How can I get a count of files in a directory using the command line?

Question

I have a directory with a large number of files. I don't see a ls switch to provide the count. Is there some command line magic to get a count of files?

Answer 1

Using a broad definition of "file"

ls | wc -l

Answer 2

set -- *; echo $#

This is obviously generalizable to any glob. The -- can be omitted if you're sure the first file doesn't begin with -.

In a script, this has the sometimes unfortunate side effect of overwriting the positional parameters. You can work around this with a function (Bourne/POSIX version, you can write it more clearly in fancy shells) (warning, typed directly into the browser):

count_words () {
  eval 'shift; '$1'=$#'
}
count_words number_of_files *
echo There are $number_of_files non-dot files in the current directory

An alternative solution is $(ls * | wc -l). If the glob is *, the command can be shortened to $(ls | wc -l). Parsing the output of ls always makes me uneasy, but here it should work as long as your file names don't contain newlines, or your ls escapes them. And $(ls * 2>/dev/null | wc -l) has the advantage of handling the case of a non-matching glob gracefully (i.e., it returns 0 in that case, whereas the set * method requires fiddly testing if the glob might be empty).

Answer 3

Here's another technique along the lines of the one Gilles posted:

word_count () { local c=($@); echo ${#c[@]}; }
file_count=$(word_count *)

which creates an array with 13,923 elements (if that's how many files there are).

Answer 4

ls -1 | wc -l

...

$ ls --help | grep -- '  -1'
    -1                         list one file per line

...

$ wc --help | grep -- '  -l'
    -l, --lines            print the newline counts

PS: Note ls -<number-one> | wc -<letter-l>

Answer 5

For narrow (I hope) definition of file:

 find -maxdepth 1 -type f | wc -l