convert Oracle DATE to linux time - linux

Question

I'm having trouble converting an Oracle DATE field to a linux time format in perl.
I'm pulling the date field like this:
my $query = "SELECT RESPONSE_DATE FROM TABLENAME";
my $sth = $dbh->prepare($query);
$sth->execute() or die "Couldn't execute statement: " . $sth->errstr;
my #results = $sth->fetchrow_array();
printf "response_date=%s",$results[0];
printf "localtime(time)=%s",localtime(time);
Output:
response_date=14-OCT-14 08.35.00.000000 PM
localtime(time)=Tue Aug 18 23:35:13 2015
I can get all the pieces of the local time like this:
my ($sec,$min,$hour,$mday,$mon,$year,$wday,$yday,$isdst)=localtime(time);
But now I need to compare those pieces with similar pieces of response_date. Any ideas?

Answer 1

I presume you need all the localtime fields for some sort of date calculation?
The Time::Piece module is the usual solution to this. It allows you to parse your database time and also overloads localtime to that it returns a Time::Piece object in scalar context
Unfortunately Time::Piece won't parse the fractional seconds field, so if it's not always zero then you will need to remove it from the date-time string
Clearly other calculations are possible once you have both the database value and the current date as Time::Piece objects
use strict;
use warnings;
use Time::Piece;
use Time::Seconds 'ONE_DAY';
my $s = '14-OCT-14 08.35.00.000000 PM';
my $dt = Time::Piece->strptime($s, '%d-%b-%y %H.%M.%S.000000 %p');
my $now = localtime;
printf "%s was %.3f days ago\n",
$dt->strftime('%a %d-%b-%Y at %H:%M:%S'),
($now - $dt)->days;
output
Tue 14-Oct-2014 at 20:35:00 was 308.190 days ago