ls command and size of files in shell script - bash

Question

count=0; #count for counting
IFS='
'
for x in `ls -l $input`; #for loop using ls command
do
a=$(ls -ls | awk '{print $6}') #print[6] is sizes of file
echo $a
b=`echo $a | awk '{split($0,numbers," "); print numbers[1]}'`
echo $b
if [ $b -eq 0 ] # b is only size of a file
then
count=`expr $count + 1` #if b is zero , the count will increase one by one
fi
echo $count
done
I want to find 0 size files . I do that using find command. The second thing is I want to count number of has 0 size of files using ls command and awk. But It doesn't true code . What is my mistake ?

Answer 1

The -s test is true if a file has non-zero size. If that test fails for file, increment your empty-file count.
empty_files=0
for f in "$input"/*; do
[ -s "$f" ] || : $(( empty_files++ ))
done

Answer 2

Your main mistake is that you're parsing ls!
If you want to find (regular) files that are empty, and if you have a version of find that supports the -empty predicate, use it:
find . -type f -empty
Note that this will recurse in subfolders too; if you don't want that, use:
find . -maxdepth 1 -type f -empty
(assuming that your find also supports -maxdepth).
If you only want to count how many empty (regular) files you have:
find . -maxdepth 1 -type f -empty -printf x | wc -m
and if you want to perform both operations at the same time, i.e., print out the name or save them in an array for future use, and count them:
empty_files=()
while IFS= read -r -d '' f; do
empty_files+=( "$f" )
done < <(find . -maxdepth 1 -type f -empty -print0)
printf 'There are %d empty files:\n' "${#empty_files[#]}"
printf ' %s\n' "${empty_files[#]}"
With Bash≥4.4, you could use mapfile instead of the while-read loop:
mapfile -t -d '' empty_files < <(find . -maxdepth 1 -type f -empty -print0)
printf 'There are %d empty files:\n' "${#empty_files[#]}"
printf ' %s\n' "${empty_files[#]}"
For a POSIX-compliant way, use test with the -s option:
find . -type f \! -exec test -s {} \; -print
and if you don't want to recurse into subdirectories, you'll have to -prune them:
find . \! -name . -prune -type f \! -exec test -s {} \; -print
and if you want to count them:
find . \! -name . -prune -type f \! -exec test -s {} \; -exec printf x | wc -m
and here, if you want to perform both operations (count them and save them in an array for later use), use the previous while-read loop (or mapfile if you live in the future) with this find:
find . \! -name . -prune -type f \! -exec test -s {} \; -exec printf '%s\0' {} \;
Also see chepner's answer for a pure shell solution (needs minor tweaking to be POSIX compliant).
Regarding your comment
I want to count and delete [empty files]. How can I do that at the same time?
If you have GNU find (or a find that supports all the goodies):
find . -maxdepth 1 -type f -empty -printf x -delete | wc -m
if not,
find . \! -name . -prune -type f \! -exec test -s {} \; -printf x -exec rm {} \; | wc -m
Make sure that the -delete (or -exec rm {} \;) predicate is at the end! do not exchange the order of the predicates!