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

With zsh you could use a glob like ?(#cN) (here the c flag requires the previous ? to match exactly N times):

setopt extendedglob
print -rl -- ?(#c12)

if you prefer ls:

ls -d -- ?(#c12)

You can also add qualifiers, e.g. search recursively for regular files with fixed name length:

print -rl -- **/?(#c12)(.)

Answer 3

With ksh93:

printf '%s\n' {12}(?)

for (non-hidden) files whose name are made of 12 characters.

Or if you prefer regular expressions:

printf '%s\n' ~(E)^.{12}$

Answer 4

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}$'

Answer 5

Another solution is to use echo command.

echo lazer_??????

Note that regular expression handling is done by whatever shell the echo command is running within.