Ubuntu – Listing files with grep

bashcommand linegrepls

I have a directory that contains the following files:

file1a file1ab file12A2 file1Ab file1ab

I want to list all files that start with file1 and followed by two letter at most!

The solution I have proposed is as follows:

ls | grep -i file1 [az] {2}

But it does not work!

I want to know why? and how to list?

Best Answer

You don't need piping, grep or ls. Just use shell globbing.

In bash, using extglob pattern (should be enabled by default in interactive sessions, if not do shopt -s extglob to set it first):

file1@(|?|??)

? matched any single character, @(||) selects any of the patterns separated by |.

If you only meant to match any characters between a-z and A-Z, use use character class [:alpha:] which denotes all alphabetic characters in the current locale:

file1@(|[[:alpha:]]|[[:alpha:]][[:alpha:]])

Example:

$ ls -1
file1
file112
file11a
file12A2
file1a
file1ab
file1Ab
file1as
file2
fileadb

$ ls -1 file1@(|[[:alpha:]]|[[:alpha:]][[:alpha:]]))
file1
file1a
file1ab
file1Ab
file1as

zsh supports this natively:

file1(|[[:alpha:]]|[[:alpha:]][[:alpha:]])

I am answering this portion very reluctantly, upon request from OP.

Any future reader, Don't parse ls, use globbing.

Using ls and grep:

ls | grep -E '^file1[[:alpha:]]{,2}$'

Example:

% ls | grep -E '^file1[[:alpha:]]{,2}$'
file1
file1a
file1ab
file1Ab
file1as
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