Get the newest directory in bash to a variables

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 &         


        

Answer 1

Well, I think this solution is the most efficient:

path="/my/dir/structure/*"
backupdir=$(find $path -type d -prune | tail -n 1)

Explanation why this is a little better:

We do not need sub-shells (aside from the one for getting the result into the bash variable). We do not need a useless -exec ls -d at the end of the find command, it already prints the directory listing. We can easily alter this, e.g. to exclude certain patterns. For example, if you want the second newest directory, because backup files are first written to a tmp dir in the same path:

backupdir=$(find $path -type -d -prune -not -name "*temp_dir" | tail -n 1)

Answer 2

To get the newest folder using ls -t you may have to differentiate files from folders if your directory doesn't have only directories. Using a simple loop you will have a safe and fast result and also allowing to easily implement different filters in the future:

while read i ; do if [ -d "${i}" ] ; then newestFolder="${i}" ; break ; fi ; done < <(ls -t)

Elaborated block:

while read currentItemOnLoop # While reading each line of the file
do 
  if [ -d "${currentItemOnLoop}" ] # If the item is a folder
  then 
    newestFolder="${currentItemOnLoop}" # Then save it into the "newestFolder" variable
    break # and stop the loop
  else
    continue # Look for the next newest item
  fi 
done < <(ls -t) # Sending the result of "ls -t" as a "file" to the "while read" loop

Beware of the continue logic on my elaborated block:

else
  continue # Look for the next newest item

You won't use it. I've put it there just for the sake of your visibility as in this case it wouldn't affect the results.