Shell脚本在有条件的情况下删除小于xMB的文件

Question

I'm trying to write a script to delete the smallest file if inside the folder has more than one file smaller than 10MB, but I did not succeed.

In my attempt

find . -type f -size -10M -exec rm {} +

Remove all less than 10 Mb, I need to remove only the smallest if inside the folder have 2 files smaller than 10MB recursively.

Any can help-me?

Answer 1

An option is to loop through the output of find and then keep track of then number of files and keep track of the smallest file, so at the end you can remove it:

#!/bin/bash

path=/path/to/dir # replace here with your path

while read -d '' -r dir;do

    files_count=0
    unset min
    unset smallest_file

    while read -d '' -r file;do

            let files_count++
            min_size="$(du -b "${file}"|cut -f1)"
            min=${min:-"$min_size"}
            smallest_file=${smallest_file:-"$file"}

            if (( min_size < min ));then
                    min=$min_size
                    smallest_file=$file
            fi

    done < <(find "${dir}" -type f -size -10M -maxdepth 1 -print0)

    if (( files_count >= 2 ));then

            echo "$smallest_file"
            #rm -v "$smallest_file"

    fi

done < <(find "${path}" -type d -print0)