How do I convert a date/time to epoch time (unix time/seconds since 1970) in Perl?

Solution 1:

If you're using the DateTime module, you can call the epoch() method on a DateTime object, since that's what you think of as unix time.

Using DateTimes allows you to convert fairly easily from epoch, to date objects.

Alternativly, localtime and gmtime will convert an epoch into an array containing day month and year, and timelocal and timegm from the Time::Local module will do the opposite, converting an array of time elements (seconds, minutes, ..., days, months etc.) into an epoch.

Solution 2:

This is the simplest way to get unix time:

use Time::Local;
timelocal($second,$minute,$hour,$day,$month-1,$year);

Note the reverse order of the arguments and that January is month 0. For many more options, see the DateTime module from CPAN.

As for parsing, see the Date::Parse module from CPAN. If you really need to get fancy with date parsing, the Date::Manip may be helpful, though its own documentation warns you away from it since it carries a lot of baggage (it knows things like common business holidays, for example) and other solutions are much faster.

If you happen to know something about the format of the date/times you'll be parsing then a simple regular expression may suffice but you're probably better off using an appropriate CPAN module. For example, if you know the dates will always be in YMDHMS order, use the CPAN module DateTime::Format::ISO8601.

For my own reference, if nothing else, below is a function I use for an application where I know the dates will always be in YMDHMS order with all or part of the "HMS" part optional. It accepts any delimiters (eg, "2009-02-15" or "2009.02.15"). It returns the corresponding unix time (seconds since 1970-01-01 00:00:00 GMT) or -1 if it couldn't parse it (which means you better be sure you'll never legitimately need to parse the date 1969-12-31 23:59:59). It also presumes two-digit years XX up to "69" refer to "20XX", otherwise "19XX" (eg, "50-02-15" means 2050-02-15 but "75-02-15" means 1975-02-15).

use Time::Local;

sub parsedate { 
  my($s) = @_;
  my($year, $month, $day, $hour, $minute, $second);

  if($s =~ m{^\s*(\d{1,4})\W*0*(\d{1,2})\W*0*(\d{1,2})\W*0*
                 (\d{0,2})\W*0*(\d{0,2})\W*0*(\d{0,2})}x) {
    $year = $1;  $month = $2;   $day = $3;
    $hour = $4;  $minute = $5;  $second = $6;
    $hour |= 0;  $minute |= 0;  $second |= 0;  # defaults.
    $year = ($year<100 ? ($year<70 ? 2000+$year : 1900+$year) : $year);
    return timelocal($second,$minute,$hour,$day,$month-1,$year);  
  }
  return -1;
}

Solution 3:

To parse a date, look at Date::Parse in CPAN.

Solution 4:

I know this is an old question, but thought I would offer another answer.

Time::Piece is core as of Perl 5.9.5

This allows parsing of time in arbitrary formats via the strptime method.

e.g.:

my $t = Time::Piece->strptime("Sunday 3rd Nov, 1943",
                              "%A %drd %b, %Y");

The useful part is - because it's an overloaded object, you can use it for numeric comparisons.

e.g.

if ( $t < time() ) { #do something }

Or if you access it in a string context:

print $t,"\n"; 

You get:

Wed Nov  3 00:00:00 1943

There's a bunch of accessor methods that allow for some assorted other useful time based transforms. https://metacpan.org/pod/Time::Piece

Solution 5:

$ENV{TZ}="GMT";
POSIX::tzset();
$time = POSIX::mktime($s,$m,$h,$d,$mo-1,$y-1900);