Re: Date Time problem.

You could also use the following:

my $SecPerDay = 24*60*60;
my ($sec,$min,$hour,$mday,$mon,$year,$wday,$yday,$isdst) = localtime(t
+ime - $SecPerDay);
[download]

This may not be accurate, of course, on days when leap-seconds are applied :)

Comment on Re: Date Time problem. Download Code

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Re^2: Date Time problem. by markkawika (Monk) on Sep 04, 2009 at 18:39 UTC
It's also inaccurate twice a year, on the day when Daylight Saving Time begins, and on the day that it ends. DateTime takes DST into account.	[reply]
Re^2: Date Time problem. by ikegami (Patriarch) on Sep 04, 2009 at 19:00 UTC
I don't think that leap seconds will affect the result. Time zone changes from DST, on the other hand, can.	[reply]