How can I search for a file with fixed name length using ls?

Question

In a directory, I have files like

lazer_100506
lazer_100707
lazer_091211
lazer_110103
lazer_100406_temp
lazer_100622#delete

etc

How can I get a listing of only the first four files?

$ ls lazer_......
ls: lazer_......: No such file or directory
$ 

Answer 1

There are multiple methods:

ls only

ls lazer_??????

ls and egrep

ls | egrep '^lazer_.{6}$'

find

find . -regextype posix-egrep -regex '^./lazer_.{6}$'

Answer 2

As pointed out by SiegeX, Shell alone does not understand regular expressions. If you want a precise filter of your files, you must use regular expressions and hence use a command like egrep.

Here, the files you want to list begin with lazer_ and are followed only by some digits (possibly more or less that 6). I would do it this way:

ls | egrep '^lazer_[[:digit:]]*$'

This regex works the same as '^lazer_[0-9]*$'.

Regular expressions with egrep also handles repetition just like in the answer of wag, if you want to restrict your list to files ending with exactly 6 digits:

ls | egrep '^lazer_[[:digit:]]{6}$'