How to keep only one largest size file and delete all other files

Question

There are 3 or 4 files in same directory as below;

AAA.360p.mp4
AAA.450p.mp4
AAA.720p.mp4
AAA.1080p.mp4

Filenames of those files are almost the same except expression of frame resolution.

(AAA is example of file name. To present that those filenames are same except frame resolution.)

And There are several dots between AAA.

For example, filename is like this;

Interesting.Comedy.E10.200406.450p.mp4
Interesting.Comedy.E10.200406.720p.mp4
Interesting.Comedy.E10.200406.1080p.mp4

Sizes of each file are different

(file size : 360p < 450p < 720p < 1080p) → It is always true.

I'd like to keep only one file, the one that is largest size and delete all other files.

Location of directory is /volume1/video/

It will be run the command only on the synology. (to use task scheduler in the control panel)

If you explain to me, please include path of directory in my case as above.

(because I can't apply the code what you recommend to me for lack of my understanding. I apologize)

Answer 1

The below code will be useful for you, however you can create a copy of files and try on that directory first.

#!/bin/bash

var=`find /volume1/video/ -printf '%s %p\n'| sort -nr | head -n 1 | awk '{ print $2 }'`
largestfilename=`echo $var | rev | cut -d"/" -f1 | rev`

for i in `ls -S /volume1/video/`
do
        if [ ! $i == $largestfilename ]
        then
                rm -f /volume1/video/$i
        fi
done

Answer 2

LARGEST=$(ls -S /volume1/video/* | head -1)
ls /volume1/video/* | grep -v "$LARGEST" | xargs -d '\n' rm

The first line runs a command to list all files in /volume1/video/ sorted by size and pipes that list into head to identify the first one. The output of that command is then stored in a variable named $LARGEST.

The second line lists all files in /volume1/video/ again, greps out the name of the largest one, and removes the rest.

Edit: Added -d flag to xargs to account for potential whitespace in filenames. See this answer for more info.