I have a delete backup files function which takes in the arguments as a directory name and to backup the files of a specific directory and specific type of file like this delete_old_backup_files $(dirname $$abc) "$abc.*"

The function body is

local fpath=$1
local fexpr=$2

   # delete backup files older than a day
   find $fpath -name "${fexpr##*/}" -mmin  1 -type f | xargs rm -f

Currently deleting files that are older than a day. Now I want to modify the function such that this function should delete all backup files of type $abc.*, except the last 5 backup files created. Tried various commands using stat or -printf but couldn't succeed.

Let me know the correct way of completing this function.

Thanks in Advance.

CodePudding user response：

Assuming the filenames do not contain newline characters, would you please try:

delete_old_backup_files() {
    local fpath=$1
    local fexpr=$2

    find "$fpath" -type f -name "${fexpr##*/}" -printf "%T@\t%p\n" | sort -nr | tail -n  6 | cut -f2- | xargs rm -f --
}

• -printf "%T@\t%p\n" prints the seconds since epoch (%T@) followed by a tab character (\t) then the filename (%p) and a newline (\n).
• sort -nr numerically sorts the lines in descending order (newer first, older last).
• tail -n 6 prints the 6th and following lines.
• cut -f2- removes the prepended timestamp leaving the filename only.

[Edit]
In case of MacOS, please try instead (not tested):

find "$fpath" -type f -print0 | xargs -0 stat -f "%m%t%N" | sort -nr | tail -n  6 | cut -f2- | xargs rm --

In the stat command, %m is expanded to the modification time (seconds since epoch), %t is replaced with a tab, and %N to be a filename.

CodePudding user response：

I would use sorting instead of find. You can use ls -t

$ touch a b c
$ sleep 3
$ touch d e f
ls -t | tr ' ' '\n' | tail -n  4
a
b
c
$ ls -t | tr ' ' '\n' | tail -n  4 | xargs rm
$ ls
d  e  f

From man ls:

-t     sort by modification time, newest first

Make sure you create backups before you delete stuff :-)