script to count files in a directory

Question

I'm looking for a script that counts files in the current directory (excluding sub directories). It should iterate through each file and if not a directory increment a count. The output should just be an integer representing the number of files.

I've come up with

find . -type f | wc -l

But I don't really think it does the whole counting bit. This is for an assignment so if you only want point me in the right direction that would be great.

Answer 1

If you want only regular files,

With GNU find:

find . -maxdepth 1 -type f -printf . | wc -c

Other finds:

find . ! -name . -prune -type f -print | grep -c /

(you don't want -print | wc -l as that wouldn't work if there are file names with newline characters).

With zsh:

files=(*(ND.)); echo $#files

With ls:

ls -Anq | grep -c '^-'

To include symlinks to regular files, change -type f to -xtype f with GNU find, or -exec test -f {} \; with other finds, or . with -. with zsh, or add the -L option to ls. Note however that you may get false negatives in cases where the type of the target of the symlink can't be determined (for instance because it lies in a directory you don't have access to).

If you want any type of file (symlink, directory, pipes, devices...), not only regular one:

find . ! -name . -prune -printf . | wc -c

(change to -print | grep -c / with non-GNU find, (ND.) to (ND) with zsh, grep -c '^-' with wc -l with ls).

That will however not count . or .. (generally, one doesn't really care about those as they are always there) unless you replace -A with -a with ls.

If you want all types of files except directories, Replace -type f with ! -type d (or ! -xtype d to also exclude symlinks to directories), and with zsh, replace . with ^/, and with ls, replace grep -c '^-' with grep -vc '^d'.

If you want to exclude hidden files, add a ! -name '.*' or with zsh, remove the D or with ls, remove the A.

Answer 2

To exclude subdirectories

find . -maxdepth 1 -type f | wc -l

This assumes that none of the filenames contain newline characters, and that your implementation of find supports the -maxdepth option.

Answer 3

To iterate over everything in the current directory, use

for f in ./*; do

Then use either

[ -f "$f" ]

or

test -f "$f"

to test whether the thing you got is a regular file (or a symbolic link to one) or not.

Note that this will exclude files that are sockets or special in other ways.

To just exclude directories, use

[ ! -d "$f" ]

instead.

The quoting of the variable is important as it would otherwise miscount if there, for example, exists a directory called hello world and a file called hello.

To also match hidden files (files with a leading . in their filenames), either set the dotglob shell option with shopt -s dotglob (in bash), or use

for f in ./* ./.*; do

It may be wise also to set the nullglob shell option (in bash) if you want the shell to return nothing when these glob patterns match nothing (otherwise you'll get the literal strings ./* and/or ./.* in f).

Note: I avoid giving a complete script as this is an assignment.

Answer 4

#!/bin/bash

# initialize counter
count=0;

# go through the whole directory listing, including hidden files
for name in * .*
do
    if [[ ! -d $name ]]
    then
        # not a directory so count it
        count=$(($count+1))
    fi
done

echo $count

Answer 5

If you want something that will handle files containing embedded newlines and other unprintable characters, this solution will work. But it's almost certainly overkill for your assignment.

find . -maxdepth 1 -type f -print0 | tr -dc '\0' | wc -c

What this does is to discard everything apart from the trailing \0 at the end of each filename. It then counts those characters, which gives the number of files.

Answer 6

Here's a bash script that will count up and then echo the number of (non-directory) files in the current directory:

#!/usr/bin/env bash
i=0
shopt -s nullglob dotglob
for file in *
do
  [[ ! -d "$file" ]] && i=$((i+1))
done
echo "$i"

Setting "nullglob" gets the count right when there are no files (hidden or otherwise) in the current directory; leaving nullglob unset would mean that the for loop would (incorrectly) see one item: *.

Setting "dotglob" tells bash to also include hidden files (those start with a .) when globbing with *. By default, bash will not include . or .. (see: GLOBIGNORE) when generating the list of hidden files. If your assignment does not want to count hidden files, then do not set dotglob.

Answer 7

I believe the simplest way to count the files and directories in the current directory is:

$ ls | wc -w

-w: is used to count words (not lines as the -l does) because ls outputs all files and directories as an empty-space separated list of words.

And to count only the files :
You can playing a little-bit with the marvelous ls flags you can get the directories out by adding a forward slash to directories ls -p make ls print the files in lines with -l and then invert grep to filter out the forward slashes:

$ ls -pl | grep -v '/' | wc -l

Answer 8

Your command will include sub-directories.

I'd go with:

count=0
for file in $(ls -a); do 
  if [ -f $file ]; then 
    count=$(expr $count + 1)
  fi
done
echo "There are $count files on $(pwd)"

Answer 9

I usually stick to the **ls** command. It gives me all kinds of possibilities.

A basic example :

ls | wc -l 

Will list out the count of all files and directories in a path

ls -R | wc -l 

Will list out the count of all files and directories in a path and its sub-directories.

ls -R *.log | wc -l

The count of only log files in the path and also all the sub-directories.

ls -la *.log | wc -l

Will give you only the log files count in the current directory and not sub-directories.

You limit here is your ls command

'ls' also gives us the ability to filter out files only by certain date, by file formats, and a lot more, using wc in par with 'ls', is possibly the best robust solution I would consider.

Have a look at 'man ls'