Compare if the newest file in the directory is different from the previous one

Question

I have the following script:

#!/bin/bash
y=$(ls -t ./pics/0/*.png | head -n 2 | tail -n 1)
new=$(ls -t ./pics/0/*.png | head -n 1)
while true
do
 if cmp --silent "$y" "$new" ; then
y=$(ls -t ./pics/0/*.png | head -n 1)
    base64 $y | tr -d '\n' | sed '$ a \'
new=$(ls -t ./pics/0/*.png | head -n 1)
 fi
done

What am I doing wrong? To be clear my goal is to compare whether the newest file is different from the previous newest file and only if it is generate a unique BASE64 to STDOUT (meaning it should be printed only ONCE).

Answer 1

Please note that the following snippet will NOT work with filenames containing spaces, tabs or newlines.

To find out more about the find-command used here, please reference https://unix.stackexchange.com/a/240424/364705

#!/bin/bash
lastrun=''
while true; do
  read -r newer older <<< \
    $(find . -type f -exec stat -c '%Y %n' {} \; \
    | sort -nr \
    | awk 'NR==1,NR==2 {print $2}'\
    | xargs )

  if [ "$newer" == "$lastrun" ]; then
    :
  else
    if ! cmp "$newer" "$older" > /dev/null ; then
      base64 "$newer"| tr -d '\n' | sed '$ a \'
      lastrun="$newer"
    fi
  fi
done

And a solution using inotify-wait:

#!/bin/bash
lastfile=''
inotifywait -m /path/to/somewhere -e close -e moved_to |
while read path action file; do
  if [ "$lastfile" != '' ];then
    if ! cmp "${path}$file" "${path}${lastfile}" > /dev/null  ; then
    base64 "${path}${file}"
    fi
  fi
    lastfile="$file"
done