Thanks but I am looking of any regexp substitution. sorry for typo : I need to change 9/9/1973 to 09/09/1973
Thanks Prad > -----Original Message----- > From: Bill -OSX- Jones [mailto:[EMAIL PROTECTED]] > Sent: Thursday, January 31, 2002 5:31 PM > To: [EMAIL PROTECTED] > Subject: Re: substitution question > > > Greetings, On Thursday, January 31, 2002, at 08:17 PM, Pradeep > Sethi wrote: > > > Hi All, > > > > I want to change date 9/9/1987 to 09/09/1973 > > You want to change 1987 to 1973 ? > > > was wondering, what is the most efficient way ? > > # Assuming: > my ($sec,$min,$hour,$mday,$mon,$year,$wday) = localtime; > > # Then - > my $vdate = sprintf("%02d", $mon + 1) . '/' . sprintf("%02d", $mday); > $vdate .= '/' . ($year += 1900); > > print $vdate; # prints 01/31/2002 or say 01/01/2002 > > and so on for whichever combinations of month/day/year... > > Or a basic program example: > http://cpan.valueclick.com/authors/id/S/SN/SNEEX/msg_log.perl_v8s > > > No, I am not ignoring the 'efficient' part, but felt this was good > starting spot; > HTH/Sx :] >
