Print an ordered list of files based on files size in bash

Question

I made the following script to find files based on a 'find' command and then print out the results:

#!/bin/bash
loc_to_look='./'

file_list=$(find $loc_to_look -type f -name "*.txt" -size +5M)

total_size=`du -ch $file_list | tail -1 | cut -f 1`

echo 'total size of all files is: '$total_size

for file in $file_list; do
    size_of_file=`du -h $file | cut -f 1`
    echo $file" "$size_of_file
done

...which give me output like:

>>> ./file_01.txt 12.0M
>>> ./file_04.txt 24.0M
>>> ./file_06.txt 6.0M
>>> ./file_02.txt 6.2M
>>> ./file_07.txt 84.0M
>>> ./file_09.txt 55.0M
>>> ./file_10.txt 96.0M

What I would like to do first, though, is sort the list by file size before printing it out. What is the best way to go about doing this?

Answer 1

Easy to do if you grab the file size in bytes, just pipe to sort

find $loc_to_look -type f -name "*.txt" -size +5M -printf "%f %s\n" | sort -n -k 2

If you wanted to make the file sizes print in MB, you could finally pipe to awk:

find $loc_to_look -type f -printf "%f %s\n" | sort -n -k 2 | awk '{ printf "%s %.1fM\n", $1, $2/1024/1024}'