bash script, list all the files over a specific size

Question

so I have a line of code like this:

result=`find . -type f -size -1000c -print0 | xargs -0 ls -Sh | head`

for i in $result; do
    item=`wc -c $i`
    echo $item1 
done

this will print out all the files in the current fold that are at most 1000bytes, it has the format like:

size_of the file ./name_of_the_file

but i want to get rid of the "./" symbol, so i try to use "cut"

i want to do something like:

for i in $result; do
    item=`wc -c $i`
    item1=`cut -f 1 $item`    // this gives me the size 
    item2=`cut -c 7- $item`   // this gives me all the character after ./ 
    echo item1, item2         // now make it print 
done

but i'm getting error like:

cut: 639: No such file or directory

can anyone please give me a hint on this? I appreciate it.

Answer 1

Depending on your system's support for POSIX, I believe this will give you the same result you are trying to achieve:

    find . -type f -size -1000c -printf '%P %s Bytes\n'

Answer 2

Look at the Synopsis of cut(1):

cut OPTION... [FILE]...

It's expecting its last argument to be a file (or list of files). What you're doing is supplying $item as the last argument. It seems that $item is 639 which would explain the error message: you're passing 639 in a place where cut is expecting a file name argument. You need to replace $item in those cut calls with $i.

I think the best way to extract the file name (minus ./) would be to use parameter expansion instead of relying on a fixed number of characters to be there every time:

item2="${i#*./}"

Answer 3

try this:

ls -l | awk '$5 <= 1000{print $5, $9}'

Answer 4

find . -type f -size -1000c -printf "%s %f\n"

%s prints File's size in bytes.
%f prints File's name with any leading directories removed (only the last element).