Re: String (date) to time (integer)

Use Time::Local for the function timelocal, which is the inverse of the builtin localtime.

For example:

use strict;
use warnings;
use Time::Local;

my $date  = "06/06/2009";

my ($m,$d,$y) = $date =~ m|(\d+)/(\d+)/(\d+)|;
my $timet = timelocal(0, 0, 0, $d, $m-1, $y);
print "Date '$date' => $timet\n"

# Updated -- check the reverse
my $ltime = localtime($timet);
print "$timet => $ltime\n";

# Results:
#   Date '06/06/2009' => 1244260800
#   1244260800 => Sat Jun  6 00:00:00 2009
[download]

Note that the first three arguments to timelocal above represent (respectively) the number of seconds, minutes, and hours; in your case they can be zero, as you're only interested in a one-day granularity.

Update: As Corion reminds me, the month passed to timelocal needs to be zero-based, not one-based the way humans represent dates. Changing "$m" to "$m-1" in my code fixed this (thanks, Corion!).

s''(q.S:$/9=(T1';s;(..)(..);$..=substr+crypt($1,$2),2,3;eg;print$..$/

Comment on Re: String (date) to time (integer) Select or Download Code

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Re^2: String (date) to time (integer) by Corion (Patriarch) on Jun 06, 2009 at 14:30 UTC
Actually not exactly: `use strict; use Time::Local; for my $date ("06/06/2009",'01/30/09') { my ($m,$d,$y) = $date =~ m\|(\d+)/(\d+)/(\d+)\|; my $timet = timelocal(0, 0, 0, $d, $m, $y); print "Date '$date' => $timet\n"; print "localtime($timet): ", scalar localtime $timet, "\n"; } __END__ Date '06/06/2009' => 1246831200 localtime(1246831200): Mon Jul 6 00:00:00 2009 Day '30' out of range 1..28 at .pl line 6` [download] As the Time::Local documentation says: The value for the day of the month is the actual day (ie 1..31), while the month is the number of months since January (0..11). So you need to pass in `$m-1` to adjust for that.	[reply] [d/l] [select]
Re^3: String (date) to time (integer) by northwestdev (Acolyte) on Jun 06, 2009 at 15:01 UTC
Thank you. I am new to Perl, so I am not clear on how to decipher ("06/06/2009",'01/30/09').	[reply]
Re^4: String (date) to time (integer) by Corion (Patriarch) on Jun 06, 2009 at 15:05 UTC
Actually, the relevant part is `for my $date ("06/06/2009",'01/30/09') { ...` [download] which is a loop that loops over the two elements `06/06/2009` and `01/30/09`. I used a loop to demonstrate the original case and a problematic case, because `timelocal(...,30,1,9)` would try to find the epoch time of the 30th of February, which doesn't exist in our calendar.	[reply] [d/l] [select]
Re^2: String (date) to time (integer) by dsheroh (Monsignor) on Jun 07, 2009 at 13:59 UTC
Personally, I tend to see Time::Local as "the hard way" (due in part to the 0-based/1-based issue) and prefer DateTime::Format::Strptime: `#!/usr/bin/perl use strict; use warnings; use DateTime::Format::Strptime; my $date_parser = DateTime::Format::Strptime->new(pattern => '%D'); while (<DATA>) { print $date_parser->parse_datetime($_)->epoch, "\n"; } __DATA__ 06/06/2009 01/30/2009` [download] This example prints the epoch values for each date, as the OP said he wanted to convert them to an integer (and epoch is what the `time` function returns), but the DateTime suite also includes tools for displaying the date in pretty much whatever format you may want if epoch isn't what you're looking for.	[reply] [d/l] [select]