Matching n letter filename with ls

Question

I am using the ls command in bash, trying to find all files or directory of length n. Let's say n=5

My command is:

ls ?????

But this would also include characters that are non letters such as period. For example, the following files would match:

ab.cd    
abd.c

I only want to match files that have 5 letter or number names:

five1
five2    
five3

But not

abc.d    
ab.cd    
a.bcd

How can I modify my command?

Answer found:

ls [a-zA-Z0-9][a-zA-Z0-9][a-zA-Z0-9][a-zA-Z0-9][a-zA-Z0-9][a-zA-Z0-9][a-zA-Z0-9]

I found the answer but how can I make this less ugly?

Answer 1

Note that it's not ls that interprets those globs. Those globs are expanded by your shell into a list of file names that is passed as arguments to ls. Different shells have different globbing capabilities. bash has a few extensions over standard globs (borrowed from ksh88 and enabled with shopt -s extglob) but is still limited compared to shells like zsh or ksh93.

With zsh:

setopt extendedglob
ls -d [[:alnum:]](#c5)

ksh93:

ls -d {5}([[:alnum:]])

or:

ls -d {5}(\w) # (\w includes underscore in addition to alnums)

or, if you wanted to use extended regular expressions:

ls -d ~(E)^[[:alnum:]]{5}$

With bash or other POSIX shells which don't have equivalent globbing operators, you'd need to do:

ls -d [[:alnum:]][[:alnum:]][[:alnum:]][[:alnum:]][[:alnum:]]

Note that [[:alnum:]] includes any alphabetic character in the current locale (not only latin alphabets let alone the English one) and 0123456789 (and possibly other types of digits). If you want the letters in the English alphabet, name characters individually:

c='[0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ]'
unset -v IFS
ls -d $c$c$c$c$c

Or use the C locale:

(export LC_ALL=C
ls -d [[:alnum:]][[:alnum:]][[:alnum:]][[:alnum:]][[:alnum:]])

Answer 2

ls -q | grep -Ex '[[:alnum:]]{5}'

Answer 3

If you're going to use ls in the way you mentioned, you should use the -d option; otherwise, it will list the contents of any directories whose names are five letters and/or digits, rather than listing the names themselves.  Also, you can do

ls -d [[:alnum:]][[:alnum:]][[:alnum:]][[:alnum:]][[:alnum:]]

but that's exactly as much typing as the answer you have now.

Also, if you don't need to use ls in your command, you could use

echo [[:alnum:]][[:alnum:]][[:alnum:]][[:alnum:]][[:alnum:]]

which will list all the matches on one line, or

printf "%s\n" [[:alnum:]][[:alnum:]][[:alnum:]][[:alnum:]][:alnum:]]

which will list them on separate lines.

Answer 4

I don't know if you'll count this as "less uglier" but you could use GNU find like so:

 find -maxdepth 1 -regextype posix-extended -regex './[[:alnum:]]{5}'

though that will put ./ in front of each entry, though you could sed that away

Depending on what you do next find may also have the advantage of helping to avoid the Parsing ls troubles

Keeping it with extglobs (shopt -s extglob) you could rewrite it a little, though again I don't know if you count it as cleaner

ls [[:alnum:]][[:alnum:]][[:alnum:]][[:alnum:]][[:alnum:]]

Answer 5

Build up the expression in a variable:

 e=""; for m in $(seq 1 5); do e="$e[A-Za-z0-9]"; done

Put it in a function to be fancy/reusable:

 alnumglob() {
    local e=""
    for m in $(seq 1 $1) ; do e="$e[A-Za-z0-9]"; done
    echo "$e"
 }

 ls -ld $(alnumglob 5)

Answer 6

If you want to safely determine the length of a string with bash, you should use parameter expansion. ls alone is only capable of glob syntax, which (likely) cannot do what you want. find has different implementations on BSD and some Linuxes and -regextype isn't necessarily a legal flag. The good news is, bash has loops, and these can give you what you want.

for filename in *              # globs all files in your directory.
do
    clip=${filename%.*}        # excludes the first extension
    if [[ ${#clip} -eq 5 ]]    # test the length of the remaining string
    then
        ls -d $filename        # call ls to show you the file or directory
    fi
done

If instead you need any filename solely with five alphanumeric chars, your method and other answers will only return those files which begin with five chars. For example:

Using ls [a-zA-Z0-9][a-zA-Z0-9][a-zA-Z0-9][a-zA-Z0-9][a-zA-Z0-9]

five1.txt      # matches
fi_ve.txt      # fails
a.pages        # fails
q_fiver.txt    # fails

To include any file with any five-character long alphanumeric string, and only those with any five-character long alphanumeric string, you can use grep's more universal regex implementation under bash. Although I wouldn't normally recommend it, the use of the \w and \W here can help readability a great deal (where \w = [[:alnum:]] and \W = [^[:alnum:]] – but they will include underscores, so use at peril).

for filename in *
do
    if (grep -qE '(^|\W)\w{5}($|\W)' <<<"$filename")
    then
        ls -d "$filename"
    fi
done

Answer 7

If all you want is to make it less ugly, then you can change [a-zA-Z0-9] into [[:alnum:]], the assign that to a variable:

a='[[:alnum:]]'

And use it five times:

$ ls $a$a$a$a$a             ### important that the vars are not quoted.  
five1 five2 five3