递归删除所有目录中除一个最新文件之外的所有文件

Question

My system depends on only having one file (PDF, DOCX) per subdirectory. There are thousands and thousands of subdirectories. Due to a permission error, in some of them, I have ended up with more than one file. In these instances, I only want to keep the one most recently modified file.

I was able to export a list of directories that contain more than one file successfully:

find . -type f -printf '%h\n' | sort | uniq -d >test.txt

So I end up with a nice list of all those directories that I need to look at. But it's rather long.

I was also able to automate the deletion of everything but the most recently modified file in a directory:

ls -t | tail -n +2 | xargs -d '\n' rm -f

That does remove all files but the most recently modified one.

The problem I am running into is that the second command only works within that directory. I have not figured out a way to apply it recursively to all directories.

I have attempted:

find /data/test/CONTAINER/SANDBOX -type f -exec sh -c 'ls -t | tail -n +2 | xargs -d '\n' rm -f ' {} \;

but that just yielded xargs: argument line too long

I have tried to adjust the xargs parameters, but I am sure there must be a better way to perform this? Perhaps a shell scrip that pipes the test.txt file fo the folders to cd into and then perform command two in each of these? Or simply a way to recursively apply command 2 to all subfolders, regardless of how many files are contained within that folder?

The last thing I was thinking of is that perhaps the command 3 I had tried applies from the main directory, where I have hundreds of thousands of directories, no wonder the argument line could be too long - but -mindepth 2 didnt change a thing.

Thank you

Answer 1

I think the following script should do the trick for you.

#!/bin/bash

DIR_TO_FIND="/path/to/dir"

find "$DIR_TO_FIND" -type d | while read -r DIR; do
    cd "$DIR"
    ls -t | tail -n +2 | xargs -d '\n' rm -f
    cd "$DIR_TO_FIND" 
done