How can I get start of month in epoch form from random epoch timestamp in Perl?

Question

I get some times from a database and I am trying to store them by month, using the epoch form. So I could have some random times like:

1354120744
1328978360
1297388332

and for each one I want to find the epoch value which represents the 1st of that month so for the last timestamp it would be:

1296518400 (using lovely epochconverter.com).

I know that I can use POSIX mktime to reduce the timestamp down to days, mins etc and change them accordingly etc as was the answer to a post I did a while ago: How to go back months in Perl taking account of different days in month?

The trouble I have with that is that while I can zero times, I'm worried that by changing the date the day of week will not match up, so essentially making an invalid date. This may be an unnecessary worry; I don't know if it would make a difference having the wrong day of week if the times are validated.

I couldn't see any other POSIX methods which seemed useful. Any help appreciated!

score 9 · Answer 1 · answered Nov 28 '12 at 17:22

9

I stopped worrying about timestamp issues since I've met DateTime:

$ perl -MDateTime -E 'say DateTime->from_epoch(epoch => 1354120744)->truncate(to => "month")->epoch;'
1351728000

answered Nov 28 '12 at 17:22

creaktive

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Not bad module but timezones are hardcoded and it can segfault on daylight times – Galimov Albert Nov 28 '12 at 17:28
DateTime is The Way, but I think we need to respect time zone differences by giving `from_epoch` a `time_zone` argument. – darch Nov 28 '12 at 17:29
seems to work well, but not as easy to understand without looking at documentation - interesting what @PSIAlt said – dgBP Nov 28 '12 at 17:30
As the DateTime->from_epoch() docs state, "By default, the returned object will be in the UTC time zone.". Of course, you could state `DateTime->from_epoch(epoch => 1354120744, time_zone => "local")` – creaktive Nov 28 '12 at 17:33
@darch epoch time(unix timestamp) is UTC so nothing to do with timezone. creaktive, i dont have code it was on my previous work. But there was something with datetime mathematics – Galimov Albert Nov 28 '12 at 17:36

score 5 · Accepted Answer · answered Nov 28 '12 at 17:13

5

My first thought would be to use Time::Local to convert the localtime results back to epoch, with the appropriate values forced to 0 or 1 as needed. Note that day of week is not among the input values it expects, so you don't need to worry about what happens if you get that wrong.

$start_of_month_epoch = timelocal(0, 0, 0, 1, $month, $year);

answered Nov 28 '12 at 17:13

Dave Sherohman

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and here I naively thought localtime and timelocal were the same... silly me! thanks :) – dgBP Nov 28 '12 at 17:17
localtime and timelocal are opposites; localtime turns an epoch integer into the 6 (or 9) broken-down fields, and timelocal turns those 6 fields back into an epoch integer. They're great for this kind of problem. – LeoNerd Nov 29 '12 at 16:13

score 1 · Answer 3 · answered Nov 29 '12 at 16:15

The trouble I have with that is that while I can zero times, I'm worried that by changing the date the day of week will not match up, so essentially making an invalid date.

mktime(3) ignores the wday, yday and isdst fields for this exact reason. You can safely adjust the mday to 1 then call POSIX::mktime() again to turn it back to an epoch value.

my @t = localtime();
$t[0] = $t[1] = $t[2] = 0; # sec/min/hour
$t[3] = 1;                 # mday
my $epoch_start_of_month = mktime @t;

Or simpler

my $epoch = mktime 0, 0, 0, 1, (localtime)[4,5];

How can I get start of month in epoch form from random epoch timestamp in Perl?

3 Answers3