I want to count empty and non empty files in a directory with use of loop, without using find command

Question

#!/bin/bash for i in manu; do if [ -f $i ] && [ -s $i ]; then echo " $(ls $i | wc -l) non-empty files" elif [ -f $i ] && [ ! -s $i ]; then echo " $(ls $i | wc -l) empty files" fi done 

Output of this script is:

1 non-empty files 1 empty files 1 non-empty files 1 empty files 1 empty files 1 empty files 

... and the output should be:

2 non-empty files 4 empty files 

I don't know where I'm going wrong. Can anybody help me?

Answer 1

I'm not quite sure why you use ls. The ls utility is usually used for listing the files in a directory. Here you want to loop over files, which is easiest done with a file globbing pattern and a for loop.

You also output something that looks like it should be a summary, but you do that in each iteration rather than at the end.

I'm assuming that manu is a directory in the current directory and that you want to count the number of empty and non-empty files in that directory. I furthermore assume that you don't want to count any other filetype than regular files.

#!/bin/bash shopt -s nullglob dotglob nonempty=0 empty=0 for name in manu/*; do if [ -f "$name" ] && [ ! -h "$name" ]; then # $name is a regular file if [ -s "$name" ]; then nonempty=$(( nonempty + 1 )) else empty=$(( empty + 1 )) fi fi done printf 'There were %d empty files and %d non-empty files\n' "$empty" "$nonempty" 

The nullglob shell option stops the loop from running even once if the pattern manu/* does not match anything, and the dotglob shell option allows the pattern to match hidden names too.

In each iteration, we then determine whether the current file is a regular file (the -h test tests for a symbolic link) and then whether it is empty or not. We update one of two counters depending on the outcome of the -s test. The result is shown at the end.

Answer 2

Loops are generally the wrong way to do things in shells.

Here, you could pipe a directory listing command (ls) to a counting command (awk):

LC_ALL=C ls -Aqn manu/ | LC_ALL=C awk ' /^-/ {if ($5 == 0) empty++; else non_empty++} END {printf "%d empty files\n%d non-empty files\n", empty, non_empty}' 

Where LC_ALL=C disables localisation, so we're sure to get a standard and consistent ls output, and any sequence of bytes (as filenames are) form valid text.

ls -n lists the entries of the manu directory in long format which includes the type and size of the files, -q guarantees one line per file even if the file names (or symlink targets) contain newline characters and -A includes all entries, even hidden ones (not . nor .. which are of no use here).

Then awk can just check the file type (first character being - means it's a regular file) and 5^th field is the size.

Add a -L option to ls for the type and size of file to be considered after symlink resolution. With GNU ls, you can add -U to disable sorting which we don't care for here.

Or you can use zsh and its advanced globs:

empty=(manu/*(NDoN.L0)) non_empty=(manu/*(NDoN.L+0)) print "$#empty empty files\n$#non_empty non-empty files" 

Where the . glob qualifier selects regular files, L0 those of length 0 and L+0 those of length greater than 0 (and N for nullglob (don't complain if there's no match), D for dotglob (includes hidden files), oN do Not order the result).