Selecting latest files and grouping by name

Question

I have a directory containing files of the following name structure.

<device>.<yyyy-mm-dd-hh-mm-ss> 

I am working on a script that will retain the the last X copies of these configuration backups per each device. What kind of select magic do I need in order to make this happen?

I was thinking BASH but I'm really language agnostic.

Answer 1

Wildcards expand file names in lexicographic order. Since your date format matches lexicographic order, your requirement boils down to retaining the last X matches for a wildcard (or all matches if there are fewer than X). I'll assume that your backups are the files matching $device.*, adjust the pattern as needed.

In zsh:

set -- $device.*
if [[ $# -gt $X ]]; then set -- $@{[-$X,-1]}; fi
cp -- $@ /retain/area/

In any Bourne-style shell (ash, bash, ksh, …):

set -- "$device".*
if [ $# -gt $X ]; then shift $(($#-$X)); fi
cp -- "$@" /retain/area

---

If what you want is in fact to remove older files, you need to act on the first matches but X (or no matches if there are fewer than X).

In zsh:

files=($device.*)
rm -f -- $files[1,-$X-1]

In other shells:

set -- "$device".*
while [ $# -gt $X ]; do
  rm -- "$1"
  shift
done

Answer 2

Taking advantage of your file names, and that ls sorts lexically by default:

ls | awk -F . '
    $1 != current_device {
        current_device = $1
        n5[$1] = n4[$1]
        n4[$1] = n3[$1]
        n3[$1] = n2[$1]
        n2[$1] = n1[$1]
        n1[$1] = $0
    }
    END {
        for (device in n1) {
            print n1[device]
            print n2[device]
            print n3[device]
            print n4[device]
            print n5[device]
        }
    }
'

I assume the device names do not contain dots. Results likely won't be sorted

Answer 3

May need some changes but:

ls <DEVICE>.* | sort -r | tail -n +<X+1> | xargs echo rm 

That will output the rm statement, remove the "echo" to just delete directly.