 file.0
 file.1
 file.t9
 file.22
 file.4444

if I wanted to list only the files that end in '.digits' (0, 1, 22, 4444) but not (t9) I could try and use wildcards such as this:

 ls tmp/file.{[0-9],[0-9][0-9],[0-9][0-9][0-9],[0-9][0-9][0-9][0-9]}

however I get the following results with the ugly error

 ls: cannot access tmp/file.[0-9][0-9][0-9][0-9]: No such file or directory
 file.0
 file.1
 file.22
 file.4444

I've also tried using {0..999} but that also results in the same sorts of errors (and a lot more of them). Any clues as how to do this without errors from the experts?

Answer 1

At least in bash 4, when the extglob shell option is enabled:

shopt -s extglob
ls /tmp/file.+([0-9])

The pattern +([0-9]) there matches one or more digits.

You can read more about this in the Pattern Matching section of man bash.

UPDATE

Actually, as @chepner pointed out, since extglob was introduced in version 2.02, this should work in pretty much every bash you come across today, unless you pulled your distro out of a rock or something.

Answer 2

You can use ls just to list all the files then filter the output of that through grep:

ls -1 | grep -E '\.[0-9]+$'

as per the following test:

pax> printf 'file.0\nfile.1\nfile.t9\nfile.22\nfile.4444\n' | grep -E '\.[0-9]+$'
file.0
file.1
file.22
file.4444

The -E gives you extended regular expressions so that the + modifier works. If that's not available to you, you can use the \.[0-9][0-9]*$ regex instead.