Bash script to find and display oldest file

Question

I\'m trying to write a script that will display the name of oldest file within the directory that the script is executed from.

This is what I have so far:

Answer 1

The ls program has an option to sort on time and you can just grab the last file from that output::

# These are both "wun", not "ell".
#             v          v
oldest="$(ls -1t | tail -1)"

If you want to avoid directories, you can strip them out beforehand:

# This one's an "ell", this is still a "wun".
              v                         v
oldest="$(ls -lt | grep -v '^d' | tail -1 | awk '{print $NF}')"

I wouldn't normally advocate parsing ls output but it's fine for quick and dirty jobs, and if you understand its limitations.

---

If you want a script that will work even for crazies who insist on putting control characters in their file names :-) then this page has some better options, including:

unset -v oldest
for file in "$dir"/*; do
    [[ -z $oldest || $file -ot $oldest ]] && oldest=$file
done

Though I'd suggest following that link to understand why ls parsing is considered a bad idea generally (and hence why it can be useful in limited circumstances such as when you can guarantee all your files are of the YYYY-MM-DD.log variety for example). There's a font of useful information over there.

Answer 2

You may use find command: find -type f -printf '%T+ %p\n' | sort | head -1

Answer 3

The following oneliner finds the oldest file in the whole subtree and reports it using the long format and relative full path. What else could you ever need?

ls -ltd $(find . -type f) | tail -1

Answer 4

You can use this function to find oldest file/directory in any directory:

oldest() { 
   oldf=
   for f in *; do
      # not a file, ignore
      [[ ! -f "$f" ]] && continue
      # find oldest entry
      [[ -z "$oldf" ]] && oldf="$f" || [[ "$f" -ot "$oldf" ]] && oldf="$f"
   done
   printf '%s\n' "$oldf"
}

And call it in any directory as:

oldest