Archive the oldest file, where files are named according to their timestamp

Question

Context

I have backup scripts running which create tar.gz files and transfer them to a dropbox folder on my headless linux VM.

Problem

I do not know how to create a script which will detect the oldest file among files named according to the scheme /root/Dropbox/apache2-backup-%Y-%m-%d-%H-%M.

I then want to move this file to a /root/Dropbox-archive/ folder, because there is only 2GB of room on Dropbox.

Answer 1

"Simple" solution

ls -1 /Dropbox/apache2-backup-* should order alphabetically by name, i.e. by date (since date is formatted with most significant digit on the left). You can then get the oldest by piping to head -1 (assuming there are no newlines in file names). Generally, piping from ls is a bad idea, but it should be fine in this case since you know that there are no special characters in the file names.

mv "$(ls -1 /Dropbox/apache2-backup-* | head -1)" /root/Dropbox-archive/

I'd probably also recommend using mv --no-clobber (or mv -n) instead, so you don't accidentally overwrite files.

"Proper" solution

If you are really worried about special characters, you can use the following instead.

to_move="$(find /Dropbox -maxdepth 1 -type f -name 'apache2-backup-*' -print0 | sort -z | tr '\0\n' '\n\0' | head -1 | tr '\0\n' '\n\0')"
mv "$to_move" /root/Dropbox-archive/

Explanation

• find /Dropbox -maxdepth 1 -type f -name 'apache2-backup-*' -print0: return normal files (-type f) in /Dropbox without descending into subdirectories (-maxdepth 1), that match the pattern -name 'apache2-backup-*'. Delimit by null character, instead of newlines -print0.
• | sort -z: sort based on null character (N.B. not all sorts can do this).
• | tr '\0\n' '\n\0': swap null and newlines for processing by head.
• | head -1: return first line.
• | tr '\0\n' '\n\0': swap null and newlines back.
• mv "$to_move" /root/Dropbox-archive/: do the move!

Or, in a single line:

mv "$(find /Dropbox -maxdepth 1 -type f -name 'apache2-backup-*' -print0 | sort -z | tr '\0\n' '\n\0' | head -1 | tr '\0\n' '\n\0')" /root/Dropbox-archive/

Answer 2

If you are only moving the oldest one, use this:

_mv(){ mv -- "$1" /root/Dropbox-archive/; }
_mv /Dropbox/apache2-backup-*

But your title doesn't say so.

---

Since processing ls output is a bad idea, here is a better approach:

if [ "$BASH" ]; then
# If you happen to use bash or any other shell that has some similar array:
move_first_things(){ mv -- "${@:0:50}" /root/Dropbox-archive/; }
else # Unfortunate POSIX way
# As long as copy-pasting is acceptable, loop-unrolling with
# `mv -- "$1" "$2" "$3"... /root/Dropbox-archive/` is better since it calls `mv` less.
move_first_things(){
    local max=50 count=0
    while [ "$count" -lt "$max" ]; do
        mv -- "$1" /root/Dropbox-archive/
        shift
        : $((count = count + 1))
    done
}
fi

And since globbing sorts the output just like ls does:

move_first_things /Dropbox/apache2-backup-*

PS: This can be indeed done with external programs (mikeserv, explanation here). With a bit more care and extra obfuscation, people can have it done safely. But I am still going to use the nice filename in this case.

PPS: don_crisst mentioned some zshism. zsh has nice support for anonymous things, like functions, param expansions and arrays. Still using the filename way (since bash doesn't have such sorting operators), the whole thing can be written as baks=( /Dropbox/apache2-backup-* ); mv "$baks" /root/Dropbox-archive/ (referencing an array like this simply gives its first member), or for multiple things, mv "${baks[@]:0:50}" /root/Dropbox-archive/. Well, this looks a bit better than those wrapper functions actually.

Answer 3

(   IFS=/                                      # split on /
    set -ef /Dropbox/apache2-backup*           # set arg array to last glob
    for    f in    $(\ls -rtd "$@";echo /)     # iterate over sorted/split array
    do     [ -z "${f##D*}" ]  ||               # ignore dirname
           mv "${1%/*}/${f%?}" \
              "/root${1%/*}-archive"           # do sorted mv
    done                                       # do && break to do only one mv
)

For whatever reason people seem to think that you can't reliably sort a list of arguments with ls. I think this is because these same people are trying to delimit pathnames with newlines. But the thing is, pathnames do not delimit on newlines - they never have. Pathnames delimit on / - that is the pathname delimiter, and so newlines only confuse things. If you want to reliably parse ls output, you've got to split it correctly.

Answer 4

The question specified oldest file. ls -rt lists the files sorted by modification time, oldest first. This answer does not depend on the filename format except for the prefix identifying a backup file. The date part of the filename is not considered here:

mv "$(ls -tr /root/Dropbox/apache2-backup-*| head -1)" /root/Dropbox-archive/

Sorting by file time is not better than sorting by filename and vice versa. It's largely personal preference and depends on the situation and culture.

In both of these cases, and most, it's safer to enclose the command expansion in double quotes " like this "$( ... )", which will handle all special characters in the backup file filenames (spaces, punctuation, tabs, etc) except for filenames containing newlines.