#!/usr/bin/perl -w use strict; use File::Find; use Date::Pcalc qw(:all); use Getopt::Std; my %opts = (); getopts('d:l:v' , \%opts); die "\nUsage: $0 [-d startdirectory] [-v verbose] age_limit_in_days \n\n" unless @ARGV; my $start_dir = $opts{d} || "."; my $verbose = $opts{v} || 0; my $day_limit = $ARGV[0] || "10"; find(\&wanted, $start_dir); sub wanted { my $filename = $_; my $age = (stat $filename)[8]; print "Processing file : " . $filename, "\n" if $verbose; return unless is_older_than($age, $day_limit); # put your delete code here - to dangerous to actually do it in demo script print "Would delete $filename, it's older than $day_limit business days ...\n\n"; } sub is_older_than { my ($age, $days) = @_; my ($sec,$min,$hour,$mday,$mon,$year,$wday,$yday,$isdst) = localtime($age); # months start at 1 in Date::Pcalc and at 0 in localtime $mon++; # year is base 1900 in localtime, base 0 in Date::Pcalc $year += 1900; if ($verbose) { print "Today is : " . join ' ', Today(), "\n"; print "File last accessed : " . join ' ', $year, $mon, $mday, "\n"; print "File age in days : " . Delta_Days($year, $mon, $mday, Today()), "\n"; print "File age in business days: " . Delta_Business_Days($year, $mon, $mday, Today()), "\n"; } return 1 if Delta_Business_Days($year, $mon, $mday, Today()) > $days; return 0; } # From the Date::Pcalc manual # 15) How can I calculate the difference in days between dates, but # without counting Saturdays and Sundays? sub Delta_Business_Days { my(@date1) = (@_)[0,1,2]; my(@date2) = (@_)[3,4,5]; my($minus,$result,$dow1,$dow2,$diff,$temp); $minus = 0; $result = Delta_Days(@date1,@date2); if ($result != 0) { if ($result < 0) { $minus = 1; $result = -$result; $dow1 = Day_of_Week(@date2); $dow2 = Day_of_Week(@date1); } else { $dow1 = Day_of_Week(@date1); $dow2 = Day_of_Week(@date2); } $diff = $dow2 - $dow1; $temp = $result; if ($diff != 0) { if ($diff < 0) { $diff += 7; } $temp -= $diff; $dow1 += $diff; if ($dow1 > 6) { $result--; if ($dow1 > 7) { $result--; } } } if ($temp != 0) { $temp /= 7; $result -= ($temp << 1); } } if ($minus) { return -$result; } else { return $result; } } # NOTE THIS: # This solution is probably of little practical value, # however, because it doesn't take legal holidays into # account.