Get the newest directory to a variable in Bash

Question

I would like to find the newest sub directory in a directory and save the result to variable in bash.

Something like this:

ls -t /backups | head -1 > $BACKUPDIR 

Can anyone help?

Answer 1

BACKUPDIR=$(ls -td /backups/*/ | head -1) 

$(...) evaluates the statement in a subshell and returns the output.

Answer 2

There is a simple solution to this using only ls:

BACKUPDIR=$(ls -td /backups/*/ | head -1) 

• -t orders by time (latest first)
• -d only lists items from this folder
• */ only lists directories
• head -1 returns the first item

I didn't know about */ until I found Listing only directories using ls in bash: An examination.