How can I find files with 7 characters (and no extension) in their names?

Question

I want to list files that contain only seven characters (without extensions) in their filename.

I have these files in a directory, for example:

1234567
bar.12
dog.cat
1234567.txt
tst.new
big.tst
abcdefg
abcd.efg

I want to list files using the command ls to find file names without extensions that have exactly 7 characters. I want the output to look something like this:

$ ls <some kind of code>
1234567
abcdefg

I prefer not to use any globs of any sort, or the find command.

Answer 1

Here's a sort-of solution with shell wildcards (exactly 7 characters that are not .)

ls [!.][!.][!.][!.][!.][!.][!.]

And here's a solution with find using regex negation (not literal . = [^\.]) and specifying how many characters {7}

find -maxdepth 1 -regextype posix-extended -regex '\./[^.]{7}'

Not what you asked for, but I doubt it can be done without some kind of wildcards or regex.

Answer 2

In Bash:

 for f in ??????? ; do 
     [[ $f = *.* ]] && continue; 
     echo $f ;                     # do something useful here
 done

With find:

find -maxdepth 1 -name "???????" ! -name "*.*"

Though since you asked for it without globs or find and used the word "professor", they maybe looking for ls | grep .... But see also a couple of reasons to avoid using ls for this.

Answer 3

With extended globs in zsh:

% ls
1234567  1234567.txt  abcdefg  abcd.efg  bar.12  big.tst  dog.cat  tst.new

then

% setopt extended_glob
% ls (?~.)(#c7)
1234567  abcdefg

• ?~. any character except .
• (#c7) a globbing flag equivalent to regular expression quantifier {7}

See zsh: 14 Expansion