Shell command to get directory with least access date/time

Question

Is there any sort option available in find command to get directory with least access date/time

Answer 1

 find . -type d -printf "%A@ %p\n" | sort -n | tail -n 1 | cut -d " " -f 2-

If you prefer the filename without leading path, replace %p by %f.

Answer 2

This sound like more of a job for ls:

ls -ultd *|grep ^d

The problem with using find, at least on my system (cygwin/bash), is that find accesses the dirs, so all access-times result in current time, defeating your apparent purpose.

Answer 3

find -type d -printf '%T+ %p\n' | sort | head -1

source

Answer 4

• the below linux command displays the access and modified time along with size

  stat -f

Answer 5

find -type d -printf '%T+ %p\n' | sort

Answer 6

A simple shell script will also do:

unset -v oldest
for i in "$dir"/*; do
    [ "$i" -ot "$oldest" -o "$oldest" = "" ] && oldest="$i"
done

note: to find the oldest directory use "$dir"/*/ above (thanks Cyrus) and -type d below with the find command.

In bash if you need a recursive solution, then you can rewrite it as a while loop with process substitution using find

unset -v oldest
while IFS= read -r i; do
    [ "$i" -ot "$oldest" -o "$oldest" = "" ] && oldest="$i"
done < <(find "$dir" -type f)