How to remove all directories except for the last one - bash

Question

I can get the amount of dirs with command
ls -dtr */ | wc -l
But how do I specifically delete the N-1 directories, leaving the most recent one?

Answer 1

As usual for any bash operation on "a bunch of files", you have to be aware of the pain that is spaces and newlines, which may legally appear in file names. These break naïve xargs / for based approaches and require some extra hoops to jump.
Many tools support -z or -0 options, which use NUL bytes as line separator instead of newlines -- NUL may never be part of a file name.
Unfortunately, ls is not one of them, so we have to go through find to get the latest directory.
find . -maxdepth 1 -mindepth 1 -type d
This gets you all the directories in the current directory. (Note: As opposed to your ls -dtr */, this will also find "hidden" directories, i.e. ones starting with a .. Add ! -name ".*" to avoid that.)
-maxdepth 1 avoids recursion, -mindepth 1 keeps the parent directory (.) out of the list.
find . -maxdepth 1 -mindepth 1 -type d -printf "%T+ %f\0"
This lists the directories and their timestamps, using NUL instead of newline for line separation. (NUL can never be part of a file name.)
find . -maxdepth 1 -mindepth 1 -type d -printf "%T+ %f\0" | sort -z
This sorts the results, using NUL instead of newline for line separation.
find . -maxdepth 1 -mindepth 1 -type d -printf "%T+ %f\0" | sort -z | head -z -n -1
This takes everything but the last entry (the latest directory and its timestamp) from the list, using NUL instead of newline for line separation.
find . -maxdepth 1 -mindepth 1 -type d -printf "%T+ %f\0" | sort -z | head -z -n -1 | cut -z -d' ' -f 2-
Using NUL instead of newline for line separation and space as field delimiter, this filters the first field (the timestamp) from the output.
find . -maxdepth 1 -mindepth 1 -type d -printf "%T+ %f\0" | sort -z | head -z -n -1 | cut -z -d' ' -f 2- | xargs -0 rm -rf
Using NUL instead of newline for line separation, this calls rm -rf for each entry.

Answer 2

Use tail:
ls -dt */ | tail -n +2 | while read dirName; do rmdir "$dirName"; done
When specifying the -n option of tail with a +, it skips the first lines.
This will list the directories sorted by date, skip the most recent one and keep the rest, which will be deleted. You can replace rmdir with rm -rf following your use case.
You can try it out:
mkdir test && cd test
for i in {1..100}; do mkdir subdir-$i; done
ls -dt */ | tail -n +2 | while read dirName; do rmdir "$dirName"; done
ls
Output:
subdir-100
EDIT: if any of the directory names contain newline characters, this approach will fail because of tail. See #DevSolar's answer for a more complete one in that case.

Answer 3

This is a pure Bash solution:
shopt -s nullglob # Globs that match nothing expand to nothing
shopt -s dotglob # Globs include files whose names start with '.'
newest_dir=
for sdir in */ ; do
dir=${sdir%/} # Remove slash to enable a symlink check
[[ -L $dir ]] && continue # Skip symlinks to directories
if [[ -z $newest_dir ]] ; then
newest_dir=$dir
elif [[ $dir -nt $newest_dir ]] ; then
echo rm -rf -- "$newest_dir"
newest_dir=$dir
else
echo rm -rf -- "$dir"
fi
done
In its current form it just prints the commands to remove the old subdirectories. Remove the echos to make it functional.

Answer 4

So many ways, here's another. :)
rm -rf `ls -1dt * | tail -n +2 | tr '\n' ' '`
First, the ls -1dt * just lists all the directories sorted by modification time, one per line. Then tail -n +2 removes the first, most recently modified. Then tr '\n ' ' puts them all into a single line. Then the rm -rf the dirs.